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 ...
21
votes
Accepted
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 $(a^2+b^2)^2 - (a^2-b^2)^2 =4 (ab)^2$.
18
votes
Accepted
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 \...
17
votes
Accepted
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 "...
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.
...
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 ...
16
votes
Accepted
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.
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, $\...
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 ...
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 ...
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)...
12
votes
Accepted
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 ...
12
votes
Accepted
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 ...
12
votes
Accepted
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 ...
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 ...
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 ...
11
votes
Accepted
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 ...
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 ...
11
votes
Accepted
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 ...
11
votes
Accepted
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\...
11
votes
Accepted
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.)
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, ...
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 ...
10
votes
Accepted
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. ...
10
votes
Accepted
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 ...
10
votes
Accepted
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 ...
10
votes
Accepted
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
$$ ...
9
votes
Accepted
Alternative description of strict henselization
(Assuming $R$ is not complete)
For 2, the answer is no, as suggested in a now-deleted comment of LSpice.
For $k$ of characteristic not $2$, with $R$ the localization of $k[t]$ at $0$, the element $\...
9
votes
Accepted
What is the intersection of all ideals whose radicals are prime?
Is the intersection of all fuzzy primes $\{0\}$?
Not always. Let me describe a commutative ring where the intersection of the fuzzy primes is nonzero.
Plan. The idea will be to construct a commutative ...
9
votes
Accepted
Can the ring of symmetric polynomials be generated by powers of symmetric polynomials of degree 1?
Yes.
For variables $x_1,\dots, x_k$, we have
$$ \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 ...
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