shuoxing/qwen-3.5-27b-absorb
Text Generation • 28B • Updated • 260
question large_stringlengths 35 749 | answer large_stringlengths 184 2.03k | primitive dict |
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State the Freudenthal Theorem. Assuming $\pi_4(S^3)$ is nontrivial, prove that it has order two. | Freudenthal: $\Sigma : \pi_q(S^n) \to \pi_{q+1}(S^{n+1})$ is an isomorphism for $q < 2n-1$ and an epimorphism for $q = 2n-1$. Take $n = 2$: $\Sigma : \pi_3(S^2) \to \pi_4(S^3)$ is onto (since $3 = 2\cdot2 - 1$). Now $\pi_3(S^2) = \mathbb{Z}$ generated by the Hopf class $z$ with $h(z) = 1$ (as $X_H = S^2 \cup_H e^4 \sim... | {
"essential_property": "Freudenthal makes Σ:π_3(S^2)→π_4(S^3) an epimorphism, and π_3(S^2)=Z (Hopf) maps onto π_4(S^3), which suspension theory pins to order two.",
"solution_principle": "Use the Freudenthal surjection from π_3(S^2)=Z onto π_4(S^3), then identify the kernel/relations forcing the image to have orde... |
Let $\mathbb{RP}^2 = S^2/\!\sim$ and $f : \mathbb{RP}^2 \to \mathbb{R}^3$, $f([x,y,z]) = \frac{1}{x^2+y^2+z^2}(yz, zx, xy)$. (a) Show $f$ is smooth. (b) Find a point where $f$ is not an immersion. | (a) Cover $S^2$ (hence $\mathbb{RP}^2$) by the six charts $S_{x,\pm} = \{x > 0\}$ etc. On $S_{x,+}$ with $\phi_{x,+}(x,y,z) = (y,z)$ (and $x = \sqrt{1 - y^2 - z^2}$), $f \circ \phi_{x,+}^{-1}(y,z) = (yz,\ z\sqrt{1-y^2-z^2},\ y\sqrt{1-y^2-z^2})$, which is smooth on $\{y^2 + z^2 < 1\}$; by symmetry $f$ is smooth in all c... | {
"essential_property": "The map is a smooth well-defined map on RP^2 (even in the coordinates), and it fails to be an immersion where its differential drops rank—found by locating a critical point of the parametrization.",
"solution_principle": "Verify smoothness in charts, then compute the differential and locate... |
Say $g : X \to Y$ has the right lifting property (RLP) with respect to a class $S$ if every square with left map in $S$ admits a diagonal lift. Prove that if $g$ has the RLP with respect to $\{S^n \hookrightarrow D^{n+1}\}_{n \ge 0}$, then it has the RLP with respect to all relative CW-pairs. | Let $(W, A)$ be a relative CW-complex with skeleta $W^n$, and a square $A \to X$, $W \to Y$. Build the lift by induction on skeleta. $W^0 = A \sqcup (\text{points})$; extending over the added $0$-cells (attached along $S^{-1} = \emptyset \hookrightarrow D^0$) is immediate. Assume a lift $W^n \to X$ is defined. The push... | {
"essential_property": "Having the RLP against sphere inclusions S^n↪D^{n+1} propagates to the RLP against all relative CW inclusions by inductively lifting over cells (each cell attachment is a pushout of a sphere inclusion).",
"solution_principle": "Build the lift skeleton-by-skeleton: extend over each attached ... |
Let $f \in L^1(\mathbb{R})$. Suppose that $\int_{[x, x+1]} f(t)\, dt = 0$ for almost every $x \in \mathbb{R}$. Prove that $f(x) = 0$ for almost every $x \in \mathbb{R}$. | Define $F(x) = \int_0^x f(t)\, dt$. By the Fundamental Theorem of Calculus (for Lebesgue integrals), $F$ is absolutely continuous on bounded intervals, continuous on $\mathbb{R}$, with $F' = f$ a.e. Let $E = \{x : \int_{[x,x+1]} f \ne 0\}$, which has measure zero, and $P = \{x + n : x \in E,\ n \in \mathbb{Z}\}$, a cou... | {
"essential_property": "If ∫_{[x,x+1]} f=0 for a.e. x, the antiderivative F satisfies F(x+1)=F(x) a.e., so F is periodic and (being absolutely continuous) constant, giving f=F'=0 a.e.",
"solution_principle": "Set F(x)=∫_0^x f; the hypothesis makes F 1-periodic and continuous, and an L^1-antiderivative that is peri... |
True or false (with justification): The $\mathbb{C}$-algebras $\mathbb{C}[x]/(x^3 - x)$ and $\mathbb{C}[x]/(x^3 + x + 1)$ are isomorphic. | True. Both polynomials have $3$ distinct roots (for $x^3 + x + 1$, $\gcd(f, f') = 1$, so it is separable). For any degree-$3$ polynomial $p(x) \in \mathbb{C}[x]$ with distinct roots $\alpha_1, \alpha_2, \alpha_3$, the map $\mathbb{C}[x] \to \mathbb{C}^3$, $x \mapsto (\alpha_1, \alpha_2, \alpha_3)$, has kernel exactly $... | {
"essential_property": "Any degree-3 polynomial over C with distinct roots yields C[x]/(p)≅C^3 (CRT/evaluation at the roots), so two separable cubics give isomorphic split algebras.",
"solution_principle": "Check both cubics are separable (distinct roots), then use CRT/evaluation to identify each quotient with C^3... |
Let $\mathbb{P}^{n^2 - 1}$ be nonzero $n \times n$ complex matrices modulo scalars, and $X = \{[A] : A \text{ nilpotent}\}$. (a) Show $X$ is a closed subvariety. (b) Show $X$ is irreducible and find its dimension. | (a) $A$ is nilpotent iff its characteristic polynomial is $T^n$, i.e. the coefficients $p_1(A), \ldots, p_n(A)$ (homogeneous polynomials in the entries) all vanish. So $X = V(p_1, \ldots, p_n)$ is a closed subvariety.
(b) Let $F$ be the flag variety of complete flags in $\mathbb{C}^n$, $\dim F = \binom n2 = \frac{n(n-1... | {
"essential_property": "The nilpotent locus is cut out by the vanishing of the characteristic-polynomial coefficients (closed), and it is irreducible of dimension n^2−n−1 via the incidence variety of nilpotents preserving a flag.",
"solution_principle": "Realize X as V(p_1,…,p_n) (closed), then fibre the incidence... |
Fix a prime $p$ and let $G$ be a finite group such that every nonidentity $p$-subgroup is contained in a unique Sylow $p$-subgroup. Let $N \trianglelefteq G$ with $p \mid |N|$. (i) If $P, Q \in \mathrm{Syl}_p(G)$, show $Q = P^n$ for some $n \in N$. (ii) Prove $G/N$ has a unique Sylow $p$-subgroup. | (i) Since $N \trianglelefteq G$, $P \cap N, Q \cap N \in \mathrm{Syl}_p(N)$; by Sylow in $N$, $Q \cap N = (P \cap N)^n$ for some $n \in N$. Now $P \cap N \ne 1$ (as $p \mid |N|$), so $(P \cap N)^n = Q \cap N \ne 1$ is a nonidentity $p$-subgroup contained in both $Q$ and $P^n$ (two Sylow $p$-subgroups of $G$). By the un... | {
"essential_property": "The hypothesis 'every nontrivial p-subgroup lies in a unique Sylow p-subgroup' plus a normal N with p | |N| makes Sylow p-subgroups of G conjugate by elements of N, and this uniqueness passes to G/N.",
"solution_principle": "Intersect Sylow subgroups with N to conjugate them by N (Sylow in ... |
Prove Taylor's theorem with the Peano remainder: if $f : \mathbb{R}\to\mathbb{R}$ has an $n$th derivative at $0$, then $f(x) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k = o(x^n)$ as $x \to 0$. (Do not assume the $n$th derivative exists in a neighborhood.) | Let $g(x) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$. Then $g(0) = g'(0) = \cdots = g^{(n)}(0) = 0$ (the Taylor polynomial matches all derivatives at $0$). We must show $g(x)/x^n \to 0$. Apply L'Hôpital's rule $n-1$ times (each numerator and denominator vanishes at $0$, and derivatives through order $n-1$ exist ne... | {
"essential_property": "The Taylor error g vanishes to order n at 0, so applying L'Hôpital n−1 times reduces g(x)/x^n to a difference quotient of f^{(n-1)} tending to 0 by the definition of f^{(n)}(0).",
"solution_principle": "Note g and its first n derivatives vanish at 0, apply L'Hôpital repeatedly, and use the ... |
Let $f : \mathbb{C}\to\mathbb{C}$ be entire with $f(z)/z \to 0$ as $|z| \to \infty$. Prove $f$ is constant. | By the Cauchy formula for the derivative, for any $R > 0$, $f'(0) = \frac{1}{2\pi i}\oint_{|z|=R}\frac{f(z)}{z^2}\,dz$. Given $\varepsilon > 0$, the hypothesis gives $R_\varepsilon$ with $|f(z)/z| < \varepsilon$ for $|z| > R_\varepsilon$. For $R > R_\varepsilon$, $\Big|\oint_{|z|=R}\frac{f(z)}{z^2}\,dz\Big| \le \frac{1... | {
"essential_property": "The Cauchy estimate for f'(a) with the sublinear bound f(z)/z→0 forces f'≡0 (translation-invariant), so f is constant.",
"solution_principle": "Bound |f'(a)| by max|f(z)/z| on large circles (which →0), giving f'(a)=0 for all a, hence constant.",
"core_concept": "Sublinear growth f(z)/z→0 ... |
Prove that $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}$ as rings. | Define $\varphi : \mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$ by $(\frac ab, \frac cd) \mapsto \frac{ac}{bd}$ (ordinary multiplication). It is $\mathbb{Z}$-balanced, so induces $f : \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Q} \to \mathbb{Q}$, a ring homomorphism. Define $g : \mathbb{Q} \to \mathbb{Q}\otimes\mathbb{Q}$ ... | {
"essential_property": "The multiplication map Q×Q→Q, (a/b,c/d)↦ac/bd, is Z-balanced and induces a ring isomorphism Q⊗_Z Q≅Q because Q is a localization of Z (idempotent under ⊗).",
"solution_principle": "Define the balanced multiplication map, check it induces a well-defined ring homomorphism Q⊗Q→Q, and exhibit t... |
Let $q$ be a positive integer, $\Omega \subset \mathbb{C}$ connected open. Suppose $f_n$ are holomorphic on $\Omega$ with $|f_n^{-1}(c)| \le q$ for every $c \in \mathbb{C}$, and $f_n \to f$ uniformly on compact subsets. Prove that either $f$ is constant, or $|f^{-1}(c)| \le q$ for every $c$. | Suppose $f$ is nonconstant but attains some value $c$ at $q+1$ distinct points $z_1,\dots,z_{q+1} \in \Omega$; we derive a contradiction. Since $f$ is nonconstant holomorphic on the connected set $\Omega$ (a uniform limit of holomorphics is holomorphic), choose $\varepsilon > 0$ so that the closed disks $\{|z - z_j| \l... | {
"essential_property": "If the limit f attained a value q+1 times, Rouché (via uniform convergence) would give f_n the same q+1 preimages, contradicting |f_n^{-1}(c)|≤q; so f is constant or has ≤q preimages.",
"solution_principle": "Assume f nonconstant with q+1 preimages of c; isolate them in disjoint disks, appl... |
Let $f$ be holomorphic in the strip $-1 < \Im z < 1$, real on $\mathbb{R}$, with positive imaginary part in $0 < \Im z < 1$. (a) Prove $\Im f < 0$ in $-1 < \Im z < 0$. (b) Prove $f'(x) \ge 0$ for real $x$. (c) Prove $f'(x) > 0$ for real $x$. | (a) Let $g(z) = \overline{f(\bar z)}$; it is holomorphic on the strip and agrees with $f$ on $\mathbb{R}$ (since $f$ is real there), so by the identity theorem $g = f$ (Schwarz reflection). For $-1 < \Im z < 0$, $\bar z$ lies in $0 < \Im < 1$, so $\Im f(z) = \Im g(z) = -\Im f(\bar z) < 0$.
(b) For real $x$, $f'(x) = \l... | {
"essential_property": "Schwarz reflection (g(z)=conj f(conj z)=f) makes Im f antisymmetric across the real axis, so Im f<0 below; the boundary behavior and open mapping then force f'(x)>0.",
"solution_principle": "Use reality on R to reflect and get Im f<0 in the lower strip, then compute f'(x) as a limit with po... |
Let $\alpha : \mathbb{C}[s, t, u] \to \mathbb{C}[x, y, z]$ be the ring homomorphism $s \mapsto xy$, $t \mapsto xz$, $u \mapsto yz$. (i) Describe the corresponding morphism $\phi$ of algebraic sets explicitly. (ii) Describe the preimage $\phi^{-1}(2,3,6)$. (iii) Describe the preimage $\phi^{-1}(0,1,2)$. (iv) Decompose t... | (i) $\phi : \mathbb{A}^3 \to \mathbb{A}^3$, $(a,b,c) \mapsto (ab, ac, bc)$.
