What is $B/A \cap B$ for $R$-submodules $A,B$ of an $R$-module $M$?

Question

There's an exercise in my book asking me to show that for $R$-submodules $A,B$ of an $R$-module $M$ the following is true

$$(A+B)/A \cong B/A \cap B$$

as $R$-modules.

It's been a while since I've done abstract algebra, so I'm a bit rusty. As I sketched out a proof for this using the fundamental theorem of $R$-homomorphisms, I began thinking about $B/A \cap B$.

Since $A\cap B \neq \emptyset$, it seems to me that $B/A \cap B = B/A$.

So then $B/A \cap B = B/A$. If this is the case, then the proof I've sketched out is correct.

Because I've been away from this material for so long, I'm doubting the validity of these arguments. If there is an error here somewhere, please point it out and explain to me why it's wrong. If I have in fact made no mistakes, then why would the book use $B/A \cap B$ instead of just $B/A$?

Answer 1

First of all, $A\cap B$ is never empty, $0$ is always in it. Secondly, you're playing rather fast and loose with your quotients, and not justifying your equalities/isomorphisms at all (for example, saying that something is equal to $B/A$, when $A$ is not even known to be a subset of $B$ - it's unjustified and worse, $B/A$ doesn't even have meaning!).

Let me give a proof of the isomorphism, checking the details: Consider the map \begin{align*} \phi : B&\to (A + B)/A\\ b&\mapsto b + A \end{align*} (recall that for a module $M$ and submodule $N$, the quotient module $M/N$ is $\{m + N\mid m\in M\}/\sim$, where $m + N\sim m' + N$ if $m - m'\in N$ as a set). We must show that this map is surjective, a homomorphism, and has kernel $B\cap A$.

1. (Surjectivity) Let $x + A\in (A + B)/A$. Then $x = a + b$ for some $a\in A$ and $b\in B$. But then $x + A = b + A$ as elements of $(A + B)/A$, because $a + b - b = a\in A$. Thus, $b\in B$ maps to $x + A$, so the map is surjective.

2. ($\phi$ is a homomorphism) To see that $\phi$ is a homomorphism, let $b,b'\in B$. Then $\phi(b + b') = (b + b') + A = (b + A) + (b' + A) = \phi(b) + \phi(b')$, so the map is additive (the second equality follows from the definition of addition in a quotient module). Let $r\in R$, $b\in B$. then $\phi(rb) = rb + A = r(b + A)$, by definition of the action on the quotient, so $\phi(rb) = r\phi(b)$, and $\phi$ is a homomorphism.

3. ($\ker\phi = A\cap B$) Lastly, we look at the kernel. Take $b\in A\cap B$. Then $\phi(b) = b + A$, but since $b\in A$, $b + A = 0 + A$, so $A\cap B\subseteq\ker\phi$. Now, take $k\in\ker\phi$. Then $k\in B$, and $\phi(k) = 0 + A$. But $\phi(k) = k + A$, so $k - 0 = k\in A$, so $k\in A$ as well. Hence, $k\in A\cap B$, so $\ker\phi\subseteq A\cap B$. It follows that $\ker\phi = A\cap B$.

Then we conclude by the first isomorphism theorem for modules that $B/(A\cap B)\cong (A + B)/A$.

(If $A\subseteq B$, then it is true that $B/(A\cap B) = B/A = (A + B)/A$, but this case is trivial: $A\cap B = A$ and $A + B = B$ if $A\subseteq B$.)

Answer 2

Note that $B/A$ doesn't generally make sense, unless $A\subseteq B$; but in this case $A+B=B$ and $A\cap B=A$, so the isomorphism is obvious.

Consider the homomorphism $$ f\colon B\to (A+B)/A, \qquad f(b)=b+A $$ Prove that $f$ is surjective and that $\ker f=A\cap B$.