# The associated primes of the dual

I was trying to prove the following formula:

$\mathrm{Ass}\;M^*=\mathrm{Ass}\;R\cap\mathrm{Supp}\;M$

(we can suppose $R$ noetherian, $M$ finitely generated. If it is useful even $R$ local)

I proved that $\mathrm{Ass}\;M^*\subset\mathrm{Supp}\;M$. But I'm having troubles with the other containments, any idea?

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What do you mean by $M^*$? – Giovanni De Gaetano Oct 15 '12 at 13:03
$M^*=\mathrm{Hom}_R(M,R)$ – Chris Oct 15 '12 at 13:08

I think I have a proof of:

If $R$ is noetherian, $M$ finitely generated and $N$ arbitrary then

$\mathrm{Ass}\;\mathrm{Hom}(M,N)=\mathrm{Supp}\;M\cap\mathrm{Ass}\;N$.

First: let's prove that $\mathrm{Ass}\;\mathrm{Hom}(M,N)\subset\mathrm{Supp}\;M$. Take $p\in\mathrm{Ass}\;\mathrm{Hom}(M,N)$, I want $\mathrm{Ann}\;M\subset p$. Take $x\in\mathrm{Ann}\;M$ and suppose $p=\mathrm{Ann}\;\varphi$. Then $x\varphi(m)=\varphi(xm)=\varphi(0)=0$ and so $x\in p$.

Second: $\mathrm{Ass}\;\mathrm{Hom}(M,N)\subset\mathrm{Ass}\;N$.

Suppose $p\in \mathrm{Ass}\;\mathrm{Hom}(M,N)$, this is equivalent to $PR_P\in\mathrm{Ass}\;\mathrm{Hom}(M_p,N_p)$ that is equivalent to $\mathrm{depth}\;\mathrm{Hom}(M_p,N_p)=0$.

In the same way $p\in\mathrm{Ass}\;N$ is equivalent to $\mathrm{depth}\;N_p=0$

So I want to prove: let $(R,m,k)$ local noetherian and $M\neq0$ finitely generated then $\mathrm{depth}\;\mathrm{Hom}(M,N)=0$ implies $\mathrm{depth}\;N=0$.

Suppose that $x\in m$ is a $N$-regular element. Take $\varphi\in\mathrm{Hom}(M,N)$, suppose $x\varphi=0$, then $x\varphi(m)=0$ for all $m$ and so $\varphi(m)=0$ for all $m$ because $x$ is $N$-regular and so $\varphi=0$, that implies $x$ is $\mathrm{Hom}(M,N)$-regular, contradiction.

Third: $\mathrm{Supp}\;M\cap\mathrm{Ass}\;N\subset\mathrm{Ass}\;\mathrm{Hom}(M,N)$.

As before I can suppose $(R,m,k)$ local noetherian, $M$ non zero finitely generated. It's enough to prove that if $m\in\mathrm{Ass}\;N$ then $m\in\mathrm{Ass}\;\mathrm{Hom}(M,N)$.

If $m\in\mathrm{Ass}\;N$ then I have an injection $k\rightarrow N$. Applying $\mathrm{Hom}(M,.)$ we obtain an injection $\mathrm{Hom}(M,k)\rightarrow\mathrm{Hom}(M,N)$. I want to prove that there is an injection $k\rightarrow\mathrm{Hom}(M,k)$. Suppose $M=<m_1,\ldots,m_l>_R$ is a minimal system of generators. Take $\bar{r}\in k$ and define $\varphi_{\bar{r}}:M\rightarrow k$ in the following way: $\varphi_{\bar{r}}(m_1)=\bar{r}$ and $\varphi_{\bar{r}}(m_i)=0$ if $i\neq1$. The function $\bar{r}\mapsto\varphi_{\bar{r}}$ is the required injection.

$\varphi_{\bar{r}}$ is well-defined: suppose $\sum_i r_im_i=0$, by Nakayama this implies $r_i\in m$. $\varphi_{\bar{r}}(\sum_i r_im_i)=r_1\bar{r}=0$ because $r_1\in m$.

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