(ii) If $ab = 2$, $ac = 3$, $bc = 6$, then $(abc)^2 = 2 \cdot 3 \cdot 6 = 36$, so $abc = \pm 6$. If $abc = 6$: $a = abc/bc = 1$, $b = 2$, $c = 3$. If $abc = -6$: $a = -1$, $b = -2$, $c = -3$. So $\phi^{-1}(2,3,6) = \{(1,2,3), (-1,-2,-3)\}$.
(ii... | {
"essential_property": "The monomial map (a,b,c)↦(ab,ac,bc) has fibers solvable by taking products: (abc)^2 equals the product of the three targets, so each fiber is a finite explicit set of points.",
"solution_principle": "Solve the monomial system by forming (abc)^2 from the coordinates, extract abc up to sign, ... |
True or false: for $n \ge 5$ and a prime $p \le n$, the group $A_n$ has at least $n$ Sylow $p$-subgroups. | True. The number $r$ of Sylow $p$-subgroups is nonzero (since $p \mid n!/2 = |A_n|$), and $A_n$ acts transitively on them by conjugation, giving a homomorphism $A_n \to S_r$. Since $A_n$ is simple (for $n \ge 5$) and $A_n \not\cong S_r$ in the relevant range, this action is a proper injective embedding, so $|A_n| = n!/... | {
"essential_property": "The conjugation action of the simple group A_n on its Sylow p-subgroups is faithful, giving an embedding into S_r; simplicity and |A_n|>|S_{n-1}| force r≥n.",
"solution_principle": "Use that A_n (simple, n≥5) acts faithfully by conjugation on its r Sylow p-subgroups, embedding into S_r, so ... |
Let $F[x,y]$ be a polynomial ring, $f(x) \in F[x]$, $g(y) \in F[y]$ of positive degree, and $I = (f(x), g(y))$. (a) Prove $I \ne F[x,y]$. (b) If $f = x - \alpha$, $g = y - \beta$, show $I$ is maximal. | (a) Let $\alpha$ be a root of $f$ and $\beta$ a root of $g$ in $\bar F$. The evaluation $\theta : F[x,y] \to \bar F$, $x \mapsto \alpha$, $y \mapsto \beta$, kills $f(x)$ and $g(y)$, so $I \subseteq \ker\theta$. Since $\theta(1) = 1 \ne 0$, $1 \notin I$, so $I \ne F[x,y]$.
(b) With $f = x - \alpha$, $g = y - \beta$: $x ... | {
"essential_property": "Evaluating at a common zero (α,β) of f and g in the algebraic closure gives a homomorphism killing I but not 1, so I is proper; when f,g are linear the evaluation lands in F, making the quotient a field and I maximal.",
"solution_principle": "Construct the evaluation homomorphism at roots o... |
Give an example of the same smooth manifold with two different Riemannian metrics: one so that the total volume is infinite, the other so that the volume is finite. | Take $M = \mathbb{R}$. With the standard metric $g(\partial_x, \partial_x) = 1$, the $1$-dimensional volume is $\int_{\mathbb{R}} dx = \infty$. With the metric $g(\partial_x, \partial_x) = e^{-x^2}$, the volume element is $\sqrt{e^{-x^2}}\,dx = e^{-x^2/2}\,dx$, so the total volume is $\int_{\mathbb{R}} e^{-x^2/2}\, dx ... | {
"essential_property": "The same manifold R carries the standard metric (infinite length) and a conformally rescaled metric like e^{-x^2} (finite length), showing volume depends on the metric, not just the smooth structure.",
"solution_principle": "Exhibit R with dx (infinite total length) versus e^{-x^2/2}dx (fin... |
Let $G \cong S_4$ be the rotational symmetry group of the cube, and $V$ its (complexified) $3$-dimensional geometric irreducible representation. Let $\pi$ be the permutation representation on the $4$-element subsets of the $8$ vertices. Write the characters of $\pi$ and $V$, and find the multiplicity of $V$ in $\pi$. | The multiplicity is $\langle \chi_V, \chi_\pi\rangle = \frac{1}{|G|}\sum_g \chi_V(g)\overline{\chi_\pi(g)}$. Conjugacy classes (by rotation type) with sizes: $C_1$ (identity, $1$); $C_2$ (order-$2$ face rotations, $3$); $C_2'$ (order-$2$ edge rotations, $6$); $C_3$ (order-$3$ vertex rotations, $8$); $C_4$ (order-$4$ fa... | {
"essential_property": "The multiplicity of V in the permutation representation π is the character inner product ⟨χ_V,χ_π⟩, computed by summing over conjugacy classes with χ_π counting fixed 4-subsets.",
"solution_principle": "Compute χ_π (fixed 4-subsets) and χ_V (1+2cosθ) on each class, then evaluate the inner p... |
Prove that the number of orbits of a finite group acting on a set equals the average number of fixed points of its elements. In how many ways, up to symmetry, can one color the $4$ corners of a square with $10$ colors? | Burnside's lemma: let $G$ act on a finite set $X$. Count pairs $(g, x)$ with $gx = x$ in two ways: by $g$ it is $\sum_{g\in G}|\mathrm{Fix}(g)|$; by $x$ it is $\sum_{x\in X}|\mathrm{Stab}(x)|$. By orbit–stabilizer, $|\mathrm{Stab}(x)| = |G|/|\mathrm{orbit}(x)|$, so $\sum_x |\mathrm{Stab}(x)| = |G|\sum_{\text{orbits}} 1... | {
"essential_property": "Burnside's lemma counts orbits as the average number of fixed points; for D_8 acting on 10^4 corner-colorings, summing fixed colorings per group element gives 1540.",
"solution_principle": "Prove Burnside by double-counting fixed pairs, then apply it to the dihedral action on colorings, com... |
State the Krull–Schmidt Theorem for modules of finite length. | Let $R$ be a ring and $V$ an $R$-module of finite length (i.e. $V$ has a composition series). Then $V$ can be written as an internal direct sum $V = V_1 \oplus \cdots \oplus V_n$ of finitely many indecomposable submodules. This decomposition is unique in the following sense: if $V = W_1 \oplus \cdots \oplus W_m$ is ano... | {
"essential_property": "Krull–Schmidt: a finite-length module decomposes into indecomposables uniquely up to isomorphism and reordering.",
"solution_principle": "State existence and uniqueness of the indecomposable decomposition for finite-length modules.",
"core_concept": "Finite length guarantees a unique inde... |
In $C([a,b])$ (complex, sup norm), show that a sequence $(f_n)$ converges weakly to $f$ if and only if $(\|f_n\|)$ is bounded and $f_n(t) \to f(t)$ for every $t \in [a,b]$. | ($\Leftarrow$) Let $\omega \in C([a,b])^*$. By Riesz representation, $\omega(g) = \int_{[a,b]} g\, d\mu$ for a complex measure $\mu$; write $d\mu = h\, d|\mu|$ with $|h| = 1$ a.e. (Radon–Nikodym). With $M = \sup_n \|f_n\|$, $|f_n h| \le M \in L^1(|\mu|)$ and $f_n h \to f h$ pointwise a.e., so by Dominated Convergence $... | {
"essential_property": "By Riesz representation, functionals on C[a,b] are integration against measures, so weak convergence is testing against all measures; boundedness plus pointwise convergence give it via dominated convergence, and the converse is immediate.",
"solution_principle": "Represent functionals by me... |
True or false (with proof or counterexample): A field extension of degree $4$ has an intermediate subfield of degree $2$. | False. Let $E/F$ be Galois with $\operatorname{Gal}(E/F) = S_4$ (e.g. $E = \mathbb{C}(x_1, x_2, x_3, x_4)$, $F = E^{S_4}$). Let $K = E^{S_3}$ be the field fixed by $S_3 \subset S_4$; then $[K : F] = [S_4 : S_3] = 4$. An intermediate $F \subset L \subset K$ with $[L:F] = 2$ would correspond to a subgroup $H$ with $S_3 \... | {
"essential_property": "A degree-4 extension need not have a quadratic subfield: realizing S_4 as a Galois group and taking the fixed field of S_3 gives a degree-4 extension whose corresponding subgroup S_3 is maximal, so no intermediate quadratic exists.",
"solution_principle": "Construct an S_4-extension, take K... |
Let $X \subset \mathbb{A}^3_{\mathbb{C}}$ be the subvariety defined by $xy = z^2$. (a) Show $X$ is not smooth and compute the dimension of the Zariski tangent space at $(0,0,0)$. (b) Show that the blow-up $Y = \mathrm{Bl}_{(0,0,0)} X$ is smooth. | (a) With $f = xy - z^2$, the Jacobian $(\partial_x f, \partial_y f, \partial_z f) = (y, x, -2z)$ vanishes only at $(0,0,0)$, so $X$ is singular exactly there. The Zariski cotangent space is $\mathfrak{m}/\mathfrak{m}^2$ where $\mathfrak{m} = (x,y,z)$ in $R = \mathbb{C}[x,y,z]/(f)$; since $f \in \mathfrak{m}^2$, the cla... | {
"essential_property": "The quadric cone xy=z^2 is singular only at the origin (Jacobian vanishes there, tangent space 3-dimensional), and blowing up the origin separates the branches, giving a smooth strict transform.",
"solution_principle": "Compute the Jacobian to locate the singularity and the 3-dimensional ta... |
Let $G$ be a finite group and $H < G$ a proper subgroup. Prove there is an element of $G$ not lying in any conjugate $xHx^{-1}$ of $H$. | The number of distinct conjugates of $H$ is $[G : N_G(H)] \le [G : H] = |G|/|H|$, since $H \le N_G(H)$. Each conjugate contains the identity and $|H| - 1$ non-identity elements, and all conjugates share the identity. So the number of non-identity elements covered by the union $\bigcup_x xHx^{-1}$ is at most $\frac{|G|}... | {
"essential_property": "The union of conjugates of a proper subgroup covers at most [G:H](|H|−1)+1<|G| elements (all sharing the identity), so some element lies outside every conjugate.",
"solution_principle": "Bound the number of conjugates by [G:H] and count the non-identity elements they cover, showing it is le... |
Let $a, b, c$ lie on the unit circle of $\mathbb{C}$ with $a + b + c = 0$. Prove $a, b, c$ are the vertices of an equilateral triangle. | Multiplying all three by $1/a$ (a rotation, preserving the shape and unit circle), assume $a = 1$. From $1 + b + c = 0$, the imaginary parts of $b, c$ sum to $0$ and their real parts sum to $-1$. Since $|b| = |c| = 1$, real parts lie in $[-1,1]$; summing to $-1$ with opposite imaginary parts forces $\Re b = \Re c = -\t... | {
"essential_property": "Three unit-modulus numbers summing to zero must be the vertices of an equilateral triangle: normalizing one to 1 forces the other two to be the primitive cube roots of unity.",
"solution_principle": "Rotate so a=1; the sum-zero and unit-modulus constraints force the real parts to −1/2 and t... |
Let $G$ be a bounded connected open subset of the plane and $f$ continuous on $\overline{G}$, analytic in $G$. Suppose there is a constant $k$ with $|f(z)| = k$ for all $z \in \partial G$. Prove that either $f$ has a zero in $G$ or $f$ is constant. | Suppose $f$ has no zero in $G$. If $k = 0$, then $|f| = 0$ on $\partial G$, and by the Maximum Modulus Principle $\max_{\overline G}|f| = \max_{\partial G}|f| = 0$, so $f \equiv 0$, constant. Assume $k > 0$. By the Maximum Modulus Principle, $\max_{\overline G}|f| = \max_{\partial G}|f| = k$. If $|f|$ is not constant, ... | {
"essential_property": "If f has no zero, then 1/f is analytic, and applying the maximum modulus principle to both f and 1/f (constant boundary modulus) forces |f| constant, hence f constant.",
"solution_principle": "Assume f nonvanishing; the maximum and minimum modulus principles both give |f| bounded by and bel... |
Let $\mu, \nu$ be positive measures on $\mathbb{R}^n$ ($n \ge 1$) and $f \ge 0$. For $1 \le p < \infty$ set $A := \int \nu(dy)\big(\int f(x,y)^p\,\mu(dx)\big)^{1/p}$ and $B := \big(\int \mu(dx)(\int f(x,y)\,\nu(dy))^p\big)^{1/p}$ (all finite). Is $A \ge B$ or $A \le B$ for all $f$? Prove your assertion. | We have $A \ge B$; this is Minkowski's integral inequality. By duality (with $q$ the conjugate exponent), $B = \big[\int \mu(dx)\,(\int f(x,y)\,\nu(dy))^p\big]^{1/p} = \sup_{\|g\|_{L^q(\mu)}\le 1} \int \mu(dx)\, g(x)\int f(x,y)\,\nu(dy)$. Interchange the order of integration (Tonelli, $f\ge 0$): this equals $\sup_g \in... | {
