Finitely generated module over PID with tensor with itself is zero 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?

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 $A = 0$.
