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Results tagged with ac.commutative-algebra
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user 750
Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
46
votes
How to memorise (understand) Nakayama's lemma and its corollaries?
The Graded Nakayama's Lemma
My intuition for Nakayama's lemma is rooted in the graded version.
(Graded Nakayama's Lemma)
Let $R$ be a $\mathbb{N}$-graded algebra, and let $R_+$ be the 'irrelevant' i …
32
votes
3
answers
5k
views
Krull dimension less or equal than transcendence degree?
Let $k$ be a field, and $A$ a $k$-domain, so that the fraction field of $A$ has transcendence degree $n$ over $k$.
If $A$ is finitely-generated over $k$, then $A$ has Krull dimension $n$ (Theorem A i …
23
votes
Accepted
How much theory works out for "almost commutative" rings?
Don't get too excited about the theory of algebraic geometry for almost commutative algebras. A ring can be almost commutative and still have some very weird behavior. The Weyl algebras (the differe …
17
votes
Accepted
When are dual modules free?
The dual module of a finitely generated module is reflexive, that is, $M^{**}=M$, and reflexives are awfully close to projectives. Specifically, if $R$ is a Noetherian domain, then a module is projec …
11
votes
1
answer
835
views
Which cluster algebras are coordinate rings of double Bruhat cells?
Background
A uselessly vague paragraph follows. A cluster algebra is a commutative algebra $A$ with a distinguished set of generators called cluster variables. These cluster variables are grouped in …
10
votes
Accepted
Different definitions of the dimension of an algebra
In a non-commutative ring, you need to be careful with what you even mean by a prime ideal, and usually there are very few two-sided ideals you might call prime. Oh, and even in the cases when there …
10
votes
1
answer
601
views
Are cluster variables prime elements?
Cluster algebras introduction
A cluster algebra is a subalgebra $A$ of $k[x_1^{\pm1},...,x_n^{\pm1}]$ generated by a set of cluster variables, which are elements which can be generated from the set $\ …
9
votes
Free resolution dimension?
When requiring finitely-generatedness of the resolution, then the free dimension of a projective module can be infinite.
As a simple example, take the ring $R=k\oplus k$, and let $e_1=(1,0)$ and $e_2= …
8
votes
Are cluster variables prime elements?
Boo. The answer is no. Consider the Markov cluster algebra, whose corresponding matrix is
$$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 & -2 \\ -2 & 2 & 0 \end{array}\right]$$
For an initial clust …
8
votes
What conditions are needed for $-\otimes_A B$ to be faithful?
A morphism $f:X\rightarrow Y$ of schemes gives a faithful pullback functor $f^*$ exactly when the morphism $f$ is surjective on underlying sets. This can be seen in several steps.
Note that the func …
5
votes
1
answer
497
views
The ring of SL_2 invariants in sums of conjugation and tautological modules
Rings of Invariants
Consider $G=SL_2(\mathbb{C})$, and let $V$ be a finite-dimensional $G$-representation. Let $\mathbb{C}[V]$ denote the ring of polynomial functions on the space $V$; it is a free p …
4
votes
In which commutative algebras does any derivation possess a flow?
My guess is this is the kind of algebra you don't care about (since they aren't subrings of real-valued functions), but algebras of the form $\mathbb{R}[x]/x^n$ will have your property. When looking …
4
votes
Accepted
The correspondence between affine vector bundles and f.g. projective modules
Given any $R$-module $M$, there is a scheme which corresponds to the 'total space' of $M$, given by
$$ Tot(M):=Spec( Sym_R(M*))$$
where $M*$ is the dual module $Hom_R(M,R)$ and $Sym_RM*$ is the symme …
4
votes
1
answer
209
views
Explicitly generating 1 in an ideal without prime support
The Question
Let $R$ be a unital commutative ring, and let $a,b_1,b_2\in R$. The following is a basic commutative algebra exercise.
Lemma. If $Ra+Rb_1=R$ and $Ra+Rb_2=R$, then $Ra+Rb_1b_2=R$.
Proof. …
3
votes
Graded or stacky Serre duality
Theres graded local duality which works just like local duality; however, it requires that $A_0$ is a field. I've had some luck making things work when A_0 is not a field, but then the local duality …