"essential_property": "Minkowski's integral inequality (A≥B) follows by duality: representing the L^p norm as a supremum over the unit ball of L^q and swapping the order of integration with Hölder.",
"solution_principle": "Write B as a sup over g in the L^q unit ball, interchange the integrals (Tonelli), and appl... |
Let $f : (a,b] \to \mathbb{R}$ be strictly increasing on $(a,b)$ and left-continuous at $b$. Show $f$ is strictly increasing on $(a,b]$. | It remains to show $f(x_1) < f(b)$ for every $x_1 \in (a,b)$. Fix such $x_1$ and choose $x_0 \in (x_1, b)$. Since $f$ is increasing on $(a,b)$, $f(x) \ge f(x_0)$ for all $x \in (x_0, b)$, so taking the left limit at $b$ (which exists and equals $f(b)$ by left-continuity), $\lim_{x\to b^-} f(x) \ge f(x_0)$, i.e. $f(b) \... | {
"essential_property": "Left-continuity at b lets the strict monotonicity on (a,b) pass to b: f(x_1)<f(x_0)≤lim_{x→b^-}f(x)=f(b).",
"solution_principle": "Pick x_0 between x_1 and b, use monotonicity to get f(x_1)<f(x_0)≤f(b) via the left limit.",
"core_concept": "The left limit at b bounds f(b) below by interio... |
Set $B = \{|z| < 2\}$. Prove there is no holomorphic $f$ on $B$ with $\left|f(z) - \frac1z\right| < 1$ for all $z$ with $|z| = 1$. | Suppose such $f$ exists. Let $g(z) = zf(z)$, holomorphic on $B$. Then for $|z| = 1$, $|g(z) - 1| = |z|\,|f(z) - 1/z| = |f(z) - 1/z| < 1$. By Rouché's theorem, $g$ and the constant function $1$ have the same number of zeros in $\{|z| < 1\}$ (since $|g - 1| < 1 = |1|$ on the boundary). The constant $1$ has no zeros, but ... | {
"essential_property": "For g(z)=zf(z), the bound |g−1|<1 on |z|=1 forces (Rouché) g and 1 to have the same zero count in the disk (zero), but g(0)=0 makes g have a zero, a contradiction.",
"solution_principle": "Set g=zf, use |g−1|<1 on the circle to apply Rouché comparing g to the constant 1 (no zeros), contradi... |
Let $f : [-1, 1] \to \mathbb{R}$ be continuous with $\int_{-1}^1 x^{2n} f(x)\, dx = 0$ for all $n = 0, 1, 2, \ldots$. Prove $f$ is odd. | Let $g(x) = f(x) + f(-x)$, a continuous even function; showing $g \equiv 0$ is equivalent to $f(-x) = -f(x)$. For each $m$, $\int_{-1}^1 x^m g(x)\, dx = 0$: automatic for $m$ odd (since $g$ is even, $x^m g$ is odd), and for $m$ even it follows from the hypothesis (as $\int x^m g = \int x^m f(x) + \int x^m f(-x) = 2\int... | {
"essential_property": "The even part g(x)=f(x)+f(−x) is orthogonal to all polynomials, so by Weierstrass approximation g⊥g in L^2, forcing g≡0, i.e. f is odd.",
"solution_principle": "Show ∫x^m g=0 for all m (odd m automatic, even m by hypothesis), use Weierstrass to approximate g uniformly by polynomials, and co... |
Let $\mathbb{H} = \{\Im z > 0\}$ and $\zeta_n = e^{2\pi i/n}$. Suppose $f : \mathbb{H}\to\mathbb{H}$ is analytic with $f(\zeta_4) = \zeta_3$. Prove $|f'(\zeta_4)| \le \sqrt3/2$. | Note $\zeta_4 = i$ and $\zeta_3 = e^{2\pi i/3} = -\tfrac12 + \tfrac{\sqrt3}{2}i$, so $\Im\zeta_4 = 1$ and $\Im\zeta_3 = \frac{\sqrt3}{2}$. Set $a = \frac{\Im\zeta_3}{\Im\zeta_4} = \frac{\sqrt3}{2}$. Choose $b \in \mathbb{R}$ so that $g(z) = az + b$ is an automorphism of $\mathbb{H}$ (real affine with positive slope) ma... | {
"essential_property": "Normalizing the fixed point via an affine automorphism g of H (with g'(ζ_4)=Im ζ_3/Im ζ_4=√3/2), the Schwarz–Pick lemma bounds |h'(ζ_4)|≤1 for h=g^{-1}∘f, giving |f'(ζ_4)|≤√3/2.",
"solution_principle": "Choose an affine automorphism g sending ζ_4↦ζ_3 with derivative √3/2, apply Schwarz–Pick... |
Let $G \ne 1$ be a (possibly infinite) group whose subgroups are linearly ordered by inclusion. (a) Prove $G$ is abelian and the orders of its elements are all powers of the same prime $p$. (b) If $G_n = \{g \in G : g^{p^n} = 1\}$, prove $|G_n| \le p^n$. | (a) For $x, y \in G$, the subgroups $\langle x\rangle, \langle y\rangle$ are comparable; say $\langle x\rangle \subseteq \langle y\rangle$, so $x$ is a power of $y$ and hence $x, y$ commute. Thus $G$ is abelian. If $o(x) = \infty$, then $\langle x^2\rangle$ and $\langle x^3\rangle$ are incomparable, a contradiction; so... | {
"essential_property": "If all subgroups are linearly ordered by inclusion, then any two cyclic subgroups are comparable, forcing every pair of elements to be powers of a common element; this rigidly constrains the group to be an abelian p-group with the ordered-subgroup structure of a Prüfer/cyclic p-group.",
"so... |
Prove that for any ring $R$, the left $R$-module $\operatorname{Hom}_{\mathbb{Z}}({}_R R, \mathbb{Q})$ is injective. | Let $V = \operatorname{Hom}_{\mathbb{Z}}({}_R R, \mathbb{Q})$ (with left $R$-action from the right $R$-action on $R$). To show $V$ is injective, it suffices to show the contravariant functor $\operatorname{Hom}_R(-, V) : R\text{-Mod} \to \mathbf{Ab}$ is exact. By the tensor–hom adjunction and the isomorphism ${}_R R \o... | {
"essential_property": "Hom_Z(R,Q) is the coinduced module from the injective (divisible) Z-module Q, and coinduction of an injective is injective, so it is an injective R-module (Baer/character-module).",
"solution_principle": "Use the adjunction Hom_R(-,Hom_Z(R,Q))≅Hom_Z(-,Q) and injectivity (exactness of Hom(-,... |
Let $M$ be a connected non-orientable manifold and $p : N \to M$ a connected $3$-fold covering. Prove that $N$ is non-orientable. | Let $\dim M = n$ and $\widetilde M = \{(x, \alpha) : \alpha \in H_n(M, M\setminus x) \text{ a generator}\}$ be the orientation double cover; $M$ is orientable iff $\widetilde M \to M$ is trivial. Since $p_* : H_n(N, N\setminus x) \to H_n(M, M\setminus p(x))$ is an isomorphism, the pullback gives a covering $\widetilde ... | {
"essential_property": "The orientation double cover of M pulls back to N; a 3-fold (odd) cover cannot trivialize the orientation Z/2-bundle, so N inherits non-orientability.",
"solution_principle": "Relate orientability to the orientation double cover being trivial; an odd-degree cover cannot split off the nontri... |
True or false (with proof or counterexample): There exists a nonzero natural transformation $\bigwedge^2 \to \mathrm{Id}$, where $\mathrm{Id}, \bigwedge^2 : \mathbf{Vec}_{\mathbb{C}} \to \mathbf{Vec}_{\mathbb{C}}$. | False. Suppose $\eta$ is a nonzero natural transformation. Pick $V$ and $x \in \bigwedge^2 V$ with $v := \eta_V(x) \ne 0$. Choose a linear map $f : V \to \mathbb{C}$ nonzero on $v$. The naturality square for $f$ gives a contradiction: one way around is $f(\eta_V(x)) = f(v) \ne 0$, while the other way factors through $\... | {
"essential_property": "A natural transformation ∧^2→Id would, by naturality against linear functionals, factor a nonzero vector through a map that must vanish, giving a contradiction.",
"solution_principle": "Assume a nonzero component, pick a functional detecting the output, and chase the naturality square to fo... |
Consider the disk $D^2 = \{x^2 + y^2 < 1\}$ with metric $g = \frac{1}{1 - (x^2+y^2)}(dx\otimes dx + dy\otimes dy)$. Compute the Levi-Civita connection. | For a diagonal metric $g = \lambda(dx^2 + dy^2)$ with $\lambda = \frac{1}{1 - r^2}$, $r^2 = x^2 + y^2$, the Christoffel symbols are $\Gamma^k_{ij} = \frac{1}{2}g^{kk}(\partial_i g_{kj} + \partial_j g_{ki} - \partial_k g_{ij})$. Computing (with $\partial_x\lambda = \frac{2x}{(1-r^2)^2}$, etc., and $g^{kk} = 1 - r^2$): $... | {
"essential_property": "For a conformal (diagonal) metric λ(dx^2+dy^2), the Levi-Civita connection's Christoffel symbols are given by the standard formula in terms of derivatives of log λ, computed directly.",
"solution_principle": "Plug the conformal factor λ=1/(1−r^2) into the Christoffel formula Γ=½g^{kk}(∂g+∂g... |
Let $R$ be a ring with center $Z$. A derivation $D : R \to R$ satisfies $D(a+b) = D(a) + D(b)$ and $D(ab) = aD(b) + D(a)b$. (a) Show $A_r(a) = ar - ra$ is a derivation. (b) Prove $D(Z) \subseteq Z$. (c) If $e \in Z$ is idempotent, prove $D(e) = 0$. | (a) $A_r(a+b) = (a+b)r - r(a+b) = A_r(a) + A_r(b)$; and $A_r(ab) = abr - rab = a(br - rb) + (ar - ra)b = aA_r(b) + A_r(a)b$. So $A_r$ is a derivation.
(b) Let $z \in Z$, $r \in R$. From $rz = zr$: $rD(z) + D(r)z = D(rz) = D(zr) = zD(r) + D(z)r$. Since $z$ commutes with $D(r)$, the $D(r)z$ and $zD(r)$ terms cancel, leav... | {
"essential_property": "Inner maps a↦ar-ra are derivations, and the Leibniz rule forces any derivation to preserve the center and to annihilate central idempotents.",
"solution_principle": "Verify the derivation axioms for the inner map, use commutativity with the center to trap D(z) in the center, and apply D to ... |
Let $(f_n)$ be holomorphic on $B_2(0)$ with $Z(n)$ the number of zeros of $f_n$ in $B_1(0)$ (with multiplicity). Suppose $f_n \to f$ uniformly on compact subsets of $B_2(0)$, no $f_n$ is identically zero, and $f(z) \ne 0$ for $|z| = 1$. Prove $\sup_n Z(n) < \infty$. | The uniform-on-compacts limit $f$ is holomorphic. Since $S = \{|z| = 1\}$ is compact and $f$ is nonvanishing there, $|f(z)| \ge r > 0$ on $S$. Choose $N$ with $\sup_{z \in S}|f_n(z) - f(z)| < r/2 \le |f(z)|$ for all $n \ge N$. By Rouché's theorem, $f_n$ and $f$ have the same number of zeros in $B_1(0)$ for $n \ge N$; i... | {
"essential_property": "By Hurwitz's theorem (Rouché applied to the uniform limit), the zero count of f_n in the disk stabilizes to that of the nonvanishing-on-boundary limit f, since |f_n−f|<|f| on the boundary circle.",
"solution_principle": "Use that f is nonvanishing on the compact boundary (|f|≥r), apply Rouc... |
What is the maximal dimension of a linear subspace of $\mathbb{R}^{2n+1}$ on which the quadratic form $x_0^2 + x_1 x_2 + x_3 x_4 + \cdots + x_{2n-1}x_{2n}$ takes only non-positive values? | The answer is $n$. The form is identically zero on the $n$-dimensional subspace $\{x_0 = x_2 = x_4 = \cdots = x_{2n} = 0\}$ (each product $x_{2i-1}x_{2i}$ has a factor set to $0$, and $x_0 = 0$), so a subspace of dimension $n$ with $Q \le 0$ exists. Conversely, the quadratic form is positive definite on the $(n+1)$-dim... | {
"essential_property": "The quadratic form has an n-dimensional totally isotropic (nonpositive) subspace and any larger subspace meets an (n+1)-dimensional positive-definite subspace nontrivially, so the maximal nonpositive subspace has dimension n (Witt index).",
"solution_principle": "Exhibit an n-dimensional su... |
Let $G$ be a group and $K \subseteq H$ subgroups with $K \trianglelefteq H$. (a) Prove $H$ normalizes $C_G(K)$. (b) If $H \trianglelefteq G$ and $C_H(K) = 1$, prove $H$ centralizes $C_G(K)$. | (a) Let $g \in C_G(K)$, $k \in K$, $h \in H$. From $gk = kg$, conjugating by $h$: $g^h k^h = k^h g^h$. Since $K \trianglelefteq H$, $k^h$ ranges over $K$, so $g^h$ centralizes $K$, i.e. $g^h \in C_G(K)$. Hence $C_G(K)^h \subseteq C_G(K)$, so $H$ normalizes $C_G(K)$.
