# Is this proof involving tensors between $R$-modules correct?

Let $$k$$ be an arbitrary field, $$R = k[X,Y]$$ a polynomial ring and $$\mathfrak{m} = (X,Y)$$ an ideal. Let us consider $$R$$ and $$\mathfrak{m}$$ as $$R$$-modules. I want to prove that if $$f:M\rightarrow M'$$ is an injective $$R$$-module homomorphism, that $$f\otimes \operatorname{Id}: M\otimes_R \mathfrak{m}\rightarrow M'\otimes_R\mathfrak{m}$$ is injective. I wrote down a proof but I feel this is incorrect, since I do not really use that we have $$\mathfrak{m}$$ but rather an arbitrary module and this statement does not hold for arbitrary modules. So here is the proof

Let $$f:M\rightarrow M'$$ be an injective $$R$$-module homomorphism. Let us consider $$f\otimes \operatorname{Id}: M\otimes_R \mathfrak{m}\rightarrow M'\otimes_R\mathfrak{m}$$. Since $$f$$ and $$\operatorname{Id}$$ are homomorphisms, $$f\otimes\operatorname{Id}$$ is also a homomorphism. By the isomorphism theorems, we have that $$\tilde{f}:M\rightarrow\operatorname{Im}f:x\rightarrow f(a)$$ is an isomorphism, so $$\tilde{f}^{-1}$$ as well. Consider now $$f^{-1}\otimes \operatorname{Id}:\operatorname{Im}f\otimes_R \mathfrak{m}\rightarrow M\otimes_R\mathfrak{m}$$. Now take an elementary tensor $$m\otimes n\in M\otimes_R \mathfrak{m}$$. Then $$(\tilde{f}^{-1}\otimes \operatorname{Id})( (f\otimes\operatorname{Id})(m\otimes n)) = (\tilde{f}^{-1}\otimes \operatorname{Id})(f(m)\otimes n) = \tilde{f}^{-1}(f(m))\otimes n = m\otimes n$$ Now since every element of $$M\otimes_R\mathfrak{m}$$ can be generated by elementary tensors, the above holds for arbitrary tensors, hence $$f\otimes\operatorname{Id}$$ is injective.

At which step am I assuming something which I cannot and why?

• I think this is false? See the comments to this answer math.stackexchange.com/questions/152596/… – user113102 Nov 26 '19 at 20:26
• and I guess the question itself – user113102 Nov 26 '19 at 20:27
• Yes, I also think this is false but what is wrong with the proof? – Jarne Renders Nov 26 '19 at 20:34

This is a well-known pitfall. The domain of the map \begin{align} \tilde{f}^{-1} \otimes \operatorname{Id} : \operatorname{Im} f \otimes_R \mathfrak{m} \to M \otimes_R \mathfrak{m} \end{align} is $$\operatorname{Im} f \otimes_R \mathfrak{m}$$, whereas the codomain of the map \begin{align} f \otimes \operatorname{Id} : M \otimes_R \mathfrak{m} \to M' \otimes_R \mathfrak{m} \end{align} is $$M' \otimes_R \mathfrak{m}$$. Thus, the composition $$\left(\tilde{f}^{-1} \otimes \operatorname{Id}\right) \circ \left(f \otimes \operatorname{Id}\right)$$ is not well-defined.
"But wait", you will say, "isn't it sufficient that the image of the map $$f \otimes \operatorname{Id}$$ is contained in the domain of $$\tilde{f}^{-1} \otimes \operatorname{Id}$$ in order for the composition to be well-defined?". Yes, it would be sufficient. But it isn't true. The image of the map $$f \otimes \operatorname{Id}$$ is spanned by tensors of the form $$f\left(m\right) \otimes n$$ with $$m \in M$$ and $$n \in \mathfrak{m}$$, but these tensors are still formed in the tensor product $$M' \otimes \mathfrak{m}$$, not in the tensor product $$\operatorname{Im} f \otimes \mathfrak{m}$$. The inclusion $$i : \operatorname{Im} f \hookrightarrow M'$$ gives rise to an $$R$$-module map $$i \otimes_R \operatorname{Id} : \operatorname{Im} f \otimes_R \mathfrak{m} \to M' \otimes_R \mathfrak{m}$$, but not (in general) to an inclusion $$\operatorname{Im} f \otimes_R \mathfrak{m} \hookrightarrow M' \otimes_R \mathfrak{m}$$; thus, tensors of the form $$f\left(m\right) \otimes n$$ with $$m \in M$$ and $$n \in \mathfrak{m}$$ lie in the image of this map $$i \otimes_R \operatorname{Id}$$, but this does not mean that they lie in its domain (or can be mapped into it in any well-defined way). Unless we know that $$\mathfrak{m}$$ is a flat $$R$$-module, we cannot guarantee that the map $$i \otimes_R \operatorname{Id} : \operatorname{Im} f \otimes_R \mathfrak{m} \to M' \otimes_R \mathfrak{m}$$ will be injective, and so we cannot identify its domain with its image.
So the confusion stems from the fact that there are two different meanings of $$f\left(m\right) \otimes n$$: one is a tensor in $$M' \otimes_R \mathfrak{m}$$, and the other is a tensor in $$\operatorname{Im} f \otimes_R \mathfrak{m}$$. They are denoted the same, but they are not equal to each other and cannot safely be identified.
The ultimate reason for the confusion is thus the notation $$a \otimes b$$ for pure tensors. If you recall how tensors are defined, you will realize that a pure tensor $$a \otimes b$$ in a tensor product $$A \otimes_R B$$ depends not only on the elements $$a \in A$$ and $$b \in B$$, but also on the ambient $$R$$-modules $$A$$ and $$B$$. Thus, denoting it by $$a \otimes b$$ is an abuse of notation. If you would instead denote it by $$\left(a, A\right) \otimes_R \left(b, B\right)$$ (thus keeping not only the values $$a$$ and $$b$$, but also the ambient $$R$$-modules $$A$$ and $$B$$ explicit in the notation), then such confusion could not happen. But of course, barely anyone wants to use this kind of clumsy notation.