- 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 $N/\ker\iota\hookrightarrow L$.