26 votes

What are some toy models for the stable homotopy groups of spheres?

My favorite warmup example to the stable homotopy groups of spheres is the following differential graded algebra. Let $A$ have the underlying ring $$ \Bbb Z[y] \otimes \Lambda[x], $$ a ring with a ...
Tyler Lawson's user avatar
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21 votes
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If some powers of polynomials are linearly independent, does it imply higher powers are also independent?

No for $N=1$ and $M=2$. For example $a^2+b^2, a^2-b^2, $ and $ab$ are linearly independent but $(a^2+b^2)^2 - (a^2-b^2)^2 =4 (ab)^2$.
Will Sawin's user avatar
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18 votes
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Is diagonalizability a local property?

Here is a simple counterexample (simplified and generalised after Mohan's comment): let $R$ be a domain with a non-free finite projective module $P$, and let $Q$ be a finite projective module with $P \...
R. van Dobben de Bruyn's user avatar
17 votes
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Is it a valuation ring?

This is (essentially) a conjecture of Krull; a counter-example was given by P. Ribenboim: Sur une note de Nagata relative à un problème de Krull, Math. Zeit. 64, 159-168 (1956). Note that "...
abx's user avatar
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17 votes

Multiply an integer polynomial with another integer polynomial to get a "big" coefficient

A similar fact is well known: A polynomial $f(x)$ with complex coefficients divides a polynomial with positive coefficients if and only $f(x)$ has no nonnegative root. A similar strategy works here. ...
Ilya Bogdanov's user avatar
16 votes

Why the stable module category?

To question 1: One big motivation for me is that two Frobenius algebras can be stable equivalent but not Morita equivalent and a classification up to stable equivalence can be very nice. For example a ...
Mare's user avatar
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16 votes
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In an abelian category with no nontrivial Serre subcategories, does every short exact sequence split?

The category of finite abelian $p$-groups (where $p$ is your favourite prime) is an abelian category with no proper nonzero Serre subcategories, but not every short exact sequence splits.
Jeremy Rickard's user avatar
15 votes

What are some toy models for the stable homotopy groups of spheres?

You could say that I've made a living out of looking at the stable module category of a finite group (or rather its slight enlargement, the homotopy category of complexes of injective modules, $\...
Dave Benson's user avatar
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14 votes

A condition that implies commutativity

Yes, it is possible to write down an equational proof for every $n$. This is covered in the preprint Equational proofs of Jacobson's Theorem, arXiv:2310.05301 [math.RA] The rough idea is to prove ...
Martin Brandenburg's user avatar
13 votes

Why is $\operatorname{Spec}(\mathbb Z)$ supposed to lie over $\operatorname{Spec}(\mathbb F_1)$ rather than the other way around?

(I'm not sure this qualifies as an answer or an extended comment, because I don't have any deep knowledge in the matter, but I hope this can at least help clear some possible confusion.) It seems to ...
Gro-Tsen's user avatar
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12 votes

Hilbert polynomials of graded algebras evaluated at negative numbers

For $n \gg 0$, we have that ${\rm dim~} R_n = {\rm dim~} H^0(X, \mathcal O(n))$, where $X = {\rm Proj}(R) \subseteq \mathbb P(R_1^{\vee})$ is the projective scheme associated to $R$ and $\mathcal O(n)...
Phil Tosteson's user avatar
12 votes
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irreducibility of the polynomial $ x^4 +1 $

Among $-1$, $2$, and $-2$, there are $0$, $1$ or $3$ squares. This determines the irreducible factorization of $X^4+1$. If none is a square, it is irreducible; If only $-1=i^2$ is a square, the ...
YCor's user avatar
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12 votes
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If $\mathbb{C}(u(x,y),v(x,y),f(x))=\mathbb{C}(x,y)$, for every $f(x) \in \mathbb{C}[x]-\mathbb{C}$, then already $\mathbb{C}(u,v)=\mathbb{C}(x,y)$?

The answer to the Question is ``no". As an example, let $u=xy$, $v=x+y$ be the elementary symmetric functions in $x$ and $y$. It is well-known that $[\mathbb{C}(x,y): \mathbb{C}(u,v)]=2$, so ...
Joachim König's user avatar
12 votes
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Prove that $\Bbb C[x,y]/(x^3+y^3-1)$ is not a UFD

Here is a proof using a bit of commutative algebra and algebraic geometry, elaborating on comments by @abx and @ChrisWuthrich. The ring $$R=\mathbb{C}[x,y]/(x^3+y^3-1)$$ is a Dedekind domain, because ...
GH from MO's user avatar
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11 votes

Are large powers of polynomials linearly independent?