(b) By (a), $\bar G = H\, C_G(K)$ is a group with $C_... | {
"essential_property": "Conjugation by H permutes K (since K◁H), so it preserves the centralizer C_G(K); when C_H(K)=1 and H◁G, H and C_G(K) intersect trivially and normalize each other, forcing H to centralize C_G(K).",
"solution_principle": "Show conjugation by h fixes the centralizing condition (so H normalizes... |
Evaluate $\int_{-\infty}^\infty \frac{\cos kx}{1 + x + x^2}\,dx$ for $k \ge 0$. | Let $f(z) = \frac{e^{ikz}}{z^2 + z + 1}$ and integrate over the contour $C_R = [-R,R] \cup \{$upper semicircle$\}$. Factor $z^2 + z + 1 = (z - \omega)(z - \bar\omega)$ with $\omega = \frac{-1 + \sqrt3 i}{2}$ (the root in the upper half-plane). The only enclosed pole is $z = \omega$, with residue $\frac{e^{ik\omega}}{\o... | {
"essential_property": "Integrating e^{ikz}/(z^2+z+1) over the upper semicircle picks up the single pole ω in the upper half-plane, and taking real parts gives the cosine integral (2π/√3)e^{−k√3/2}cos(k/2).",
"solution_principle": "Close in the upper half-plane, compute the residue at ω=−1/2+√3 i/2, bound the arc ... |
(a) Compute $H^k_{dR}(\mathbb{R}^n \setminus \{0\})$ for all $k$. (b) Show $\eta = \frac{1}{\|x\|^n}\sum_{i=1}^n (-1)^{i-1}x_i\, dx_1 \wedge \cdots \widehat{dx_i} \cdots \wedge dx_n$ is closed on $\mathbb{R}^n \setminus \{0\}$ with $\int_{S^{n-1}}\eta = \operatorname{Vol}(S^{n-1})$. (c) Deduce $[\eta]$ generates $H^{n-... | (a) $\mathbb{R}^n \setminus \{0\}$ deformation retracts onto $S^{n-1}$ (radially), so $H^k_{dR}(\mathbb{R}^n\setminus\{0\}) \cong H^k_{dR}(S^{n-1})$, which is $\mathbb{R}$ for $k = 0, n-1$ and $0$ otherwise.
(b) Let $\Omega = dx_1\wedge\cdots\wedge dx_n$ and $R = \sum_i x_i\partial_i$ the radial field; then $\eta = \|x... | {
"essential_property": "R^n∖{0} deformation retracts to S^{n-1}, so its de Rham cohomology is R in degrees 0 and n−1; the form η is the closed generator of H^{n-1} (the solid-angle form), not exact.",
"solution_principle": "Use the radial retraction to compute H^*_{dR}(R^n∖0)=H^*(S^{n-1}), then verify η is closed ... |
Let $L = \mathbb{C}(x)$, $K_1 = \mathbb{C}(x^2)$, $K_2 = \mathbb{C}((x-1)^2)$. Show $[L : K_1] = [L : K_2] = 2$ but $[L : K_1 \cap K_2] = \infty$. | $x$ is a root of $T^2 - x^2 \in K_1[T]$, so $[L : K_1] \le 2$; and $x \notin K_1 = \mathbb{C}(x^2)$ (elements of $K_1$ are rational functions in $x^2$, invariant under $x \mapsto -x$, whereas $x$ is not), so $[L : K_1] = 2$. Similarly $x$ satisfies a degree-$2$ polynomial over $K_2$ and $x \notin K_2$, so $[L : K_2] = ... | {
"essential_property": "x satisfies degree-2 relations over K_1 and K_2 (so [L:K_i]=2), but K_1∩K_2 consists of functions invariant under both reflections x↦−x and x↦2−x, which generate an infinite (dihedral) group, forcing K_1∩K_2=C and [L:C]=∞.",
"solution_principle": "Show [L:K_i]=2 via quadratic relations, the... |
Give examples of two non-isomorphic non-abelian groups of order $8$, and explain why they are not isomorphic. | Take the dihedral group $D_4$ (symmetries of a square) and the quaternion group $Q_8 = \{\pm1, \pm i, \pm j, \pm k\}$, both non-abelian of order $8$. They are not isomorphic because they have different numbers of elements of order $4$: in $D_4$ the elements of order $4$ are the two nontrivial rotations $r, r^3$ (just $... | {
"essential_property": "D_4 and Q_8 are non-isomorphic non-abelian order-8 groups distinguished by element orders: D_4 has two order-4 elements while Q_8 has six.",
"solution_principle": "Exhibit D_4 and Q_8 and count elements of order 4 (2 vs 6) to show they are not isomorphic.",
"core_concept": "The number of ... |
Let $E$ be a complex Banach space, $\xi_1, \ldots, \xi_n \in E$ linearly independent, and $\eta_1, \ldots, \eta_n \in E$. Prove there is a bounded linear map $T : E \to E$ with $T\xi_k = \eta_k$ for all $k$. | For each $k$ let $Z_k = \operatorname{span}\{\xi_j : j \ne k\}$, a finite-dimensional (hence closed) subspace not containing $\xi_k$ (by linear independence). By Hahn–Banach there is a bounded functional $\omega_k \in E^*$ with $\omega_k(\xi_k) = 1$ and $\omega_k|_{Z_k} = 0$, so $\omega_k(\xi_j) = \delta_{jk}$. Define ... | {
"essential_property": "Linear independence lets Hahn–Banach produce dual functionals ω_k with ω_k(ξ_j)=δ_{kj}, so T=Σ ω_k(·)η_k is a bounded operator sending ξ_k to η_k.",
"solution_principle": "Construct a biorthogonal system of bounded functionals via Hahn–Banach, then define T as the finite-rank operator Σω_k(... |
Let $X, Y$ be Banach spaces and $\varphi : X \times Y \to \mathbb{C}$ such that for each fixed $y$, $x \mapsto \varphi(x,y)$ is a continuous linear functional, and for each fixed $x$, $y \mapsto \varphi(x,y)$ is a continuous linear functional. Prove there is $C > 0$ with $|\varphi(x,y)| \le C\|x\|\|y\|$ for all $x, y$. | For each $y \in Y$, write $\varphi_y(x) = \varphi(x,y)$, a bounded functional, so $|\varphi(x,y)| \le \|\varphi_y\|\,\|x\|$. For each $x$ with $\|x\| \le 1$, the map $y \mapsto \varphi(x,y) = \varphi_x(y)$ is a bounded functional. Consider the family $\{\varphi_x : \|x\| \le 1\} \subset Y^*$. For each fixed $y$, $\sup_... | {
"essential_property": "Separate continuity of a bilinear form on Banach spaces upgrades to joint boundedness by the uniform boundedness principle applied to the family of partial functionals.",
"solution_principle": "Fix x in the unit ball to get a family of bounded functionals φ_x, use pointwise boundedness in y... |
Let $X$ be a normed space. Show that for any bounded sequence $\{x_n\}$ with $\|x_n\| \le M$, there is a subsequence $\{x_{n_k}\}$ and $x_\infty$ such that (i) $x_{n_k} \to x_\infty$ weakly, and (ii) $\|x_\infty\| \le M$. | View $\{x_n\} \subset X \hookrightarrow (X^*)^* = X^{**}$ via the canonical isometric embedding; they are bounded by $M$. By the Banach–Alaoglu theorem, the closed ball of radius $M$ in $X^{**}$ is weak*-compact; if $X^*$ is separable this gives a weak*-convergent subsequence, and in general one still extracts a subnet... | {
"essential_property": "Viewing a bounded sequence in the bidual, Banach–Alaoglu makes the ball weak*-compact, yielding a weakly convergent subsequence with limit of norm ≤M (weak lower semicontinuity of the norm).",
"solution_principle": "Embed the sequence into the bidual, use Banach–Alaoglu (with separability) ... |
Let $R$ be a ring, and let $M$ be a Noetherian left $R$-module. Let $f : M \to M$ be a surjective homomorphism. Prove that $f$ is an isomorphism. | The ascending chain $\ker(f) \subset \ker(f^2) \subset \ker(f^3) \subset \cdots$ stabilizes because $M$ is Noetherian. Assume $\ker(f^n) = \ker(f^{n+1})$. Since $f$ is surjective, so is $f^n$; the surjection $\ker(f^{n+1}) \to \ker(f)$, $x \mapsto f^n(x)$, has kernel $\ker(f^n) = \ker(f^{n+1})$, hence is the zero map, ... | {
"essential_property": "For a Noetherian module the ascending chain of kernels of powers of a surjection f stabilizes, and stabilization forces ker f=0, so f is injective and hence an isomorphism.",
"solution_principle": "Use the Noetherian ascending chain condition on ker f^n to stabilize, then exploit surjectivi... |
Find the number of conjugacy classes of $4\times4$ complex matrices $A$ with $A^4 = I$. | The polynomial $x^4 - 1 = (x-1)(x+1)(x-i)(x+i)$ has distinct roots, so any $A$ with $A^4 = I$ is diagonalizable with eigenvalues among $\{1, -1, i, -i\}$. Up to conjugacy, $A$ is determined by the multiplicities $(a, b, c, d)$ of these four eigenvalues, which are nonnegative integers summing to $4$. The number of such ... | {
"essential_property": "x^4−1 has distinct roots so A is diagonalizable with eigenvalues in {1,−1,i,−i}, and the conjugacy class is determined by the four multiplicities summing to 4, giving C(4+3,3)=35 classes.",
"solution_principle": "Diagonalize A (distinct roots), classify by eigenvalue-multiplicity vectors (a... |
Let $\pi : \mathbb{R}^n \times [0,1] \to \mathbb{R}^n$ be projection and $i_t(x) = (x,t)$. Writing $\omega = \tilde\omega + dt \wedge \eta$ (neither containing $dt$), define $G(\omega) = \sum \int_0^1 \eta_{j_1\cdots j_{k-1}}(x,t)\, dt\ dx^{j_1}\wedge\cdots\wedge dx^{j_{k-1}}$. Prove $dG(\omega) + G(d\omega) = i_1^*\om... | All operators are linear, so verify on the two types of terms. (i) $\omega = \tilde\omega = a(x,t)\, dx^1\wedge\cdots\wedge dx^k$ (no $dt$): here $G(\omega) = 0$, so $dG(\omega) = 0$; $d\omega = \partial_t a\, dt\wedge dx^1\wedge\cdots\wedge dx^k + (\text{no-}dt\text{ terms})$, so $G(d\omega) = \left(\int_0^1 \partial_... | {
"essential_property": "The fiber-integration operator G is a chain homotopy between the two inclusions i_0,i_1, satisfying dG+Gd=i_1^*−i_0^*, proving the homotopy invariance of de Rham cohomology.",
"solution_principle": "Check the homotopy formula dG+Gd=i_1^*−i_0^* on the two types of forms (with and without dt)... |
Let $V$ be finite-dimensional over an algebraically closed field $K$ and $T : V \to V$ linear. Show that $V$ has a basis of eigenvectors of $T$ if and only if $\ker(\lambda I - T)^2 = \ker(\lambda I - T)$ for all $\lambda \in K$. | ($\Rightarrow$) Let $\{v_i\}$ be an eigenbasis with $Tv_i = \lambda_i v_i$. For $v = \sum_i k_i v_i$ and any $\lambda$, $(\lambda I - T)v = \sum_i k_i(\lambda - \lambda_i)v_i$ and $(\lambda I - T)^2 v = \sum_i k_i(\lambda - \lambda_i)^2 v_i$. Since the $v_i$ are independent, $v \in \ker(\lambda I - T)$ iff $k_i = 0$ wh... | {
"essential_property": "Diagonalizability is equivalent to the absence of generalized eigenvectors, i.e. ker(λI-T)^2 = ker(λI-T) for every λ, which says the minimal polynomial is squarefree.",
"solution_principle": "Translate 'eigenbasis exists' into 'no Jordan block of size ≥2', which is exactly the stabilization... |
Prove that $\ell^1(\mathbb{Z})$ is not a Hilbert space: there is no scalar product on $\ell^1(\mathbb{Z})$ whose norm is the usual $\ell^1$ norm. | $\ell^1(\mathbb{Z})$ is separable, and by the Riesz representation for $\ell^1$, its dual is $\ell^\infty(\mathbb{Z})$, which is not separable. But the dual of a separable Hilbert space is separable (the map $\eta \mapsto \langle\cdot, \eta\rangle$ is an isometric bijection $H \to H^*$). So $\ell^1(\mathbb{Z})$ cannot ... | {
"essential_property": "A separable Hilbert space has a separable dual, but ℓ^1 is separable with non-separable dual ℓ^∞, so ℓ^1's norm cannot come from an inner product.",
"solution_principle": "Contrast separability: ℓ^1 is separable while its dual ℓ^∞ is not, whereas a Hilbert space's dual (isometric to itself)... |
Let $(H^n, g_n)$ be hyperbolic space with sectional curvature $-1$ ($n \ge 2$). For the product $H^m \times H^n$ with product metric, compute the sectional, Ricci, and scalar curvatures in an orthonormal frame. Find $\lambda > 0$ making $(H^m \times H^n, g_m \oplus \lambda g_n)$ Einstein. | Let $\{e_1, \ldots, e_m\}$, $\{f_1, \ldots, f_n\}$ be orthonormal frames on the factors; together they frame $T(H^m \times H^n)$. For a product metric $\nabla_{e_i}f_j = 0$, so mixed sectional curvatures vanish: $K(e_i, f_j) = 0$, while $K(e_i, e_j) = -1$ and $K(f_k, f_l) = -1$. Ricci: $\operatorname{Ric}(e_i, e_i) = \... | {
"essential_property": "For a product of hyperbolic factors, mixed sectional curvatures vanish and intra-factor ones are −1, so Ricci and scalar curvature are computed by summing; a suitable scaling makes the product Einstein.",
"solution_principle": "Use product connection formulas (mixed curvature zero) to compu... |
Let $A$ be a finite-dimensional algebra over $\mathbb{C}$ with center $Z(A)$. (a) Suppose $L$ is a finitely generated irreducible $A$-module. Prove any $z \in Z(A)$ acts on $L$ by a scalar $\lambda \cdot \mathrm{id}_L$. (b) For $z \in Z(A)$ acting by $\lambda$ on $M$ and by $\mu$ on $N$ with $\lambda \ne \mu$, prove an... | (a) A finitely generated module over a finite-dimensional $\mathbb{C}$-algebra is finite-dimensional, so $\operatorname{End}_A(L)$ is finite-dimensional; by Schur's lemma it is a division algebra over $\mathbb{C}$, hence $\operatorname{End}_A(L) = \mathbb{C}$. Since $z \in Z(A)$ commutes with the $A$-action, it lies in... | {
"essential_property": "A finitely generated irreducible module over a finite-dimensional C-algebra has endomorphism ring C (Schur over an algebraically closed field), so central elements, commuting with the action, act as scalars.",
"solution_principle": "Use finite-dimensionality and Schur's lemma to get End=C, ... |
Show that the differential equation $f''(z) = z f(z)$, $f(0) = 1$, $f'(0) = 1$, has a unique entire solution. | Seek $f(z) = \sum_{n\ge0} a_n z^n$. Matching $f'' = zf$ gives $a_0 = 1$, $a_1 = 1$, $a_2 = 0$, and the recurrence $k(k-1)a_k = a_{k-3}$ for $k \ge 3$. This determines all coefficients uniquely: $a_{3k} = \prod_{j=1}^k \frac{1}{3j(3j-1)}$, $a_{3k+1} = \prod_{j=1}^k \frac{1}{3j(3j+1)}$, and $a_{3k+2} = 0$. Since $\lim_{k... | {
"essential_property": "The equation f''=zf forces a three-term power-series recurrence k(k−1)a_k=a_{k-3} that uniquely determines all coefficients from the initial data, with infinite radius of convergence (entire).",
"solution_principle": "Substitute a power series, derive the recurrence, solve it uniquely from ... |
Let $f(z) = \log|z - a|$ with $a \in \mathbb{C}$, $|a| \ne 1$. Compute the average value of $f$ over the unit circle $|z| = 1$. | The average is $\log|a|$ if $|a| > 1$, and $0$ if $|a| < 1$. Reason: $f(z) = \log|z - a| = \Re\log(z - a)$ is harmonic away from $z = a$.
