Finitely generated module over PID with tensor with itself is zero

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?

• You keep using the word "next" in a strange way – Ben Grossmann Mar 5 '20 at 17:49
• Hint at an alternative strategy: the tensor product distributes over direct sums. It therefore suffices to prove the statement for $A=\Lambda$ and $A=\Lambda/a\Lambda$ for some $a\in A$. But what do you get when you tensor a cyclic module with itself? – Douglas Molin Mar 5 '20 at 17:52
• Well if $A \neq 0$ then you have to have a nonzero projection map $A^k \rightarrow A$ which clearly isn't possible since $A^k = 0$ so $A$ has to be the $0-$module. More generally in a category with $0-$objects if $A$ is a nonzero object then $\Pi_{i=1}^{i=k} A \neq 0$ since there is a nonzero morphism $A \rightarrow \Pi_{i=1}^{i=k} A$ given on coordinates by the identity map, which isn't zero since $id_A \neq 0$ because $A \neq 0$. Same thing applies to $\sqcup_{i=1}^{i=k} A$ but we consider the map $\sqcup_{i=1}^{i=k} A \rightarrow A$ – Noel Lundström Mar 5 '20 at 17:54
• The "obvious" approach to me would be to show that if $A \neq 0$, then $A \otimes A\neq 0$, which (by the universal property defining a tensor product) is equivalent to saying that there exists a non-zero bilinear map $f:A \times A \to \Lambda$. To build such a map, let $a_1 \in A$ be any non-zero element and let $\{a_1,\dots,a_n\}$ be a minimal generating set. Define $f$ so that $f(a_1,a_1) = 1$ and extend appropriately. – Ben Grossmann Mar 5 '20 at 17:59
• @Omnomnomnom: That is not equivalent; $A\otimes A$ might be nonzero but have no nonzero homomorphism to $\Lambda$ (indeed, this will be the case if $A$ is torsion). – Eric Wofsey Mar 5 '20 at 18:04

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 $$A = 0$$.

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).