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
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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
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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
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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 $n$
$$z_n \not\in \...
7
votes
Accepted
Subalgebras of quadratic algebras that are not quadratic
Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and ...
7
votes
Concept associated to the Eudoxus reals
This method can be used to construct the fields $\mathbb{Q}_p$ and the ring $\mathbb{A}_{\mathbb{Q}}$ of adeles over $\mathbb{Q}$. See T.D.J. Hermans' Bachelor's thesis: https://www....
1
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What is a fat point?
The definition I know is definition 1A.2 in "Lecture Notes on Motivic Cohomology" by Mazza, Voevodsky, Weibel.
2
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Embedding noetherian domains in a PID with finite index
The answer to the first question (and therefore also to the second question) is negative, for an elementary reason: If the ring of integers $O_K$ of a number field $K$ has finite index (as an additive ...
6
votes
Accepted
If $(f,g)$ and $(f,h)$ are maximal ideals, then $ag+bh=P(f)$ for some $a,b \in k, P(t) \in k[t]$?
The answer is no. Let
$$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
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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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