Case $|a| > 1$: $f$ is harmonic on the closed unit disc, so by the mean value property the average over $|z| = 1$ equals $f(0) = \log|{-a}| = \log|a|$.
Case $|a| < 1$: on the unit ci... | {
"essential_property": "log|z−a| is harmonic away from a, so its average over the unit circle equals its value at the center (mean value property): log|a| for |a|>1, and 0 for |a|<1 (using |z−a|=|1−āz| on the circle).",
"solution_principle": "Apply the mean value property for harmonic functions; for |a|<1 replace ... |
True or false: $\mathbb{Q}[x, x^{-1}]$ is a projective $\mathbb{Q}[x]$-module. | False. If it were projective it would be a direct summand of a free module, and any $\mathbb{Q}[x]$-module map to $\mathbb{Q}[x]$ would be determined nicely. Consider evaluation-at-$1$ maps $\mathbb{Q}[x] \to \mathbb{Q}$ and $\mathbb{Q}[x, x^{-1}] \to \mathbb{Q}$. There is no $\mathbb{Q}[x]$-module map $\mathbb{Q}[x, x... | {
"essential_property": "The localization Q[x,x^{-1}] is not projective over Q[x] because a projective would be a summand of a free module, but the localization's structure (no Q[x]-module map onto Q[x] detecting it) obstructs this.",
"solution_principle": "Show that projectivity would force a splitting/retraction ... |
State the Jacobson Density Theorem. | Let $R$ be a ring and $V$ an irreducible $R$-module. Let $D = \operatorname{End}_R(V)$, which is a division algebra by Schur's lemma. If $v_1, \ldots, v_n \in V$ are linearly independent over $D$, then for arbitrary $w_1, \ldots, w_n \in V$ there exists $r \in R$ such that $r v_i = w_i$ for all $i = 1, \ldots, n$. | {
"essential_property": "The Jacobson Density Theorem says an irreducible module makes the ring act densely as D-linear transformations: any D-independent vectors can be sent to arbitrary targets by a ring element.",
"solution_principle": "State that for irreducible V with D=End_R(V), R acts densely on V as a D-vec... |
Let $M$ be a compact connected $n$-manifold (a finite CW complex). (a) Prove that if $M$ is orientable then $H_{n-1}(M)$ is torsion-free. (b) For $n = 4$ with $\pi_1(M) \cong \mathbb{Z}/3$ and $\chi(M) = 4$, compute all $H_*(M)$. | (a) $M$ orientable gives $H_n(M) = \mathbb{Z}$, and $M$ is $\mathbb{Z}/k$-orientable for all $k$, so $H_n(M; \mathbb{Z}/p) = \mathbb{Z}/p$. By UCT, $H_n(M; \mathbb{Z}/p) = (H_n(M)\otimes\mathbb{Z}/p) \oplus \operatorname{Tor}(H_{n-1}(M), \mathbb{Z}/p)$; if $H_{n-1}(M)$ had $p$-torsion both summands would be nonzero, co... | {
"essential_property": "Orientability gives H_n=Z and, via Poincaré duality plus UCT, H_{n-1} torsion-free; with π_1=Z/3 and χ=4, duality and the known H_1 pin down all homology of the 4-manifold.",
"solution_principle": "Use Poincaré duality and UCT to kill torsion in H_{n-1}, then combine π_1=Z/3 (giving H_1) wi... |
True or false (with justification): Every torsion-free $\mathbb{C}[x]$-module is free. | False. The $\mathbb{C}[x]$-module $\mathbb{C}(x)$ (the field of rational functions) is torsion-free but not free. Any two elements of $\mathbb{C}(x)$ are linearly dependent over $\mathbb{C}[x]$, so a basis could have at most one element; but $\mathbb{C}(x)$ is not generated by a single element over $\mathbb{C}[x]$. Hen... | {
"essential_property": "Torsion-free need not mean free over C[x]: the field of fractions C(x) is torsion-free but not free, since any two elements are C[x]-dependent yet no single element generates it.",
"solution_principle": "Exhibit C(x) as a torsion-free but non-free C[x]-module by noting its rank-1 dependence... |
(a) Prove there is a simply-connected CW complex $X$ with $\tilde H_2(X; \mathbb{Z}) \cong \mathbb{Q}$ and $\tilde H_i(X; \mathbb{Z}) = 0$ for $i \ne 2$. (b) For this $X$, compute $\tilde H_*(X; \mathbb{Z}/2\mathbb{Z})$. | (a) Choose generators $\{r_i\}$ of $\mathbb{Q}$ as a $\mathbb{Z}$-module (e.g. $r_i = 1/i$). The map $\mathbb{Z}^\infty = \bigoplus_i \mathbb{Z} \to \mathbb{Q}$, $e_i \mapsto r_i$, is onto; its kernel is free abelian with a basis $\{s_j\}$. Build $X$ with $2$-skeleton $X^2 = \bigvee_i S^2_i$, and for each $s_j = (a_1, ... | {
"essential_property": "A Moore space M(Q,2) can be built as a mapping telescope/2-complex realizing H_2=Q; its Z/2-homology is computed by universal coefficients, where Q⊗Z/2=0 and Tor(Q,Z/2)=0 give vanishing reduced homology.",
"solution_principle": "Construct the 2-complex realizing Q as H_2 from a free present... |
Let $A_\bullet$ be a chain complex of abelian groups and $f : A_\bullet \to A_\bullet$ a chain map. Assume $s$ is a chain homotopy between $f^2$ and $\mathrm{id}$. Prove that $G : H_n(A) \to H_{n+1}(A)$, $[a] \mapsto [fsa - sfa]$, is a well-defined homomorphism. | First, for a cycle $a$ ($da = 0$): $d(fsa - sfa) = f(dsa) - ds(fa) = f(f^2 a - sda) - (f^2 - sd)(fa) = f^3 a - f^3 a = 0$ (using $df = fd$, $da = 0$, and $ds + sd = f^2 - \mathrm{id}$). So $[fsa - sfa]$ is a homology class. Next, if $a = dx$ is a boundary: $fsa - sfa = fsdx - sfdx = f(f^2 - \mathrm{id} - ds)x - sd(fx) ... | {
"essential_property": "Given a chain homotopy s between f^2 and id, the assignment G([a])=[fsa−sfa] is well-defined on homology because the bracketed expression is a cycle and changes by a boundary under boundaries.",
"solution_principle": "Verify d(fsa−sfa)=0 on cycles using df=fd and ds+sd=f^2−id, then check in... |
Let $V$ be finite-dimensional over $F$ and $T : V \to V$ linear, with $F[T]$ the ring of polynomials in $T$. Assume no nonzero proper subspace is $T$-invariant. (a) If $0 \ne S \in F[T]$, show $\{v : vS = 0\} = 0$. (b) Prove $F[T]$ is a field. (c) Show $[F[T]:F] = \dim_F V$. | (a) $W = \{v : vS = 0\}$ is a subspace $\ne V$ (as $S \ne 0$). Since $S \in F[T]$ commutes with $T$, $vS = 0 \Rightarrow (vT)S = (vS)T = 0$, so $WT \subseteq W$; by the irreducibility hypothesis $W = 0$.
(b) $F[T]$ is commutative and acts faithfully on $V$. For $0 \ne S \in F[T]$, (a) shows $S$ is injective, hence inve... | {
"essential_property": "Irreducibility of the T-action makes every nonzero element of the commutative algebra F[T] injective (its kernel is a T-invariant subspace), so F[T] is a field acting on V, and V is 1-dimensional over F[T], forcing [F[T]:F]=dim_F V.",
"solution_principle": "Show that any nonzero S in F[T] h... |
State the Homotopy Extension Property for a pair $(X, A)$. Let $Z$ be path-connected with $\pi_1(Z) = \pi_2(Z) = 0$ and $T$ the torus. Prove every map $T \to Z$ is homotopic to a constant. | HEP for $(X, A)$: for any $Z$ and maps $f : X \to Z$, $h : A \times I \to Z$ with $h_0 = f|_A$, there is $H : X \times I \to Z$ with $H_0 = f$ and $H|_{A\times I} = h$. Now let $f : T \to Z$. Give $T$ its usual CW structure with $1$-skeleton $S^1 \vee S^1$. Since $\pi_1(Z) = 0$, $f|_{S^1 \vee S^1}$ is null-homotopic vi... | {
"essential_property": "A map T→Z with π_1(Z)=π_2(Z)=0 is null-homotopic: obstruction theory over the CW skeleta of T vanishes because the relevant homotopy groups of Z are zero.",
"solution_principle": "Extend a nullhomotopy cell-by-cell using the HEP; the obstructions live in π_1(Z) and π_2(Z), both zero, so the... |
(a) Let $t : \mathbb{CP}^2 \times \mathbb{CP}^2 \to \mathbb{CP}^2 \times \mathbb{CP}^2$ be the twist $t(x, y) = (y, x)$. Prove any map homotopic to $t$ has a fixed point. (b) Is this true with $\mathbb{CP}^2$ replaced by an arbitrary space $X$? | (a) $H^*(\mathbb{CP}^2 \times \mathbb{CP}^2) = \mathbb{Z}[z, w]/(z^3, w^3)$ ($\deg z = \deg w = 2$). For $f \simeq t$, $f^* = t^*$ swaps $z \leftrightarrow w$. Compute traces: on $H^0, H^8$ trace $1$; on $H^2 = \langle z, w\rangle$ (swapped) trace $0$; on $H^6$ trace $0$; on $H^4 = \langle z^2, zw, w^2\rangle$, $t^*$ s... | {
"essential_property": "The Lefschetz number of the twist (and anything homotopic to it) on CP^2×CP^2 is nonzero: swapping the two cohomology generators contributes a nonzero trace, forcing a fixed point.",
"solution_principle": "Compute the Lefschetz number using the trace of the swap on H^*(CP^2×CP^2), find it n... |
Let $f_n$ be continuous on $[0,1]$. (a) If $f_n \to f$ pointwise, must $\int_0^1 f_n \to \int_0^1 f$? (b) If $f_n \to f$ uniformly, must $\int_0^1 f_n \to \int_0^1 f$? Prove or give a counterexample. | (a) False. Let $f_n$ be a continuous 'spike': $f_n(x) = 0$ for $x = 0$ or $x \ge 1/n$, rising to height $2n$ at $x = 1/(2n)$ so that $\int_0^1 f_n = 1$. Then $f_n \to 0$ pointwise (for each fixed $x > 0$, $f_n(x) = 0$ once $1/n < x$), but $\int_0^1 f_n = 1 \not\to 0 = \int_0^1 f$.
