$\newcommand{\Supp}{\mathrm{Supp}}$ $\newcommand{\Ann}{\mathrm{Ann}}$

Let $X$ be an affine algebraic variety (over a field $K$, can assume it is algebraicaly closed), $M$ a finitely generated $\mathcal{O}(X)$-module. We define the support of $M$ to be the subset $\Supp(M) \subset X$ defined by the ideal $I = \Ann(M) \subset \mathcal{O}(X)$.

Denote $A = \mathcal{O}(X)$ and turn it into a graded algebra with $K = A_0$. Note that $A$ is finitely-generated over $K$ and is Noetherian, but not local. Let $m_1,...,m_r$ be the generators of $M$ over $K$. Then there is $M$ is a graded $M$-modules defined as follows: $M_n := \sum_{i=1}^r A_{n-\deg(m_i)} m_i$. Then we define $f_M(n) = \dim_K(M_n)$ and $d(M) = \deg f_M$. Now, $$ \Ann(M) = \left\{ a \in A \mid \forall 1\le i \le r : a m_i = 0 \right\} $$ Since $\Supp(M) = \mathcal{V}(\Ann M)$ we have $\dim \Supp(M) = \dim (A/\Ann(M))$.

I want to show that $d(M)$ equals $\dim \Supp(M)$.

I know there is a Theorem for local Noetherian algebra $A$ that $\dim_{\mathrm{Krull}}A = d(A)$ and that if $M$ is finitely-generated $A$-module then $\dim \Supp M = \deg HS$ where $HS$ is the Hilbert-Samuel polynomial $HS(n) = \ell(M/M_n)$ where $M_n$ is a filtration of $M$ and $\ell(N)$ is the length of $N$ (see Atiyah & MacDonald, Commutative Algebra, ch.6,p.76), $$ \ell(N) = \sup\{ n \mid N = N_0 \supset M_1 \supset ... \supset N_n = 0 \} \ .$$ However, in our case $A$ is NOT local (it has many maximal ideals) and we work with $d(M) = \deg ( \dim_K M_n)$ instead of $HS = \ell(M/M_n)$.

  • $\begingroup$ math.stackexchange.com/q/57598/3217 $\endgroup$ – Georges Elencwajg Jan 24 '13 at 6:48
  • $\begingroup$ @Georges Thanks, but I am looking for a more elementary proof, without the usage of sheaves and without using projective varieties but for the case of affine variety. Sorry for the late reply but I was away for several days. $\endgroup$ – LinAlgMan Jan 27 '13 at 11:59
  • $\begingroup$ I am now trying to prove it by induction on the number of generators of $A/\mathrm{Ann}(M)$. $\endgroup$ – LinAlgMan Jan 27 '13 at 14:42
  • $\begingroup$ ... but so far unsuccessful. $\endgroup$ – LinAlgMan Jan 29 '13 at 11:29

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