We have used this problem for our Student Olympiad in Algebra at Moscow State University (in Russian, Пятнадцатая олимпиада, задача 8). So, here is a completely elementary solution. Exercise 1. Show ...
Anton Klyachko's user avatar
11 votes

A game on Noetherian rings

This exact game was studied and published in 1970: Henson's "Winning Strategies for the Ideal Game". In particular, if $R$ is a single-variable polynomial over any PID then the first player ...
Chris Gerig's user avatar
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11 votes
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Hilbert polynomials of graded algebras evaluated at negative numbers

This answer is essentially the same as that of Phil Tosteson, written before I saw that post. I also mention a non-Cohen-Macaulay example at the end. If $R$ is Cohen-Macaulay (but not necessarily ...
Richard Stanley's user avatar
11 votes

Why the stable module category?

One reason is just that $\text{stab}_{kG}(k,M)_*=\widehat{H}^{-*}(g;M)$ (the Tate cohomology of $G$ with coefficients in $M$). I think that Tate invented Tate cohomology for applications in Class ...
Neil Strickland's user avatar
11 votes
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Discovery of Hilbert polynomial

In the 1st chapter of Eisenbud's book that you had mentioned, he discusses four fundamental theorems of Hilbert that appeared in 1890 and 1893 (see p. 26): the basis theorem, the Nullstellensatz, the ...
KConrad's user avatar
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11 votes
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Is every character of the algebra of continuous functions on a locally compact space some evaluation?

If $X=\omega_1$ with the order topology, then $X$ is locally compact, but whenever $(Y,d)$ is a metric space, every continuous function $f:X\rightarrow Y$ is eventually constant, so the function $f:X\...
Joseph Van Name's user avatar
11 votes
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Original proof of Hilbert irreducibility theorem

See Hilbert's Proof of His Irreducibility Theorem by Mark B. Villarino, Bill Gasarch, and Kenneth Regan. (Also available as an arXiv preprint.)
Peter Mueller's user avatar
10 votes

Excellent property of rings

On page 260 of Matsumura, "Commutative Ring Theory" (CUP, 1986), he states that "One can prove that the classes of rings satisfying (1), (2) and (3) are closed under localisation, ...
Dave Benson's user avatar
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10 votes

Is the spectrum of this ring Noetherian?

In general, if $R$ is a ring of characteristic $p > 0$, write $F_*^n R$ for the ring $R$ but viewed as $R$-algebra via the $p^n$-power Frobenius $R \to F_*^nR$, $x \mapsto x^{p^n}$. Define the ...
R. van Dobben de Bruyn's user avatar
10 votes
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Ring in which $x^n-x$ is central for every $x$

Herstein, "A generalization of a theorem of Jacobson" Amer. J. Math. 73 (1951), 756–762 proves this. Also part III, Amer. J. Math. 75 (1953), 105–111 proves something a bit more general. ...
Dave Benson's user avatar
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10 votes
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Alterations and smooth complete intersections

As you guessed, there are cohomological obstructions. Indeed, if $f \colon Y \twoheadrightarrow X$ is a surjective morphism of smooth projective varieties and $H$ is a Weil cohomology theory (with ...
R. van Dobben de Bruyn's user avatar
10 votes
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A question on algebraic independence

This is true, but it's actually a bit subtle, because the completion $\hat A \to \hat B$ of an injective local homomorphism $A \to B$ of local domains need not be injective. Given $f_1,\ldots,f_n \in ...
R. van Dobben de Bruyn's user avatar
10 votes
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A polynomial identity for $\displaystyle \sum_{k=0} ^m (-1)^ka^{m-k}b^k$

If we introduce the elementary symmetric polynomials $e_1 = a+b$, $e_2 = ab$ and the power sums $p_q = a^q + b^q$, then from the Newton identities we have $$ p_1 = e_1; p_2 = e_1 p_1 - e_2$$ and $$ ...
Terry Tao's user avatar
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9 votes
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Alternative description of strict henselization

(Assuming $R$ is not complete) For 2, the answer is no, as suggested in a now-deleted comment of LSpice. For $k$ of characteristic not $2$, with $R$ the localization of $k[t]$ at $0$, the element $\...
Will Sawin's user avatar
  • 135k
9 votes
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What is the intersection of all ideals whose radicals are prime?

Is the intersection of all fuzzy primes $\{0\}$? Not always. Let me describe a commutative ring where the intersection of the fuzzy primes is nonzero. Plan. The idea will be to construct a commutative ...
Keith Kearnes's user avatar
9 votes
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Can the ring of symmetric polynomials be generated by powers of symmetric polynomials of degree 1?

Yes. For variables $x_1,\dots, x_k$, we have $$ \sum_{J \subseteq \{1,\dots, k \} } (-1)^{ k- |J|} \left(\sum_{j \in J} x_j\right)^k = k! x_1 \dots x_k $$ so that $$ e_k = \sum_{(i_1; \dots; i_k )\in ...
Will Sawin's user avatar
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