(b) True. Given $\varepsilon > 0$, uni... | {
"essential_property": "Pointwise convergence does not preserve integrals (a mass-1 spike converges pointwise to 0 with integral 1), but uniform convergence does (|∫f_n−∫f|≤∫|f_n−f|→0).",
"solution_principle": "Give the shrinking spike counterexample for (a) and the uniform bound for (b).",
"core_concept": "Unif... |
Prove that there is no map $\mu : \mathbb{RP}^5 \times \mathbb{RP}^5 \to \mathbb{RP}^5$ giving $\mathbb{RP}^5$ the structure of a topological group. | Apply $H^*(-; \mathbb{Z}/2)$: $H^*(\mathbb{RP}^5; \mathbb{Z}/2) = \mathbb{Z}/2[x]/(x^6)$, and by Künneth $H^*(\mathbb{RP}^5 \times \mathbb{RP}^5; \mathbb{Z}/2) = \mathbb{Z}/2[x, y]/(x^6, y^6)$. For a topological group with identity $p$, the standard argument shows $\mu^*(x) = x \otimes 1 + 1 \otimes x$: writing $\mu^*(... | {
"essential_property": "A topological group structure would make H^*(RP^5;Z/2) a Hopf algebra, but Z/2[x]/(x^6) is not a Hopf algebra (the coproduct would force incompatible relations), obstructing the group structure.",
"solution_principle": "Assume a multiplication with identity, deduce H^* is a Hopf algebra, an... |
Let $h : S^3 \to S^2$ be the Hopf map and $c : T^3 \to S^3$ collapse the complement of a ball to a point. Prove $h \circ c : T^3 \to S^2$ induces the trivial map on homology and homotopy but is not null-homotopic. | Homotopy: $(h\circ c)_* : \pi_q(T^3) \to \pi_q(S^3) \to \pi_q(S^2)$. Since $\pi_q(T^3) = 0$ for $q > 1$ and $\pi_1(S^3) = 0$, $c_*$ is trivial on all $\pi_q$, so $(h\circ c)_* = 0$. Homology: $\tilde H_q(T^3) \to \tilde H_q(S^3) \to \tilde H_q(S^2)$; the only common nonzero is impossible ($\tilde H_3(S^3) = \mathbb{Z}$... | {
"essential_property": "h∘c is trivial on homology and homotopy (c kills everything above degree 1 and h is into simply connected S^2 range structure), yet is not null-homotopic because it has nonzero Hopf invariant / cohomology-operation content.",
"solution_principle": "Show (h∘c)_* vanishes on all homotopy and ... |
Let $V$ be finite-dimensional over $F$, $T : V \to V$ linear with minimal polynomial $f$. (a) If $f$ has a nonconstant factor of degree $m$, show $V$ has a nonzero subspace $W$ of dimension $\le m$ with $T(W) \subseteq W$. (b) Conversely, if $V$ has a nonzero $T$-invariant subspace $W$ of dimension $n$, show $f$ has a ... | (a) Write $f = gh$ with $\deg g = m > 0$. Since $\deg h < \deg f$ and $f$ is minimal, $h(T) \ne 0$, so choose $v$ with $w = h(T)v \ne 0$. Let $W = \operatorname{span}\{w, Tw, \ldots, T^{m-1}w\}$, of dimension $\le m$, nonzero. To show $T$-invariance, it suffices that $T^m w \in W$: dividing $X^m = g(X)q(X) + r(X)$ ($r ... | {
"essential_property": "A degree-m factor of the minimal polynomial yields a cyclic subspace of dimension ≤m generated by h(T)v (where f=gh), which is T-invariant; conversely a small invariant subspace forces a low-degree minimal-polynomial factor.",
"solution_principle": "Construct the cyclic subspace on h(T)v an... |
Let $X \subset \mathbb{P}^n$ be an irreducible projective variety of dimension $k$, and $G(\ell, n)$ the Grassmannian of $\ell$-planes with $\ell < n - k$. Let $C(X) \subset G(\ell,n)$ be the $\ell$-planes meeting $X$. Prove $C(X)$ is irreducible and find its dimension. | Let $H = \{(h,x) : h \in G(\ell,n),\, x \in h\} \subset G(\ell,n)\times\mathbb{P}^n$ be the universal $\ell$-plane, and $V = H \cap (G(\ell,n)\times X)$ with projections $\mathrm{pr}_1 : V \to G(\ell,n)$, $\mathrm{pr}_2 : V \to X$. For fixed $x \in X$, the fiber $V_x = \{h : x \in h\}$ is the set of $\ell$-planes throu... | {
"essential_property": "The set of ℓ-planes meeting X is the image of the incidence variety of (point of X, ℓ-plane through it), fibered over X with Grassmannian fibers, hence irreducible; a dimension count gives dim C(X)=k+ℓ(n−ℓ).",
"solution_principle": "Form V={(x,h):x∈h∩X} fibered over X (fibers ≅ ℓ-planes thr... |
Prove that an element $g$ of a finite group $G$ is conjugate to $g^{-1}$ if and only if $\chi(g)$ is a real number for every character $\chi$. | The irreducible characters form a basis of the space of class functions $\mathbb{C}(G)$, as do the indicator functions of the conjugacy classes. Hence two elements $g, h$ are conjugate iff $\chi(g) = \chi(h)$ for all irreducible characters $\chi$. Now $\chi(g^{-1}) = \overline{\chi(g)}$ for every character. So $g$ and ... | {
"essential_property": "Characters distinguish conjugacy classes, and χ(g)=χ(g^{-1}) always holds via complex conjugation, so g∼g^{-1} iff all χ(g) are real (equal to their conjugates).",
"solution_principle": "Use that irreducible characters separate classes and χ(g^{-1})=conj χ(g); realness of all χ(g) means χ(g... |
Let $X = \{(x,y,z) \in \mathbb{R}^3 : xy = 0\}$ (the union of the $xz$- and $yz$-planes). Prove that for any neighborhood $U \subset X$ of the origin, $U \setminus 0$ retracts onto a set with non-abelian fundamental group, and use this to show $X$ is not a manifold. | Choose $\epsilon > 0$ with the ball of radius $2\epsilon$ about $0$ inside $U$. Let $Y = X \cap \{|v| = \epsilon\}$: this is two points joined by four arcs, homotopy equivalent (contracting one arc) to a wedge of three circles, so $\pi_1(Y)$ is free on three generators, non-abelian. The map $r : U \setminus 0 \to Y$, $... | {
"essential_property": "The link of the origin in the union of two planes is two points joined by four arcs, homotopy equivalent to a wedge of three circles, whose fundamental group (free on 3 generators) is non-abelian.",
"solution_principle": "Take a small sphere's intersection with X as the link, identify it as... |
Let $R$ be a ring with $1$ having a nil ideal $N$ such that $R/N$ has no zero divisors. (a) Show the only idempotents of $R$ are $0$ and $1$. (b) If $R/N$ is a division ring, prove every zero divisor of $R$ is nilpotent. | (a) Let $\bar{}: R \to R/N$. If $e^2 = e$, then $e(1-e) = 0$, so $\bar e(1 - \bar e) = 0$ in the domain $R/N$, giving $\bar e = 0$ or $\overline{1 - e} = 0$, i.e. $e \in N$ or $1 - e \in N$. Both $e$ and $1 - e$ are idempotents; an idempotent $f \in N$ satisfies $f = f^k = 0$ (N nil). So $e = 0$ or $1 - e = 0$, i.e. $e... | {
"essential_property": "A nil ideal N with R/N a domain forces idempotents to reduce to 0 or 1 mod N, and since idempotents in a nil ideal are 0, the only idempotents are 0 and 1; when R/N is a division ring, zero divisors are exactly the elements reducing to 0, i.e. nilpotents.",
"solution_principle": "Reduce ide... |
Let $A$ be an $n \times n$ complex matrix of rank $1$. (a) What are the possible Jordan canonical forms of $A$? (b) For each, compute the characteristic and minimal polynomials. | (a) The Jordan form $B$ also has rank $1$, so it has exactly one nonzero Jordan block and all others zero. A single Jordan block of rank $1$ is either $(\lambda)$ with $\lambda \ne 0$ (size $1$) or $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ (size $2$, eigenvalue $0$). Hence either $B = \operatorname{diag}(\lambda, 0, \ldot... | {
"essential_property": "A rank-1 matrix has a single nonzero Jordan block, so its Jordan form is either diag(tr A,0,…,0) with nonzero trace or a single 2×2 nilpotent block with the rest zero, according to whether the trace is nonzero.",
"solution_principle": "Use rank preservation under similarity to limit the Jor... |
Let $f$ be continuous on the closed strip $\{a \le x \le b\}$ ($z = x + iy$), holomorphic on the interior, with $|f(z)| = O(e^{\varepsilon|y|})$ for every $\varepsilon > 0$ as $|y| \to \infty$. If $|f| \le M$ on the two boundary lines and on $[a, b]$, prove $|f| \le M$ on the whole strip. | This is a Phragmén–Lindelöf argument. Fix $\varepsilon > 0$ and let $g_\varepsilon(z) = e^{\varepsilon i z}f(z)$. On the vertical boundaries, $|g_\varepsilon(a + iy)| = e^{-\varepsilon y}|f(a+iy)|$; using the growth bound $|f(x+iy)| \le C_\varepsilon e^{(\varepsilon/2)|y|}$, for $y \ge T_\varepsilon$ (large) we get $|g... | {
"essential_property": "Phragmén–Lindelöf: multiplying f by e^{εiz} tames the growth so the maximum modulus principle applies on the strip, and letting ε→0 gives |f|≤M throughout.",
"solution_principle": "For each ε, apply the maximum modulus principle to g_ε=e^{εiz}f on large rectangles (the growth bound controls... |
True or false: if $R$ is an Artinian ring with no non-zero nilpotent elements, then $R$ is a direct sum of division rings. | True. Since $R$ is Artinian, its Jacobson radical $J(R)$ is nilpotent; as $R$ has no nonzero nilpotent elements, $J(R) = 0$, so $R$ is semisimple. By the Wedderburn–Artin theorem, $R$ is a finite direct sum of matrix rings $M_{n_i}(D_i)$ over division rings. But any $M_n(D)$ with $n > 1$ contains nonzero nilpotents (e.... | {
"essential_property": "An Artinian ring has nilpotent Jacobson radical; no nilpotents forces J=0, so it is semisimple, and no nilpotents also excludes matrix blocks larger than 1×1, leaving a product of division rings.",
"solution_principle": "Use J nilpotent and no nilpotents to get J=0 (semisimple), then Wedder... |
Let $M$ be a smooth manifold and $L \to M$ a smooth real line bundle. (1) Show $L$ is trivial iff it has a smooth nowhere-vanishing section. (2) For $M = S^1$ and $L = [0,1]\times\mathbb{R}/((0,t)\sim(1,-t))$ (the Möbius bundle), show $L$ is nontrivial. (3) Show $L \otimes L$ is trivial. | (1) If $L \cong M \times \mathbb{R}$ is trivial, then $\sigma(x) = (x, 1)$ is a nowhere-vanishing section. Conversely, given a nowhere-vanishing section $\sigma$, define $M \times \mathbb{R} \to L$ by $(x, t) \mapsto t\,\sigma(x)$; this is smooth, linear and nonzero on each fiber (since $\sigma(x) \ne 0$), hence a bund... | {
"essential_property": "A line bundle is trivial iff it has a nowhere-vanishing section; the Möbius bundle's sections must vanish (a sign flip forces a zero by the IVT), so it is nontrivial, but L⊗L admits the constant section (sign flips cancel), so it is trivial.",
"solution_principle": "Prove triviality⇔nowhere... |
Let $V$ be a vector space over $\mathbb{Q}$ and $v_1, \ldots, v_n \in V$. Show there exist $w_1, \ldots, w_m \in V$, linearly independent over $\mathbb{Q}$, with $\sum_{i=1}^n v_i \mathbb{Z} = \sum_{j=1}^m w_j \mathbb{Z}$ (equality of additive subgroups). | Let $A = \sum_{i=1}^n v_i \mathbb{Z}$, a finitely generated additive abelian subgroup of $V$ (generated by $v_1, \ldots, v_n$). It is torsion-free (a subgroup of the $\mathbb{Q}$-vector space $V$), so by the structure theorem it is free: $A = \bigoplus_{j=1}^m w_j\mathbb{Z}$ for some $w_j \in V$. It remains to show the... | {
"essential_property": "A finitely generated subgroup of a rational vector space is torsion-free, hence free abelian by the structure theorem, and a free basis inside a Q-vector space is automatically Q-linearly independent.",
"solution_principle": "Recognize the additive span as a finitely generated torsion-free ... |
Let $K$ be a field and $R \subset K[X]$ the subring of polynomials with $X$-coefficient $0$. (a) Prove $X^2$ and $X^3$ are irreducible but not prime in $R$. (b) Show $R$ is Noetherian, and the ideal $I$ of polynomials in $R$ with constant term $0$ is not principal. | (a) $X^2, X^3 \in R$ (no $X$-term). If $X^3 = fg$ in $R \subseteq K[X]$, then in the UFD $K[X]$, up to constants $\{f, g\} = \{X, X^2\}$ or one is constant; but $X \notin R$, so one factor is a constant (a unit of $R$): $X^3$ is irreducible. Similarly $X^2$. Not prime: $X^3 \mid (X^2)(X^4)$ in $R$ (as $X^6 \in R$) but ... | {
"essential_property": "In the subring missing the linear term, X^2 and X^3 are irreducible (their K[X]-factors leave R) but not prime (X^3 | X^2·X^4 without dividing either), yet R is Noetherian while the augmentation ideal is non-principal.",
"solution_principle": "Use the UFD structure of K[X] to show the only ... |
Let $R$ be commutative, $P$ a prime ideal, $V$ a right $R$-module, and $W = \{v \in V : va = 0 \text{ for some } a \in R \setminus P\}$. (i) Show $W$ is a submodule. (ii) If $R$ is Noetherian and $V$ finitely generated, prove $Wb = 0$ for some $b \in R \setminus P$. (iii) If $V$ is simple and $W = 0$, prove $P$ is maxi... | (i) $0 \in W$. If $v_1, v_2 \in W$ with $v_i a_i = 0$, $a_i \notin P$, then $(v_1 + v_2)a_1 a_2 = 0$ and $a_1 a_2 \notin P$ (P prime), so $v_1 + v_2 \in W$; and $(v_1 r)a_1 = 0$ gives $v_1 r \in W$. So $W$ is a submodule.
