# homological algebra – finally generated module via PID with tensor with itself is zero

Show that if $$A$$ is a module finally created via a PID and $$A otimes _ { Lambda} A = 0$$, then $$A = 0$$.

I've done the next one, I'm looking at the next exact order

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

We have that $$A / Goal (A)$$ is therefore a finally created torsion-free module via a PID $$A / Goal (A)$$ is a free module and that implies that the short exact sequence is split up.

That's why I have a morphism $$A / goal (A) right arrow A$$ so that $$A / goal (A) right arrow A right arrow A / goal (A)$$ is identity.

Now when I'm using tensor $$A$$ I have the next composition

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

It follows $$(A / Tor (A)) otimes A = 0$$.

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

However, I don't know how to proceed and I got stuck with it, so any hint?