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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 …

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 …

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 …

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. …

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 …

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 $\ …

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 …

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 …

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= …

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 …

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 …

Let $p$ be a prime, and let $R(p)$ be the residue field at $p$. If $R \to R(p)$ is not a surjection, then then $R(p)$ is an $R$ module whose endomorphism ring is $R(p)$, but such that the image of $R …

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 …

I have a general question, and then the specific version of that question I need for research. All vector spaces over $\mathbb{C}$. Grassmanians of planes The $(2,n)$-Grassmannian, denoted $Gr(2,n) …

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 …