(ii) $V$ Noetherian, so $W = w_1 R + \cdots + w_n R$ is finitely generated. Choose $b_i \in R\setm... | {
"essential_property": "The set W of elements annihilated by something outside the prime P is a submodule (P-torsion), and finite generation lets a single element outside P annihilate all of W; simplicity plus W=0 forces P to be maximal.",
"solution_principle": "Use primeness to combine annihilators multiplicative... |
Let $V$ be a right $R$-module with $V = X \oplus Y$. For an $R$-homomorphism $\theta : X \to Y$, let $W_\theta = \{x - \theta(x) : x \in X\}$. (a) Show $W_\theta$ is a submodule and $V = W_\theta \oplus Y$. (b) Conversely, if $U$ is a submodule with $V = U \oplus Y$, prove $U = W_\theta$ for some $\theta$. | (a) $W_\theta$ is the image of the module map $x \mapsto x - \theta(x)$, hence a submodule. For $x \in X$, $x = (x - \theta(x)) + \theta(x) \in W_\theta + Y$, so $X, Y \subseteq W_\theta + Y$ and $V = W_\theta + Y$. If $y \in W_\theta \cap Y$, write $y = x - \theta(x)$; then $x = y + \theta(x) \in Y$, so $x \in X \cap ... | {
"essential_property": "Complements of a fixed summand Y are parametrized by homomorphisms X→Y via their graphs, so each complement is the graph W_θ of a unique θ.",
"solution_principle": "Construct the graph submodule W_θ={x-θ(x)} as a complement, and conversely recover θ from any complement U by projecting X int... |
True or false: if $F, G : \mathcal{A} \to \mathcal{B}$ are isomorphic functors and $f, g \in \operatorname{Hom}_{\mathcal{A}}(A_1, A_2)$, then $F(f) = F(g)$ if and only if $G(f) = G(g)$. | True. Let $\alpha : F \to G$ be a natural isomorphism. Naturality gives $Gf = \alpha_{A_2} \circ Ff \circ \alpha_{A_1}^{-1}$ and $Gg = \alpha_{A_2} \circ Fg \circ \alpha_{A_1}^{-1}$. Hence $F(f) = F(g) \iff \alpha_{A_2}Ff\alpha_{A_1}^{-1} = \alpha_{A_2}Fg\alpha_{A_1}^{-1} \iff G(f) = G(g)$. | {
"essential_property": "A natural isomorphism conjugates F(h) to G(h) by fixed isomorphisms, so equality of morphisms is preserved: F(f)=F(g) iff G(f)=G(g).",
"solution_principle": "Use Gh=α_{A_2}Fh α_{A_1}^{-1} to conjugate, so the two functors agree on morphism equalities.",
"core_concept": "Isomorphic functor... |
Prove that there is no infinitely differentiable function on $\mathbb{R}$ all of whose derivatives are $0$ at $0$ and positive everywhere else. (Hint: show $f^{(n)}(1) \ge n!\,f(1)$.) | Suppose such $f$ exists. By Taylor's formula with remainder, expanding at $a = 0$ to $b = 1$: for each $n$ there is $\xi_n \in (0,1)$ with $f(1) = \sum_{k=0}^{n-1} f^{(k)}(0)\frac{1}{k!} + f^{(n)}(\xi_n)\frac{1}{n!} = \frac{f^{(n)}(\xi_n)}{n!}$ (all lower derivatives at $0$ vanish). Each $f^{(n)}$ has positive derivati... | {
"essential_property": "If all derivatives vanish at 0 and are positive elsewhere, Taylor's remainder forces f^{(n)}(1)≥n!f(1), so f(2)>n f(1) for every n, making f(2) unbounded—a contradiction.",
"solution_principle": "Use Taylor's formula at 0 to get f(1)=f^{(n)}(ξ_n)/n! and monotonicity of derivatives to bound ... |
Prove that the group $\mathrm{GL}_6(\mathbb{F}_2)$ has an element of order $63$. | The field $\mathbb{F}_{64}$ is a $6$-dimensional vector space over $\mathbb{F}_2$. Multiplication by a fixed element of $\mathbb{F}_{64}$ is an $\mathbb{F}_2$-linear map, giving an embedding of groups $\mathbb{F}_{64}^\times \hookrightarrow \mathrm{GL}_6(\mathbb{F}_2)$. The group $\mathbb{F}_{64}^\times$ is cyclic of o... | {
"essential_property": "F_64 is a 6-dimensional F_2-vector space, and multiplication by a generator of the cyclic group F_64^× (order 63) embeds into GL_6(F_2), producing an element of order 63.",
"solution_principle": "Use the regular representation of the field F_64 over F_2 to embed F_64^× into GL_6(F_2), then ... |
The period of a rational $q$ in base $n$ is the least length of the eventually-repeating block of its base-$n$ expansion. For $p$ prime and $n > 1$ with $p \nmid n$, show the period of $1/p$ in base $n$ is at most $p - 1$, and this maximum is achieved for some $n$. | A positive integer $m$ is a multiple of the period of $1/p$ in base $n$ iff $n^m \cdot \frac1p$ has the same fractional part as $\frac1p$, i.e. iff $\frac{n^m - 1}{p}$ is an integer, i.e. iff $p \mid (n^m - 1)$, i.e. iff $n^m \equiv 1 \pmod p$. By definition of multiplicative order, this holds iff $m$ is a multiple of ... | {
"essential_property": "The period of 1/p in base n equals the multiplicative order of n modulo p, since m is a period-multiple iff p | n^m−1; this order divides p−1 (at most p−1) and is attained for a primitive root n.",
"solution_principle": "Show m is a multiple of the period iff n^m≡1 (mod p), identifying the ... |
Let $A$ be an abelian group and $B \le A$. (a) If $B$ is a direct factor of $A$, show $B$ is a direct factor of every subgroup $C$ with $B \subseteq C \subseteq A$. (b) Conversely, if $B$ is a direct factor of every $C$ with $B \subseteq C \subseteq A$ and $C/B$ cyclic, and $|A:B| < \infty$, show $B$ is a direct factor... | (Additive notation.) (a) Say $A = B \oplus X$. For $B \subseteq C \subseteq A$, the modular law gives $C = A \cap C = (B + X) \cap C = B + (X \cap C)$, and $B \cap (X \cap C) \subseteq B \cap X = 0$, so $C = B \oplus (X \cap C)$.
(b) $\bar A = A/B$ is finite abelian, so $\bar A = \bar C_1 \oplus \cdots \oplus \bar C_n$... | {
"essential_property": "A direct factor is inherited by intermediate subgroups via the modular law; conversely, being a factor of every intermediate subgroup with cyclic quotient lets one lift a cyclic decomposition of A/B to a complement of B.",
"solution_principle": "Use the modular law for the forward inheritan... |
Let $F \subseteq E$ be an algebraic field extension. Call $\alpha \in E$ abelian if $F[\alpha]$ is Galois over $F$ with abelian Galois group. Prove that the set of abelian elements of $E$ is a subfield of $E$ containing $F$. | Every element of $F$ is abelian (it generates $F$, trivially Galois abelian), so the set contains $F$. Let $\alpha, \beta$ be abelian. $F[\alpha]$ and $F[\beta]$ are splitting fields of separable polynomials $f, g \in F[x]$, so $F[\alpha, \beta]$ is the splitting field of the separable $fg$, hence Galois over $F$. Let ... | {
"essential_property": "The compositum of two Galois extensions with abelian Galois groups is again Galois, and its Galois group embeds into the product of the two abelian groups, hence is abelian.",
"solution_principle": "Show the abelian-element set is closed under field operations by passing to the compositum: ... |
Let $K = \mathbb{Q}(\alpha)$ where $\alpha = e^{2\pi i/12}$. (a) Describe the Galois group of $K/\mathbb{Q}$ and its action. (b) Find the minimal polynomial of $\alpha$. (c) Describe the intermediate fields strictly between $\mathbb{Q}$ and $K$, each as $\mathbb{Q}(\sqrt d)$. | (a) The conjugates of $\alpha$ are the primitive $12$th roots of unity $\alpha^r$ for $r \in \{1, 5, 7, 11\}$ (those coprime to $12$). An automorphism is determined by $\alpha \mapsto \alpha^r$, and each has order $2$ since $r^2 \equiv 1 \pmod{12}$ for these $r$. So $\operatorname{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/1... | {
"essential_property": "The Galois group of Q(ζ_12) is (Z/12)^×≅C_2×C_2, with each nontrivial automorphism an involution, minimal polynomial Φ_12=x^4−x^2+1, and three quadratic subfields Q(√{−1}), Q(√{−3}), Q(√3).",
"solution_principle": "Identify Gal=(Z/12)^×, note every element squares to 1 (so C_2×C_2), compute... |
Let $k$ be a field, and let $R$ be a simple non-commutative $k$-algebra such that $\dim_k R = 4$. Prove that the center of $R$ is $k$. | By the Artin–Wedderburn theorem, $R = M_n(D)$, where $D$ is a division ring over $k$. If $n = 2$ then $D = k$ (since $\dim_k R = n^2 \dim_k D = 4$), and the center of $R = M_2(k)$ is $k$. Assume now $n = 1$, so $R = D$, and let $C \subset D$ be the center. Since $D$ is noncommutative, we only have to exclude the case $... | {
"essential_property": "By Artin–Wedderburn a simple 4-dimensional k-algebra is M_2(k) or a 4-dimensional division algebra; in either case, noncommutativity forces the center to be exactly k.",
"solution_principle": "Classify R via Artin–Wedderburn as M_n(D) with n^2 dim_k D=4, and exclude the possibility of a cen... |
Working over an algebraically closed field $F$, prove that the circle $x^2 + y^2 = 1$ and $\mathbb{A}^1$ are isomorphic as algebraic sets if and only if $\operatorname{char} F = 2$. | If $\operatorname{char}F = 2$: $x^2 + y^2 - 1 = (x + y + 1)^2$, so $V(x^2+y^2-1) = V(x+y+1)$ is a line, isomorphic to $\mathbb{A}^1$. If $\operatorname{char}F \ne 2$: $x^2 + y^2 - 1$ is irreducible, so $F[S^1] = F[x,y]/(x^2+y^2-1)$; we show it is not isomorphic to $F[T]$. Suppose $\phi : F[x,y]/(x^2+y^2-1) \xrightarrow... | {
"essential_property": "In characteristic 2, x^2+y^2−1=(x+y+1)^2, so the circle degenerates to a line ≅A^1; in other characteristics x^2+y^2−1 is irreducible and its coordinate ring is not a polynomial ring, so no isomorphism.",
"solution_principle": "Use the perfect-square factorization in characteristic 2 to ide... |
Let $M = (-1,1)^n \subset \mathbb{R}^n$ and $\Omega^p_c(M)$ the compactly supported $p$-forms, with compactly supported de Rham cohomology $H^p_c(M)$. Prove that $\mathbb{R} \subseteq H^n_c(M)$ (i.e. $H^n_c(M) \ne 0$). | Define $E : H^n_c(M) \to \mathbb{R}$ by $E(\eta) = \int_M \eta$. This is well-defined on cohomology: if $\eta = d\omega$ with $\omega \in \Omega^{n-1}_c(M)$, then by Stokes' theorem $\int_M d\omega = \int_{\partial} \omega = 0$ (the support of $\omega$ is compact inside $M$, so the boundary term over a large surface ne... | {
"essential_property": "Integration ∫_M gives a well-defined nonzero functional on H^n_c(M) (Stokes kills exact compactly supported forms), so H^n_c(M)≠0.",
"solution_principle": "Define E(η)=∫_M η on compactly supported top cohomology, use Stokes to see it vanishes on exact forms, and exhibit a bump form with non... |
