1
$\begingroup$

Let $X = SpecA$ and $M$ be a finitely generated $A$-module, where $A$ is a Noetherian Ring.Define $Supp(M) = \{p\in SpecA \vert M_p \neq 0 \}$ to be support of $M$.Let $Ass_A(M)$ denote the set of associated points of $M$.Let the components $Z =V(p) = \bar{\{p\}}$ of is of dimension $d$, for all $p \in Ass_A(M)$.

How to prove that for every proper submodule $N$ of $M$ $dim(supp(N)) = dim(supp(M))$?

  • 0
    Hint: Minimal primes in Supp and Ass are the same. – 2017-01-19
  • 0
    @user26857 Yes, that I know. But can you explain how to use it to prove? – 2017-01-20
  • 0
    It implies that $$\dim \operatorname{supp} M = \max_{p \in \operatorname{Ass}_A M} \dim V(p)$$ holds for any finitely generated $A$-module $M$. Use it for both $M$ and $N$ and you will find the same numbers. – 2017-01-20
  • 0
    One last note: Of course one should assume $N \neq 0$. I am pretty sure, this is meant by *proper*, though I would think of a *proper* submodule to be a module $N \subset M$ with $N \neq M$, i.e. I would think of zero being a possible case. – 2017-01-20
  • 0
    @MooS yeah sure – 2017-01-20

0 Answers 0