If $I$ is an ideal and $M$ is a (finitely generated) module then $\text{Supp}(M/IM)\subseteq_{(=)}\text{Supp}(M)\cap V(I)$ 
Let $R$ be a commutative unital ring, $I\subseteq R$ be an ideal and $M$ an $R$-module.
Show that $\text{Supp}(M/IM)\subseteq\text{Supp}(M)\cap V(I)$, and if $M$ is finitely generated then an equality holds.


Recall:

*

*$\text{Supp}(M)=\{p\in\text{Spec}(R):M_p\ne 0\}$

*$V(I)=\{p\in\text{Spec}(R):I\subseteq p\}$

I succeed to show that $\text{Supp}(M/IM)\subseteq V(I)$:
If $p\in\text{Supp}(M/IM)$ then $(M/IM)_p\ne 0$. Thus $$
\exists IM\ne {m+IM\over t_1}\in (M/IM)_p
$$
Suppose in contradiction $\exists t_2\in I\setminus p$. Then
$$
IM\ne{t_2m+IM\over t_1t_2}
$$
But this is a contradiction because $t_2\in I, m\in M$.

Attempt $\text{Supp}(M/IM)\subseteq\text{Supp}(M)$:
Let $p\in\text{Supp}(M/IM)$. Then $\exists IM\ne {m+IM\over t}\in(M/IM)_p$. Thus ${m\over t}+(IM)_p\ne (IM)_p\Rightarrow {m\over t}\notin (IM)_p\Rightarrow m\notin IM$.
 A: So first let us show that $\mathrm{supp}(M/IM)\subset \mathrm{supp}(M)$ :
Let $p \in \mathrm{supp}(M/IM)$. There exists $\bar{m} \in M/IM$ such that $\forall r \in R\setminus p : r\bar{m} \neq 0$. Let $m \in M$ be a representative of $\bar{m}$. Then $\forall r \in R\setminus p : rm \neq 0$. Thus the image of $m$ in $M_p$ is non zero.
Let us now move on to the case of the equality when $M$ is of finite type. I detailed it (maybe to much) so it may seem a bit long, but the ideas are elementary. The idea is to make an induction of the number of generators.
If $M$ is generated by one element $e_1$, let $p\in \mathrm{supp}(M)\cap V(I)$. This implies that : $\forall r \in R\setminus p : re_1 \neq 0$ (1). 
If we had that $p \notin \mathrm{supp}(M/IM)$ then there would exist $ r \in R\setminus p $ such that $re_1 \in IM$. So there would exist $a \in I$ such that $re_1=ae_1$, implying that $(r-a)e_1=0$. But $r-a \in R\setminus p$ (else we would have $r \in p$ since $a \in I \subset p$, which is false by hypothesis).
So finally we have a contradiction with (1) and $p \in supp (M/IM)$ which concludes the proof in the case where $M$ has one generator only.
For $n\geqslant 1$, set $P(n)$ the proposition : For all $R$-module $N$ generated by $n$ elements $(e_1,\dots,e_n)$, for all ideal $J$ of $N$ and $p \in \mathrm{supp}(N)\cap V(J)$ : ($\forall i, 1\leqslant i\leqslant n$ there exists $r_i \in R\setminus p$ such that $r_ie_i \in JN \Rightarrow \forall i, 1\leqslant i\leqslant n$ there exists $a_i \in R\setminus p$ such that $a_ie_i = 0$).
We have already proved $P(1)$. In order to do an induction, let us now suppose that $P(n-1)$ is true (n > 1) and prove $P(n)$.
Let $N$ be an $R$-module generated by elements $(e_1,\dots,e_n)$, $J$ an ideal of $N$ and $p \in \mathrm{supp}(N)\cap V(J)$. For all $1\leqslant i\leqslant n$ we will denote $Re_i \subset N$ the module generated by $e_i$. Suppose that $\forall i, 1\leqslant i\leqslant n$ there exists $r_i \in R\setminus p$ such that $r_ie_i \in JN$. Then $(\bar{e_2},\dots,\bar{e_n})$ (the projections of $(e_2,\dots,e_n)$) generates $N/Re_1$ and $\forall i, 1\leqslant i\leqslant n, r_i\bar{e_i} \in J(N/Re_1)$. Applying $P(n-1)$ we have that there exist $a_2,\dots, a_n \in R\setminus p$ and $b_2,\dots,b_n \in R$ such that for all $ 2\leqslant i \leqslant n $ : $a_ie_i=b_ie_1$ . 
Moreover $r_1e_1 \in JN$ so there are $\gamma_1,\dots,\gamma_n \in J$ such that $r_1e_1 = \sum_{1\leqslant i \leqslant n} \gamma_i e_i$. Finally (setting $b_1 = 1$): $\left((\prod_{2\leqslant i\leqslant n} a_i)r_1 - \sum_{1\leqslant i \leqslant n} \gamma_i b_i \prod_{j\neq i}a_j\right)e_1=0 $. Notice following the argument in the case of one generator (and using the stability under multiplication of $R\setminus p$, true since $p$ is  a prime ideal) that $c_1 := \left((\prod_{2\leqslant i\leqslant n} a_i)r_1 - \sum_{1\leqslant i \leqslant n} \gamma_i b_i \prod_{j\neq i}a_j\right) \in R\setminus p$.
Repeating this argument for each $2\leqslant i \leqslant n$ (by considering the quotient module $N/Re_j$), we construct a family of elements $c_1,\dots,c_n \in R\setminus p$ such that for all $1\leqslant i \leqslant n$ : $c_ie_i = 0$. So we have proved $P(n)$.
By induction $P(n)$ is true for all $n \in \mathbb{N}^*$.
So (back to your notations) suppose that $M$ is generated by the $n$ elements $(e_1,\dots,e_n)$ and let $p \in \mathrm{supp}(M)\cap V(I)$.  This implies that : $\exists i, 1\leqslant i \leqslant n \text{ s.t. } \forall r \in R\setminus p : re_i \neq 0$ (2). Suppose that $p\notin \mathrm{supp}(M/IM)$. So $\forall i, 1\leqslant i\leqslant n$ there exists $r_i \in R\setminus p$ such that $r_ie_i \in IM$. By $P(n)$ this implies immediately a contradiction with (2). So $p \in \mathrm{supp}(M/IM)$.
Finally what we have is the result we wanted : If M is finitely generated $\mathrm{supp}(M/IM)= \mathrm{supp}(M)\cap V(I)$.
