On showing that $\varphi(N)=(N+K)/K$, for modules $N,K<M$, $\varphi$ natural map.

Question

In this question modules are A-modules, where A is a unitary commutative ring

This seems trivial, but beginning with module theory, I was trying to see why $\varphi(N)=(N+K)/K$, for the natural A-module homomorphism $\varphi:M\to M/K$, and submodules $N,K\subset M$.

In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:

$N+K=\{n+k:n\in N, k\in K\}$, and $(N+K)/K=\{(n+k)+K:(n+k)\in (N+K)\}$, $N/K=\{n+K:n\in N\}$, as sets.

And isn't $(n+k)+K=n+K$, for some $n\in N, k\in K$?

And then, for the quotients we have equivalence relations, so that

$N/K \ni\bar{n_1}=\bar{n_2}\Leftrightarrow n_1 + K=n_2+K\Leftrightarrow (n_1-n_2)\in K$.

Well, we also have

$(N+K)/K \ni\overline{n_1+k_1}=\overline{n_2+k_2}\Leftrightarrow n_1 + k_1 + K=n_2+k_2 +K\Leftrightarrow (n_1-n_2)+(k_1-k_2)\in K \Leftrightarrow (n_1-n_2)\in K$.

Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.

Now,

Question1: Where have I gone wrong, misunderstood or forgotten something?

Question2: Will someone provide a proof sketch of the statement in the title?

Answer 1

The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.

Proof that $\varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)

For $m\in M,$ $\varphi(m) = m+K.$ Thus, $\varphi(N) = \{n + K\mid n\in N\subseteq M\}.$ You've already noted that $(N + K)/K = \{(n + k) + K\mid n\in N, k\in K\},$ and that for any $(n + k) + K\in (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $\varphi(N) = (N+K)/K.$

Essentially, the argument you've made just shows that the module map \begin{align*} f : \varphi(N)&\to (N+K)/K\\ n + K&\mapsto (n + 0) + K, \end{align*} is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $\varphi(N)$ and $(N+K)/K$ inside $M/K.$