# Understanding homomorphism on tensor product of module

I'm trying to convince myself one part of the proof for a theorem in chapter 10 in DUmmit and Foote.

Let $$R$$ be a subring of $$S$$, let $$N$$ be a left $$R$$-module and let $$\iota:N \to S\otimes_RN$$ be the $$R$$-module homomorphism defined by $$\iota(n)=1\otimes n$$. Suppose that $$L$$ is any Left $$S$$-module and that $$\phi:N\to L$$ is an $$R$$-module homomorphism. Then there is a unique $$S$$-module homomorphism $$\Phi: S\otimes_R N \to L$$ such that $$\phi$$ factor through $$\Phi$$.

In the proof, $$\Phi$$ is given by universal property on free $$\mathbb{Z}$$-module $$S\otimes_R N$$ as $$\Phi(s\otimes n)=s\phi(n)$$. But the book just mentioned that $$\Phi$$ is additive. I guess that is because $$\phi$$ is additive. But how would I go to prove it? I let $$s_1 \otimes n_1,s_2\otimes n_2\in S \otimes_R N$$. Then show that $$\Phi(s_1\otimes n_1+ s_2\otimes n_2)=\Phi(s_1\otimes n_1)+\Phi(s_2\otimes n_2)$$ But I can't really come up anything logical that can verify $$\Phi$$ is additive.

Am I not understanding tensor product of modules?

• Maps out of a tensor product like $S \otimes_R N$ are always constructed by starting off with suitable maps out of $S \times N$. Doing that will automatically make the induced map out of the tensor product additive. Don't try to show directly that a function on a tensor product is additive by trying to evaluate it on a sum of two elementary tensors. Rely on the universal mapping property.
– KCd
Commented Feb 12 at 2:52

## 1 Answer

The proof in the book goes like this:

Consider the map $$\phi' \colon S \times N \to L$$ given by $$\phi'(s,n) = s\phi(n)$$. Then, if $$F$$ is the free $$\Bbb Z$$-module on $$S \times N$$, there is$$^*$$ a $$\Bbb Z$$-module homomorphism (= additive map) $$\phi'' \colon F \to L$$ such that $$\phi''(s,n) = \phi'(s,n)$$ for all $$(s,n) \in S \times N$$. Next, if $$H$$ is the subgroup generated by all elements of the form (10.3), it is easy to check that $$H \subseteq \ker \phi''$$. Therefore, there is$$^{**}$$ a $$\Bbb Z$$-module homomorphism (= additive map) $$\phi''' \colon F/H \to L$$ such that $$\phi'''(x+H) = \phi''(x)$$ for all $$x \in F$$. Since $$S \otimes_R N = F/H$$ and $$s \otimes n = (s,n)+H$$, $$\phi'''$$ is the desired map $$\Phi$$. Note that $$\Phi$$ is additive, trivially.

$$^*$$This is the universal property of the free module on a set: If $$X$$ is any set and $$F$$ is the free module on $$X$$, then for any module $$A$$ and any map $$f \colon X \to A$$, there exists a (unique) module homomorphism $$f' \colon F \to A$$ such that $$f'(x)=f(x)$$ for all $$x \in X$$.

$$^{**}$$This is the universal property of the quotient module: If $$g \colon M \to N$$ is a module homomorphism and $$S$$ is a submodule of $$M$$ such that $$S \subseteq \ker g$$, then there exists a (unique) module homomorphism $$g' \colon M/S \to N$$ such that $$g'(x+H)=g(x)$$ for all $$x \in M$$.