Let $F \subseteq E$ be an algebraic field extension. (a) If every $f \in F[X]$ splits over $E$, prove $E$ is algebraically closed. (b) If every $f \in F[X]$ has a root in $E$ and $\operatorname{char} F = 0$, prove $E$ is algebraically closed. | (a) Let $g \in E[X]$ be irreducible; we show $g$ is linear. Adjoin a root $\alpha$ to get $K = E[\alpha]$, algebraic over $E$ hence over $F$. Let $f \in F[X]$ be the minimal polynomial of $\alpha$ over $F$. By hypothesis $f$ splits over $E$, so all its roots lie in $E$; in particular $\alpha \in E$. Thus $g$ has a root... | {
"essential_property": "If every polynomial over F splits over the algebraic extension E, then E is algebraically closed; in characteristic 0, even 'every polynomial has a root' suffices, via separability and adjoining roots to reach a full splitting.",
"solution_principle": "Show an arbitrary irreducible over E i... |
Let $F$ be nondecreasing on $[a,b]$. Show $F' \in L^1([a,b])$ and $\int_a^b F'\, dm \le F(b) - F(a)$, with strict inequality if $F$ is discontinuous at some $t \in (a,b)$. | Extend $F$ by $F(a)$ on $[a-1, a]$ and $F(b)$ on $[b, b+1]$. Set $\varphi_n(x) = \frac{F(x + 1/n) - F(x)}{1/n} \ge 0$; these are integrable, and $\int_a^b \varphi_n = n\left(\int_b^{b+1/n} F - \int_a^{a+1/n} F\right) \le F(b) - F(a)$ (using monotonicity). Since $F$ is differentiable a.e. (monotone), $\varphi_n \to F' \... | {
"essential_property": "Difference-quotient approximations of F' are nonnegative with uniformly bounded integrals (telescoping to F(b)−F(a)), so Fatou gives F'∈L^1 with ∫F'≤F(b)−F(a); jumps lose mass, giving strict inequality.",
"solution_principle": "Apply Fatou to the nonnegative difference quotients φ_n whose i... |
Prove that for any $n$-manifold $M$, the total space $TM$ of the tangent bundle is orientable. | At any point $(p, v) \in TM$, given a chart $\phi : U \to \phi(U) \subseteq \mathbb{R}^n$ around $p$, the induced chart on $TU$ is $\Phi : TU \to \phi(U) \times \mathbb{R}^n$, $(q, w) \mapsto (\phi(q), d\phi_q(w))$. These form an atlas for $TM$. On an overlap $U \cap V$, the transition map is $\Phi_2 \circ \Phi_1^{-1}(... | {
"essential_property": "The transition maps of TM act on fibers by the derivative dφ and on the base, so the total Jacobian is a block matrix whose determinant is (det of base Jacobian)^2>0, making TM orientable.",
"solution_principle": "Compute the induced chart transitions on TM and show their Jacobian determina... |
Let $\mathbb{F}_q$ be a finite field with $q$ elements and $V$ an $n$-dimensional $\mathbb{F}_q$-vector space. (1) Number of elements of $V$. (2) Order of $GL_n(\mathbb{F}_q)$. (3) Order of $SL_n(\mathbb{F}_q)$. | (1) $|V| = q^n$. (2) $|GL_n(\mathbb{F}_q)|$ equals the number of ordered bases of $V$: the first vector can be any of $q^n - 1$ nonzero vectors, the second any of $q^n - q$ (outside the span of the first), and so on, giving $|GL_n(\mathbb{F}_q)| = (q^n - 1)(q^n - q)\cdots(q^n - q^{n-1})$. (3) The determinant map $GL_n(... | {
"essential_property": "GL_n(F_q) is counted as ordered bases (∏(q^n−q^i)), SL_n is its kernel under the surjective determinant to F_q^× (order q−1), and |V|=q^n.",
"solution_principle": "Count ordered bases for |GL_n|, divide by |F_q^×|=q−1 for |SL_n| via the determinant surjection, and note |V|=q^n.",
"core_co... |
Use contour integration to prove that for real $a > b > 0$, $\int_0^\pi \frac{d\theta}{a - b\cos\theta} = \frac{\pi}{\sqrt{a^2 - b^2}}$. | By symmetry about $\theta = \pi$, $\int_0^\pi \frac{d\theta}{a - b\cos\theta} = \frac12\int_0^{2\pi}\frac{d\theta}{a - b\cos\theta}$. Substitute $z = e^{i\theta}$, $\cos\theta = \frac12(z + z^{-1})$, $d\theta = \frac{dz}{iz}$, over $|z| = 1$: the integral becomes $\frac12\oint_{|z|=1}\frac{1}{a - \frac b2(z + z^{-1})}\... | {
"essential_property": "Substituting z=e^{iθ} turns the real integral into a contour integral with a single simple pole inside the unit circle, whose residue gives π/√(a^2−b^2).",
"solution_principle": "Symmetrize to [0,2π], substitute z=e^{iθ}, identify the interior root of the quadratic denominator, and take its... |
Let $X, Y$ be closed connected oriented surfaces and $f : Y \to X$ a branched double cover, branched over $x_1, \ldots, x_n$ (where $f$ looks like $z \mapsto z^2$), an honest $2$-sheeted cover over $U = X \setminus \{x_i\}$. Assume $n \ge 1$. Prove that $f_* : \pi_1(Y) \to \pi_1(X)$ is surjective. | Fix a branch point $x_1$ with disk neighborhood $V$ over which $f$ is $z\mapsto z^2$ and $f^{-1}(V)$ is a disk. Pick basepoint $x \in V \setminus x_1$ with $f^{-1}(x) = \{y, y'\}$, and a path $\alpha$ in $f^{-1}(V)$ from $y'$ to $y$; then $f\circ\alpha$ is a null-homotopic loop in $X$ (it stays in the contractible $V$)... | {
"essential_property": "A branched double cover is an honest 2-sheeted cover away from the branch points, where local monodromy is z↦z^2; the Riemann–Hurwitz formula relates the Euler characteristics/genera through the branch data.",
"solution_principle": "Analyze the local z↦z^2 model at branch points to fix the ... |
Let $G$ be a finite group of automorphisms of a ring $A$, and $R = A^G = \{a \in A : ga = a\ \forall g \in G\}$ the subring of invariants. Prove that the extension $R \subseteq A$ is integral. | For any $f \in A$, consider the polynomial $P_f(t) = \prod_{g \in G}(t - g\cdot f) \in A[t]$. Its coefficients are the elementary symmetric functions of the elements $\{g\cdot f : g \in G\}$, which are permuted by every element of $G$; hence the coefficients are $G$-invariant, i.e. lie in $R = A^G$. So $P_f$ is a monic... | {
"essential_property": "Each element f∈A satisfies the monic polynomial ∏_{g∈G}(t−g·f) whose coefficients are G-invariant (symmetric functions), hence lie in R=A^G, so A is integral over R.",
"solution_principle": "Construct the G-orbit polynomial of f; its coefficients are symmetric in the orbit, hence G-invarian... |
Let $\Omega \subset \mathbb{C}$ be open and bounded, and $C(\overline{\Omega})$ the continuous functions on $\overline{\Omega}$. Show that $\{f \in C(\overline{\Omega}) : f \text{ holomorphic on } \Omega\}$ is a closed subspace of $C(\overline{\Omega})$. | Let $\{f_n\}$ be a sequence of functions in the set converging uniformly (in the sup norm of $C(\overline\Omega)$) to $f \in C(\overline\Omega)$. We must show $f$ is holomorphic on $\Omega$. Fix $z \in \Omega$ and a disk $D(z, r) \subset \Omega$. For any triangle $\Delta \subset D(z,r)$, each $f_n$ is holomorphic so $\... | {
"essential_property": "Uniform limits of holomorphic functions are holomorphic: the vanishing of triangle integrals passes to the limit, so Morera's theorem makes the limit holomorphic, giving a closed subspace.",
"solution_principle": "Take a uniformly convergent sequence of holomorphic functions, pass the zero ... |
Let $k$ be a field and $a, b \in k^\times$. Let $A_{a,b}$ be the non-commutative $k$-algebra generated by $x, y, z$ modulo the relations $zy = yz + az$, $yx = xy + bx$, $zx = xz + y$. (a) Prove that $A_{a,b}$ has a $k$-basis $\{x^k y^\ell z^m\}_{k,\ell,m \ge 0}$ if and only if $a = b$; and if $a \ne b$, prove $A_{a,b} ... | (a) Apply the Bergman diamond lemma. Order the generators $x < y < z$ and use deglex order (satisfies the descending chain condition). The relations become reduction rules $zy \mapsto yz + az$, $yx \mapsto xy + bx$, $zx \mapsto xz + y$. A monomial is irreducible iff it contains none of $zy, yx, zx$ as a subword, i.e. i... | {
"essential_property": "By the Bergman diamond lemma, the PBW monomials form a basis exactly when all overlap ambiguities are resolvable, and this confluence holds iff the parameters satisfy a=b.",
"solution_principle": "Set up the reduction rules from the relations, check the overlap (diamond) ambiguities, and se... |
Let $f$ be entire with $\int_0^{2\pi} |f(re^{i\theta})|\,d\theta \le r^{17/3}$ for all $r > 0$. Prove $f \equiv 0$. | Write $f(z) = \sum_{n\ge0} a_n z^n$. By the Cauchy integral formula for coefficients, $a_n = \frac{1}{2\pi}\int_0^{2\pi}\frac{f(re^{i\theta})}{(re^{i\theta})^n}\,d\theta$ for every $r > 0$, so
$$|a_n| \le \frac{1}{2\pi r^n}\int_0^{2\pi}|f(re^{i\theta})|\,d\theta \le \frac{r^{17/3 - n}}{2\pi}.$$
For $n \le 5$: $17/3 - n... | {
"essential_property": "Cauchy's coefficient estimate bounds |a_n|≤r^{17/3−n}/(2π); letting r→0 for n≤5 and r→∞ for n≥6 kills every coefficient, so f≡0.",
"solution_principle": "Bound each Taylor coefficient using the L^1 growth hypothesis, then take the appropriate limit in r for small and large n to force a_n=0.... |
End of preview. Expand in Data Studio
math-phd-qual-709
709 open-ended proof problems from mathematics Ph.D. qualifying exams (1991–2026), each with a human-authored proof and a Mathematical Primitive: the essential conceptual observation that reveals why the problem can be solved. This is the training set of Absorb, the primitive-privileged self-distillation method released in Math-Primitive.
Every row has only question, answer, and primitive. No other metadata is included.
Schema
{
"question": str, # qualifying-exam problem
"answer": str, # human-authored proof (median 118 words); these are proof problems,
# so there is no short final answer
"primitive": {
"essential_property": str,
"solution_principle": str,
"core_concept": str, # the one-sentence primitive (median 18 words)
}
}
Sources: University of Oregon 332, UW Madison 131, Harvard 126, UC Berkeley 120. Domains: Algebra 318, Analysis 206, Topology 97, Geometry 82, Number Theory 5, Logic 1.
How it is used
- Absorb conditions the teacher (the same model) on
core_conceptas privileged information and transfers its guidance to the student, which sees only thequestion.core_conceptis exactly the primitive used to train the releasedqwen-3.5-{4,9,27}b-absorbmodels. - The OPSD baseline uses
answer(the full proof) as the privileged context instead.
Construction
Each primitive was extracted from the problem and its human-authored proof with the same
curation prompt as the Prim benchmark. The training primitive is the core_concept field
verbatim.
Usage
from datasets import load_dataset
ds = load_dataset("shuoxing/math-phd-qual-709", split="train")
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