Questions related to exterior powers I have two problems related to exterior powers, and I have little background on the exterior algebra of an $R-$mod.
Problem 1. For any ideal $J_i\subset R$, which is a commutative ring, consider $M=R/J_1\oplus\cdots\oplus R/J_k$ as an $R-$mod.  
Does the exterior power $\wedge^nM=0$ implies $k<n?$ If Yes, Why?

I ask it because I want to solve the following problem:
Prove if $M$ is a finitely generated module over a PID, and $\wedge^{r+1}M=0$, then $M$ can be generated by $r$ elements. 
To prove this, I used $\wedge^{r+1}(N_1\oplus N_2)=\bigoplus_{i+j=r+1}\wedge^{i}(N_1)\otimes \wedge^{j}(N_2)$(I think this is right) to reduce to the above problem.

Problem 2. Let $I_1\subset I_2\subset\cdots I_p$ and  $J_1\subset J_2\subset\cdots J_p$ be chains of proper ideals in a commutative ring $A$. Prove that if $$A/I_1\oplus \cdots \oplus A/I_p\cong A/J_1\oplus \cdots \oplus A/J_q,$$ then $p=q$ and $I_r=J_r$ for all $r$.

Of course, when Problem 1 has a positive answer, then $p=q$, but why $I_r=J_r$ for all $r$?
 A: P1. Let $M$ be generated by $k$ elements, say $M = R/J_1 \oplus \dotsc \oplus R/J_k$ with $J_1 \subseteq \dotsc \subseteq J_k \subseteq R$, and $\bigwedge^n(M)=0$. We claim that $M$ can be generated by $<n$ elements. So assume $k \geq n$. For $p>1$ we have $\bigwedge^p(R/J_i)=0$, since this is a quotient of $\bigwedge^p(R)=0$. Thus, we have
$$0 = \bigwedge^{n}(M) = \bigoplus_{p_1+\dotsc+p_k=n,~p_i \leq 1} \bigwedge^{p_1}(R/J_1) \otimes \dotsc \otimes \bigwedge^{p_k}(R/J_k),$$
and all summands vanish. For $p_1=\dotsc=p_n=1, p_{n+1}=\dotsc=p_k=0$ we get $0 = R/J_1 \otimes \dotsc \otimes R/J_n=R/(J_1+\dotsc+J_n)=R/J_n$, i.e. $J_n=R$. It follows that $M=R/J_1 \oplus \dotsc \oplus R/J_{n-1}$ is generated by $<n$ elements.
P2. Hint: Think of annihilators and induction.
A: The answer to Problem 1 is "no": if  $M=R/J_1\oplus\cdots\oplus R/J_k$,
then  $\wedge^nM=0$ does not  imply $k<n$.
For example take $R=\mathbb Z$ and $M=\mathbb Z/(2) \oplus \mathbb Z/(3)$.
Then $\bigwedge_\mathbb Z^2 M=\mathbb Z/(2) \otimes_\mathbb Z \mathbb Z/(3)=0$ but we cannot conclude that $2\lt 2$.
