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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): enter image description here

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.

enter image description here

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  • 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 $N/\ker\iota\hookrightarrow L$.

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  • $\begingroup$ Can the theorem be rephrased as saying that if $N/\ker \varphi \hookrightarrow L$, then we have $N/\ker \varphi \hookrightarrow N/\ker \iota \hookrightarrow L$? $\endgroup$ – Al Jebr May 9 '20 at 19:40
  • $\begingroup$ @AlJebr: Good question! I suspect the answer is no, but I do not have a counterexample; see my updated answer. $\endgroup$ – Alex Ortiz May 10 '20 at 0:44
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(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.)

$\hspace{4.25cm}$enter image description here

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.

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