# Understanding a corollary of the universal property of tensor products of modules

This is a follow-up question of a previous one of mine.

The following theorem is from the Abstract Algebra by Dummit and Foote (in the section 10.4 tensor products of modules):

Here are my questions:

• What does "largest quotient of $N$" mean in the proof? (Being "largest" in what sense?)
• How does one get "It follows that $N/\hbox{ker}\iota$ is the unique largest quotient of $N$ that can be embedded in any $S$-module"?
• Would anyone elaborate how "the last statement in the corollary follows immediately"?

Here is the reference for Theorem 8 mentioned above.

• What does "largest quotient of $$N$$" mean in the proof? (Being "largest" in what sense?)

By largest quotient of $$N$$, we mean that if $$N/A$$ is any other quotient of $$N$$, then $$\ker \iota \subset A$$. Note that any such $$A$$ arises as the kernel of some module homomorphism.

• How does one get "It follows that $$N/\ker\iota$$ is the unique largest quotient of $$N$$ that can be embedded in any $$S$$-module"?

At this point in the proof, we have already shown that if $$\varphi:N\to L$$ is any $$R$$-module homomorphism of $$N$$ into an $$S$$-module $$L$$, then $$\ker\iota\subset\ker\varphi$$ by using the universal property of the tensor product. In particular, if $$\iota(n) = 0$$, then $$\varphi(n)= \Phi(\iota(n))=\Phi(0) = 0$$.

Thus if $$\varphi:N\to L$$ is any $$R$$-module homomorphism of $$N$$ into an $$S$$-module, then there is an isomorphism theorem that tells us that $$N/\ker\varphi$$ can be embedded into $$L$$. Since $$\iota$$ is an $$R$$-module homomorphism, we can embed $$N/\ker\iota$$ into $$S\otimes_R N$$, and we have already shown it is the largest such quotient.

• Would anyone elaborate how "the last statement in the corollary follows immediately"?

If $$N$$ can be embedded as an $$R$$-submodule of some left $$S$$-module, then there is an $$R$$-module homomorphism $$\varphi:N\to L$$ such that $$\ker\varphi = 0$$. Since $$N/\ker\iota$$ is the largest quotient of $$N$$, $$\ker\iota\subset 0$$, hence $$\ker\iota = 0$$, so $$\iota$$ is injective.

If $$\iota$$ is injective, then $$N/\ker\iota = N/0 \cong N$$ can be embedded into $$S\otimes_RN$$.

In response to @AlJebr's question:

Can the theorem be rephrased as saying that if $$N/\ker\varphi\hookrightarrow L$$, then we have $$N/\ker\varphi\hookrightarrow N/\ker\iota\hookrightarrow L$$?

As far as I can see, the answer to expect is "no," but I do not have a counterexample; see below. By Theorem 8, in the situation $$N/\ker\varphi\hookrightarrow L$$, all we know is that $$N/\ker\varphi$$ is a quotient of $$N/\ker\iota$$. Part of what you are asking is that we be able to identify $$N/\ker\varphi$$ as a submodule of $$N/\ker\iota$$, and this is not possible in general.

The following is not a counterexample since $$L$$ is not an $$S$$-module, but it might give some ideas for how to generate a counterexample (assuming one exists, and there is not something special about the structure that makes the rephrasing true). Set

• $$N = R = \mathbb Z$$,
• $$S = \mathbb Q$$,
• $$\ker \varphi = n\mathbb Z$$, where $$n > 1$$,
• $$L = \mathbb Z/n\mathbb Z$$.

Then the natural map $$\iota : \mathbb Z\to \mathbb Q\otimes_\mathbb Z\mathbb Z$$ is an injection since $$1\otimes a = 0$$ implies in particular that the $$\mathbb Z$$-bilinear map $$B(\frac{r}{s},a) = \frac{r}{s}\cdot a$$ vanishes at $$1\otimes a$$, so $$a = 0$$. In this example, $$N/\ker\varphi = L$$.

Now, it is clear that we are not able to identify $$\mathbb Z/n\mathbb Z$$ with a submodule of $$\mathbb Z$$ since $$\mathbb Z$$ has no nonzero torsion elements. This example shows explicitly that if $$N/\ker\varphi\hookrightarrow L$$, then we may not have either $$N/\ker\varphi\hookrightarrow N/\ker \iota$$, nor $$N/\ker\iota\hookrightarrow L$$.

• Can the theorem be rephrased as saying that if $N/\ker \varphi \hookrightarrow L$, then we have $N/\ker \varphi \hookrightarrow N/\ker \iota \hookrightarrow L$? May 9, 2020 at 19:40
• @AlJebr: Good question! I suspect the answer is no, but I do not have a counterexample; see my updated answer. May 10, 2020 at 0:44

(1) I take the definition of "largest quotient" to be the sentence before. I.e., if $N/\ker(\varphi)$ is another quotient of $N$ that can be embedded $N/\ker(\varphi) \hookrightarrow L$ into an $S$-module $L$, then there is a unique $R$-linear surjection $\pi: N/\ker(\iota) \to N/\ker(\varphi)$ such that the following diagram commutes. (If $A$ surjects onto $B$, I usually think of $A$ as being bigger, which is at least true in terms of cardinality.)

$\hspace{4.25cm}$

As mentioned in the corollary, this realizes $N/\ker(\varphi)$ as a quotient of $N/\ker(\iota)$: $$N/\ker(\varphi) \cong \frac{N/\ker(\iota)}{\ker(\varphi)/\ker(\iota)} \, .$$

You can also think dually, in terms of ideals. $N/\ker(\iota)$ being the largest means that $\ker(\iota)$ is the smallest: if $\varphi: N \to L$ is an $S$-module homomorphism, then $\ker(\iota) \subseteq \ker(\varphi)$. This is because we "throw out" the least when quotienting by the smallest possible ideal.

Now that I think about it, I think the authors mean "largest" in terms of posets. The ideals that induce embeddings form a poset under inclusion. If we instead look at the quotients by these ideals, we get a "dual" poset where the ordering is reversed, as inclusion maps are replaced by quotient maps.

(2) now just follows from the definition of "largest."

(3) If $\iota$ is injective, then $\ker(\iota) = 0$, so $N/\ker(\iota) \cong N$ injects into $S \otimes_R N$, which is an $S$-module. Conversely, suppose that $\varphi: N \hookrightarrow L$ is an injection of $N$ into an $S$-module $L$. Then there exists a $\Phi$ as in the theorem such that $\varphi = \Phi \circ \iota$, so the injectivity of $\varphi$ implies that $\Phi$ and $\iota$ must both be injective, too.

Actually, I guess we can appeal directly to the corollary for the reverse implication. Suppose $\varphi: N \to L$ is an injection, so $\ker(\varphi) = 0$. Then we get a map $\pi: N/\ker(\iota) \to N/\ker(\varphi)$, so we must have $\ker(\iota) \subseteq \ker(\varphi) = 0$, hence $\iota$ is injective.