New answers tagged

0 votes

Gorenstein projective module over commutative local algebras

Regarding Question 2, if a local ring has finitely many indecomposable Gorenstein projective modules, and at least one non-free, then the ring must have isolated singularity and must be an abstract ...
sdey's user avatar
  • 612
5 votes

Equivalences of categories of complexes of modules

The answer is yes by the same type of Morita theory, namely $Z(Ch(R))\cong R$, where $Z(A)$ is $End(id_A)$, the ring of endomorphisms of an abelian category EDIT : sorry, I hadn't seen that you were ...
Maxime Ramzi's user avatar
  • 11.7k
3 votes
Accepted

Topological modules over a locally compact ring

Not necessarily. Start from $A=\mathbf{R}[t]/(t^2)$ and $R$ its closed cocompact unital subring $\{a+bt:a\in\mathbf{Z},b\in\mathbf{R}\}$. Write $J=tA$. Then every additive subgroup of $R^2$ contained ...
YCor's user avatar
  • 57.8k
1 vote

Is a valuation domain PID when its maximal ideal is principal?

$\def\bbZ{\mathbb{Z}} \def\frm{\mathfrak{m}} \def\bbQ{\mathbb{Q}}$(For me, a DVR is never a field.) As a complement to Pete L. Clark's answer, here's a recipe to construct a valuation ring with $\bbZ\...
Elías Guisado Villalgordo's user avatar
1 vote

Completion of $\mathbb F_q(T)$

Have you tried using Hensel's lemma to show $|T - \alpha|_P < 1$ for some $\alpha$ in $\mathbf F_{q^d}$ and then show $T-\alpha$ is a uniformizer in the completion? Once you have a uniformizer $t$ ...
KConrad's user avatar
  • 48.5k
7 votes
Accepted

Are topological PID's Noetherian?

We can study the conjecture for locally compact rings as follows. In [Kap] a subset $S$ of a topological ring is called algebraically nilpotent if for some $n$, $S^n=0$. Lemma 1. [Kap, Theorem 2] A ...
Alex Ravsky's user avatar
  • 3,627
6 votes
Accepted

GCD in $\mathbb{F}_3[T]$ with powers of linear polynomials

The smallest counterexample is $n=1+(3^{16}-1)/32$. It seems that SageMath (which uses the flint library as a backend for univariate polynomials over finite fields) is far superior than Gap. Indeed, ...
Peter Mueller's user avatar
8 votes

Are topological PID's Noetherian?

I think the answer to your question is negative. Let $\Omega$ be a non-empty open subset of $\mathbb{C}$. Choose $(z_n)_n$ a sequence of points in $\Omega$ such that, for all $n$ $$z_n \not\in \...
Romain Gicquaud's user avatar
7 votes
Accepted

Subalgebras of quadratic algebras that are not quadratic

Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and ...
Vladimir Dotsenko's user avatar
7 votes

Concept associated to the Eudoxus reals

This method can be used to construct the fields $\mathbb{Q}_p$ and the ring $\mathbb{A}_{\mathbb{Q}}$ of adeles over $\mathbb{Q}$. See T.D.J. Hermans' Bachelor's thesis: https://www....
Jesse Elliott's user avatar
1 vote

What is a fat point?

The definition I know is definition 1A.2 in "Lecture Notes on Motivic Cohomology" by Mazza, Voevodsky, Weibel.
T. Wildwolf's user avatar
2 votes
Accepted

Embedding noetherian domains in a PID with finite index

The answer to the first question (and therefore also to the second question) is negative, for an elementary reason: If the ring of integers $O_K$ of a number field $K$ has finite index (as an additive ...
GNiklasch's user avatar
  • 2,222
6 votes
Accepted

If $(f,g)$ and $(f,h)$ are maximal ideals, then $ag+bh=P(f)$ for some $a,b \in k, P(t) \in k[t]$?

The answer is no. Let $$f=x, \quad g=y+xy^2, \quad h=y+1+xy^3.$$ Then $$ (f,g)=(x,y)\quad\text{and}\quad (f,h)=(x,y+1) $$ are maximal ideals of $k[x,y]$, but for all $a,b\in k$ not both zero, you find ...
Jérémy Blanc's user avatar
3 votes
Accepted

When is the Tor-dimension of $R/(r)$ strictly smaller than that of $R$?

There is no difference between flat (Tor) dimension and global dimension for Noetherian rings. Now, it is well known that for an arbitrary local $R$ and a nonzerodivisor $f\in\mathfrak{m}$ either $\...
Anton Fonarev's user avatar
4 votes
Accepted

Is the projective dimension of finite torsion-free modules over regular ring of dimension $n$ smaller that $n$?

