Finitely generated module over PID with tensor with itself is zero

Question

Hello I have the next doubt about this problem:

Show that if $A$ is a finitely generated module over a PID and $A\otimes_{\Lambda}A=0$, then $A=0$.

I have done the next thing, I consider the next exact sequence

$0\rightarrow Tor(A)\rightarrow A\rightarrow A/Tor(A)\rightarrow 0$

We have that $ A/Tor(A)$ is a finitely generated torsion free module over a PID, therefore $ A/Tor(A)$ is a free module and that implies that the short exact sequence split.

Therefore I have a morphism $ A/Tor(A)\rightarrow A$ such that $ A/Tor(A)\rightarrow A\rightarrow A/Tor(A)$ is the identity.

Now if I tensor with $A$ I have that the next composition

$ (A/Tor(A))\otimes A\rightarrow 0\rightarrow (A/Tor(A))\otimes A$ is also the identity

Thus it follows that $(A/Tor(A))\otimes A=0$.

Since $A/Tor(A)\cong\Lambda^{k}$ I have that $A^{k}=0$

However I do not how to continue with this and I am stucked with this so any hint?

Answer 1

Could you not use the structure theorem for PIDs? If $A$ is a finitely generated $R$-module, then $A \cong \bigoplus_{i=1}^n R/(d_i)$, for $(d_1) \supseteq \ldots \supseteq (d_n)$ a sequence of proper ideals of $R$. Then

\begin{align*} 0 = A \otimes_R A & \cong \left( \bigoplus_{i=1}^n R/(d_i) \right) \otimes_R \left( \bigoplus_{j=1}^n R/(d_j) \right) \\ & \cong \bigoplus_{i,j = 1}^n \Bigl(R/(d_i) \otimes_R R/(d_j)\Bigr) \\ & \cong \bigoplus_{i,j=1}^n R/((d_i) + (d_j)) \\ & = \bigoplus_{i,j = 1}^n R/(d_{\min\{i,j\}}) \end{align*} But this implies that each $R/(d_i) = 0$, and so $A = 0$.

Answer 2

Note that the statement is true without assuming that $\Lambda$ is a PID. I give a proof of this in my answer to this question.

As explained in @QuarkAntiquark's answer, the PID assumption gives an easy proof via the structure theorem. Here's how to think about this proof: the structure theorem (and the fact that tensor products distribute over direct sums) means that when you're proving statements of the sort in this question, it suffices to prove them for modules of the form $\Lambda /I$, where $I$ is an ideal (possibly the zero ideal).