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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)

Thanks in advance.

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1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – user108580
    May 8, 2023 at 16:14
  • $\begingroup$ @user108580 I add some details showing that it is indeed an isomorphism. I think it should answer your question now. $\endgroup$
    – Ja_1941
    May 9, 2023 at 2:02
  • $\begingroup$ @user108580 By the way, I made a typo at first. The isomorphism should be with $M$. It's now fixed. $\endgroup$
    – Ja_1941
    May 9, 2023 at 2:03

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