The following is the theorem regarding the universal property of tensor products of modules in Dummit and Foote:

Let $R$ be a subring of $S$, let $N$ be a left $R$-module and $\iota:N\to S\otimes_RN$ be the $R$-module homomorphism defined by $\iota(n)=1\otimes n$. Suppose that $L$ is any left $S$-module (hence also an $R$-module) and that $\varphi:N\to L$ is an $R$-module homomorphism from $N$ to $L$. Then there is a unique $S$-module homomorphism $\Phi:S\otimes_RN\to L$ such that $\varphi$ factors through $\Phi$, i.e., $\varphi=\Phi\circ\iota$ and the diagram enter image description here

commutes. Coversely, if $\Phi: S\otimes_RN\to L$ is an $S$-module homomorphism then $\varphi=\Phi\circ\iota$ is an $R$-module homomorphism from $N$ to $L$.

The proof goes as follows

  • Define $\psi:S\times N\to L$ with $\psi(s,n)=s\varphi(n)$.
  • By the universal property of free $\mathbf{Z}$-module on the set $S\times N$, denoted as $F(S\times N)$, there exists a $\mathbf{Z}$-module homomorphism $\Psi:F(S\times N)\to L$ such that $\Psi(s,n)=\psi(s,n)=s\phi(n)$.
  • Since $\varphi$ is an $R$-module homomorphism, the generators of the subgroup $H$ in the following equations all map to zero in $L$, where the generators are given by elements of the form $$ (x+y,n)-(x,n)-(y,n),\quad (x,m+n)-(x,m)-(x,n),\quad (sr,n)-(s,rn),\\ x,y\in S,\,m,n\in N,\, r\in R. $$
  • $\color{blue}{\textrm{Hence}}$, $\Psi$ factors through $H$, i.e., there is a well defined $\mathbf{Z}$-module homomorphism $\Phi$ from $F/H=S\otimes_RN$ to $L$ satisfying $\Phi(s\otimes n)=s\varphi(n)$.
  • The rest of the proof can be read below.

Here is my question:

Could anyone elaborate how the "Hence" step is implied by the third bullet point, from which I can only see that $H\subset \ker\Psi$?

Here is the original proof in the book: enter image description here

  • $\begingroup$ What is $H$? $\,$ $\endgroup$ – D_S May 13 '17 at 1:27
  • $\begingroup$ @D_S: It is the subgroup generated by "the following equations". $\endgroup$ – Jack May 13 '17 at 2:04
  • $\begingroup$ @Jack, equations cannot generate a subgroup. $\endgroup$ – Mariano Suárez-Álvarez May 13 '17 at 3:23
  • $\begingroup$ @MarianoSuárez-Álvarez: Thanks for your comment. I have edited the post according to what is originally in the book. $\endgroup$ – Jack May 13 '17 at 11:44

$H \subseteq \textrm{ker } \Psi$ is all you need for $\Psi$ to factor through the quotient $F/H$.

In general, let $\delta: A \rightarrow B$ be a homomorphism of abelian groups ($\mathbf Z$-modules, same thing), and let $A_0$ be a subgroup of $A$. To say that $\delta$ factors through $A_0$ (or $A/A_0$) is to say that there exists an abelian group homomorphism $\bar{\delta}: A/A_0 \rightarrow B$ such that $\bar{\delta}(a + A_0) = \delta(a)$ for all $a \in A$.

If such a homomorphism exists, it is clearly unique. The only issue barring the existence of such a homomorphism is the possibility that it is not well defined. And you can check that it is well defined if and only if $A_0$ is contained in the kernel of $\delta$.

  • $\begingroup$ Thanks for your answer. I would be interested in seeing a reference for the "factor through" theorem. $\endgroup$ – Jack May 13 '17 at 2:05
  • $\begingroup$ I think every abstract algebra textbook should have some version of it. But it is easy to prove yourself: to say that $\bar{\delta}$ is well defined is to say that if $a + A_0 = a' + A_0$ (that is, $a - a' \in A_0$), then $\delta(a) = \delta(a')$. $\endgroup$ – D_S May 13 '17 at 2:13
  • $\begingroup$ Hmm, it looks very much similar to the first isomorphism theorem. Maybe it is called the universal property of quotient groups. $\endgroup$ – Jack May 13 '17 at 2:19
  • $\begingroup$ It is easy to check that if $\bar{\delta}$ is well defined (that is, $A_0 \subseteq \textrm{ker } \delta$), then $\textrm{ker } \bar{\delta} = \textrm{ker } \delta/A_0$ (the image of $\textrm{ker } \delta$ under the homomorphism $A \rightarrow A/A_0$). The first isomorphism theorem is a special case of this, when $A_0 = \textrm{ker } \delta$. Here the kernel of $\bar{\delta}$ is $\textrm{ker } \delta/\textrm{ker } \delta$, which is the trivial subgroup of $A/\textrm{ker } \delta$. Thus $\bar{\delta}$ is injective in this case (and an isomorphism if and only if $\delta$ is surjective). $\endgroup$ – D_S May 13 '17 at 2:22

On page 100 of Dummit and Foote, there is a remark about the "universal property" of quotient groups, which explains the "Hence" step (in particular it explains what "factors through" means).

One can also read this note: https://ocw.mit.edu/courses/mathematics/18-703-modern-algebra-spring-2013/lecture-notes/MIT18_703S13_pra_l_9.pdf


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