# $1\otimes_A x = 0$ in $A\otimes_A M$ then $x=0$.

I have a homework problem requiring me to show the following without using the universal property of tensor product. It asks me to use the bilinear construction directly.

• Prove that $$1\otimes_A x = 0$$ in $$A\otimes_A M$$ implies $$x=0$$, where $$A$$ is a commutative ring with identity and $$M$$ an $$A$$-module.

The construction I am using is the quotient of free module $$A^{M\times N}$$ by the bilinear submodule. However, all the bilinear properties help me transform some tensor to another. For this question $$x\in A$$ is not really a tensor form. I understand I should show work for this question, but I only got this far: Suppose $$x\neq 0$$ and $$1\otimes_A x = 0$$, we might deduce some contradiction?...

I think I might find some clues in the proof of the universal property of tensor products.(I am reading it now and I'll add later thoughts to the question body)

There is an isomorphism $$A\otimes_A M\cong M$$ via $$a\otimes m\mapsto am$$. The inverse is given by $$m\mapsto 1\otimes m$$.

First, we check the well defineness. That is, whether this map respects the relation that defines tensor prouduct as below, an easy exercise.

$$(a+b)\otimes m=a\otimes m + b\otimes m$$, $$a\otimes (m+n)=a\otimes m + a\otimes n$$, $$aa'\otimes m=a\otimes a'm$$.

It is also not hard to check that this is a module homomorphism. It remains to verify the two maps above are indeed inverse to each other.

$$a\otimes m\mapsto am\mapsto 1\otimes am=a\otimes m,$$

$$m\mapsto 1\otimes m\mapsto m.$$

This is indeed an isomorphism.

Sometimes I also get stuck on proofs like this. I think manipulating tensor product completely elementwise is hard. Although we know $$1\otimes m=0$$, it is hard to see how that happens. Maybe it follows from some kind of weird combination of the relations that define tensor product. So usually I consider some homomorphisms, where I don't have to worry about how the relations combine and only check each of them separately like how I check the well defineness above.

• The hard part, in my opinion, is to verify the map $a\otimes m\mapsto am$ is well-defined. Actually, as a special case, to verify "$a\otimes m = (a-1)\otimes m$ implies $am = (a-1)m$" is exactly to verify "$1 \otimes m = 0$ implies $m = 0$", which is just my question. May 8, 2023 at 16:14
• @user108580 I add some details showing that it is indeed an isomorphism. I think it should answer your question now. May 9, 2023 at 2:02
• @user108580 By the way, I made a typo at first. The isomorphism should be with $M$. It's now fixed. May 9, 2023 at 2:03