## New answers tagged ac.commutative-algebra

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 ...

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 ...

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 ...

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\...

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 $t$ ...

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 ...

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, ...

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 \...

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 ...

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....

1
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### What is a fat point?

The definition I know is definition 1A.2 in "Lecture Notes on Motivic Cohomology" by Mazza, Voevodsky, Weibel.

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 ...

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 ...

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 $\...

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 ...

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 ...

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 $(a^2+b^2)^2 - (a^2-b^2)^2 =4 (ab)^2$.

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. ...

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 \...

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 ...

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 ...

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[...

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 $...

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 $(1) = \mathbb{C}[x, y]$.

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 ...

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 $\...

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-\...

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