Yes. If $\operatorname{pd}_A(M)=n $, there exists a maximal ideal $\mathfrak{m}$ of $A$ such that $\operatorname{pd}_{A_{\mathfrak{m}}}(M_{\mathfrak{m}})=n $. By the Auslander-Buchsbaum theorem, this ...
abx's user avatar
  • 36.5k
6 votes
Accepted

Representation of a number as a product of $\sqrt{n^2 + 1} + n$

$\def\supp{\mathop{\mathrm{supp}}}$ Surely, in the ``Almost equivalent question'' you assume that the $d$'s are square-free. Denote $R=\mathbb Z[\sqrt{p_1},\dots,\sqrt{p_n}]$. We prove that in its ...
Ilya Bogdanov's user avatar
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 $(a^2+b^2)^2 - (a^2-b^2)^2 =4 (ab)^2$.
Will Sawin's user avatar
  • 129k
4 votes
Accepted

$K_0((k[x]/(x^2))[y])$

Let $P$ be a finitely generated projective module over your ring $S=R[y]=k[x,y]/y^2$. Then $P/yP$ is a finitely generated projective module over $k[x]$, and so is free by the Quillen-Suslin theorem. ...
Neil Strickland's user avatar
0 votes

Must 'special' $u,v \in \mathbb{C}[x,y]$ be symmetric polynomials?

A counterexample to the edit: $u=x^4+y^2$, $v=x^5+y^3$. However, it is not a counterexample to the original question before the edit, since for example $\mathbb{C}(x^4+y^2,x^5+y^3,y^6) \subsetneq \...
user237522's user avatar
  • 2,715
3 votes

Are large powers of polynomials linearly independent?

$\require{AMScd} \require{enclose}$EDIT : As noted by Zach Teitler, the argument below only proves that for $m\gg0$, the family $\left\{P_1^{\otimes m}, \dotsc, P_k^{\otimes m} \right\}$ is a free ...
Libli's user avatar
  • 6,960
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
9 votes
Accepted

When $R $ is a cusp then $K_0(R) \ncong K_0(R[s])$

Presumably you are looking at the conductor square on the left below, where $\epsilon^2=0$: $$\matrix{k[t^2,t^3]&\rightarrow& k[t]\cr \downarrow&&\downarrow\cr k&\rightarrow &k[...
Steven Landsburg's user avatar
6 votes

Is a proper map of varieties which is a bijection on points an isomorphism?

This is true. By Zariski's main theorem, we know that $f$ is finite, since it is proper and quasi-finite [Stacks, Tag 02LS]. We actually have the following: Lemma. Let $Y$ be a reduced scheme and let $...
R. van Dobben de Bruyn's user avatar
1 vote
Accepted

$k(F_i)_{i=1}^{n}=k(G_j)_{j=1}^{m}$ iff there exist $a_i,b_j \in k$ such that $\langle F_i-a_i \rangle_{i=1}^{n} = \langle G_j-b_j \rangle_{j=1}^{m}$

If $F_1 = 1$ and $G_1 = x,G_2 = x+1$, we have $(F_1 - 0) = (G_1 - 0, G_2 - 0)$ are both the trivial ideal $(1) = \mathbb{C}[x, y]$.
Justin Bloom's user avatar
7 votes

Double dual of free $\mathbb{Z}_{(p)}$-modules

E. E. Enochs, “A note on reflexive modules”, Pacific J. Math. 14 (1964), 879–881 shows that, over a discrete valuation ring $R$, the free module with denumerable base is reflexive (meaning that the ...
Gro-Tsen's user avatar
  • 27.9k
9 votes
Accepted

Double dual of free $\mathbb{Z}_{(p)}$-modules

There is at least one proof of Specker's theorem that can be adapted in an obvious way. I believe that the first half of this proof is due to Sąsiada, and the second half to Łoś. Let $A$ be a free $\...
Jeremy Rickard's user avatar
5 votes
Accepted

Ideals: If $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$, then $\langle f_1-\lambda,f_2-\mu \rangle = \langle g_1-\delta,g_2-\epsilon \rangle$?

The answer to your questions is no. The ideals $\langle x, y(1-xy) \rangle$ and $\langle x, y \rangle$ are equal, and maximal; but $$ \langle x-\lambda, y(1-xy)\rangle \neq \langle x-\delta, y-\...
Zach Teitler's user avatar
  • 5,732

Top 50 recent answers are included