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Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
21
votes
4
answers
2k
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The first female algebraist in US/Britain?
Recently I dug up some biographical details of Lindsay Burch, of Hilbert-Burch Theorem fame, whose few papers have had quite an impact on commutative algebra. This made me curious about the first wome …
20
votes
Accepted
Can a module be an extension in two really different ways?
It is worth noting some very interesting cases when the answer is yes. An amazing result by Miyata states that if $R$ is Noetherian and commutative, $M,N$ are finitely generated and $E \cong M\oplus …
15
votes
Accepted
Maximal Ideals in Formal Laurent Series Rings?
Your ring $L$ is a localization of the power series rings $R= k[[x_1,\cdots,x_n]]$ at the multiplicative set $M$ of monomials in $R$.
So the prime ideals of $L$ correspond to prime ideals in $R$ whi …
15
votes
The first female algebraist in US/Britain?
I followed the reference suggested by KConrad in the comments and found perhaps the answer to Question 1:
Annie MacKinnon, who got her PhD from Cornell in 1894 with the thesis "Concomitant Binary Form …
12
votes
Upper bound to the number of generators
Understanding the number of generators is a very subtle problem. I will focus on your second question on ideals, since the first one is a bit broad. By a theorem of Foster-Swan, the problem is local.
…
10
votes
A geometric reference for (affine) Gorenstein varieties and singularities
Sorry, tough luck, but most first (and second) algebraic geometry courses don't even touch Cohen-Macaulay rings, let alone Gorenstein. Look, for example, at Definition 4.2 here. So it is unlikely you …
10
votes
Accepted
Examples of one-dimensional non-Cohen Macaulay rings
In dimension $1$, Cohen-Macaulay just mean unmixed, so all the associated primes have the same dimension. Thus the easiest way to cook up a non-CM ring of dimension $1$ is: Pick your favorite regular …
8
votes
Square of primary ideals
What a fun problem for the holiday season! Yes, an example is $R=\mathbb C[a,b,c,d]/(a^4-bd, b^3-cd, c^2-ad, a^3b^2c-d^3) \cong \mathbb C[t^9,t^{13},t^{16},t^{23}]$. Let $I=(a,b,c)$ and $P=(a,b,c,d)$ …
5
votes
Accepted
Depth Zero Ideals in the Homogenized Weyl Algebra
I am not sure I understand the analogue correctly, but in the commutative case, one can get to depth zero with 3 generators. That is because any second syzygy of a module of depth at least $1$ is isom …
4
votes
$K_{0}(R) =\mathbb{Z}$ but some f.g. projective not stably free?
Under the mild condition that the rank of a free module is an invariant, the free modules form a monoid isomorphic to $\mathbb N$. This induces an injection:
$$f: \mathbb Z \to K_0(R) $$
Then every …
4
votes
0
answers
179
views
Global dimensions of orders over non-Gorenstein centers
This question concerns the following Lemma 4.2 in this paper by Van den Bergh:
Lemma: Let $R$ be local, normal Gorenstein ring of dimension $d$. Suppose $M$ is a reflexive $R$ module such that $A=\te …
4
votes
Dimension 1 prime ideals in the intersection of two maximal ideals
This reminds me of what someone once said to me:
The geometers can always take a hyperplane section. We can't!
The purpose of this post is to analyze the question to see how close it is to Bert …
4
votes
Ideals invariant under ring automorphisms
One can see that $I=(f_1,\dots, f_s)$ is $G$-invariant iff $\alpha\cdot f_i \in I$ for each $\alpha \in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and …
3
votes
Familiar equations in more general settings
Your question reminds me of the following quote from my advisor, which I can't resist posting here:
"Finally I want to remark that the treatment of big Cohen-Macaulay modules here serves as a remind …
2
votes
Locally square implies square
I would try something like this: $R=\mathbb C[x,y,u,v]/(f,g)$ with $f=x(y-u^2)$ and $g=(1+x)(y-v^2)$
When you localize at any prime ideal, you have to invert either $x$ or $1+x$. Either way, $y$ beco …