Quotient Modules

I'm certain that this is a stupid question, but I'll ask anyway. In one of my courses, we let $M$ be a module, and $L,N \subseteq M$ be submodules. The professor then used the quotient module $(L+N)/N$, but this is weird, since if we have $x+y$, $y \in N$, then $q(x+y) = q(x)$ where $q:(L+N) \to (L+N)/N$ is the quotient map. So we'd just have $(L+N)/N = L/N$, so why would he bother to use $(L+N)/N$ instead of $L/N$?

The same set up as in class is in Lang's Algebra, chapter X section 1, where he shows that if $N$ and $M/N$ are noetherian then $M$ is noetherian.

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It doesn't make sense to talk about the quotient module $L/N$ unless $L\supseteq N$, whereas it always makes sense to talk about $(L+N)/N$ because $L+N\supseteq N$.

By the way, it is a great question!

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This was helpful. Thanks! –  nigelvr Feb 4 '13 at 3:36
Glad I could help! –  Zev Chonoles Feb 4 '13 at 3:38
Just to add, once we see that L/N doesn't always make sense, we have to think of a natural way to form the quotient we want. One of them in to increase L so it is just large enough to include N (and L+N is the smallest submodule of M containing both L and N) which is what you saw in the question. But there is also another way: Decrease N to fit inside L, so that we would consider the quotient $L/(L\cap N).$ The second isomorphism theorem says theses two quotients are isomorphic. –  Ragib Zaman Feb 4 '13 at 23:09

I had the same question as you when I was first learning about modules (and I also think it's a great question by the way) and saw the second isomorphism theorem stating $(L+N)/N \cong L/(L \cap N)$.
The confusion arises because when we quotient out $L+N$ by $N$ we are identifying elements in $L+N$ when their difference lies in $N$, and because $(l_1 + n_1) - (l_2 + n_2) = (l_1-l_2) + (n_1 - n_2)$ and $n_1 - n_2 \in N$ this is the same as saying the "$L$-component" (not necessarily unique) of the elements are congruent modulo $N$. For example, how I thought of it was this: you could imagine all the elements of $L+N$ as being arranged in a (possibly infinite) array, first listing all the elements of $L$ as a row, then above each element $l \in L$ in this row listing all elements $l+n$ for $n \in N$ one by one as a column. Then all the elements in a particular column are congruent modulo $N$ to the element of $L$ at the bottom of the column, and then the result which we argued above that identifying elements in $L+N$ modulo $N$ is the same as identifying elements in $L$ is clearly seen here in that we are just identifying elements in the bottom row. So we feel that that we must essentially just be quotienting $L$ out by $N$.
This then leads us to consider whether $L/N$ even makes sense. We quickly realise that $L/N$ only make senses as a module if $N$ is a submodule of $L$, but in this case the second isomorphism theorem is trivial because then $L+N = L$ and $L \cap N = N$. So in this case everything is just obvious, but regardless of whether $N$ is a submodule of $L$ or not, because of the thoughts above, we still feel intuitively that there is some similarity in quotienting out $L+N$ by $N$ and quotienting out $L$ by $N$ - and there is.
The equivalence relation on $L$ of being congruent modulo $N$ makes sense regardless of whether $N$ is a submodule of $L$ or not, so in any case we can always form the quotient of equivalence classes $L/N$ - what we lose if $N$ is not a submodule of $L$ is that there won't be any natural way to make $L/N$ a module. What the similarity between quotienting out $L+N$ by $N$ and quotienting out $L$ by $N$ is that the cardinality of $(L+N)/N$ and $L/N$ (considered just as sets of equivalence classes) will be the same (that is, there will be the same number of equivalence classes), because, as the above argument shows (this is easiest to see with the array visual) identifying elements of $L+N$ modulo $N$ is essentially the same as identifying elements of $L$ modulo $N$. There might be some confusion here in that if $L$ and $N$ are finite, one might think the cardinality of $L/(L \cap N)$ should be larger than that of $L/N$ if $L \cap N$ is properly contained within $N$ (i.e. if $N$ is not a submodule of $L$) because $L/N$ is a smaller set. But this is not the case because in $L$ modding out by $N$ is the same as modding out by $L \cap N$ because the differences which we check to see if they lie in $N$ always necessarily lie in $L$ anyway.