I'm having trouble understanding the proof of this theorem, which can be found on page 460 of Dummit & Foote
To prove 1), we may without loss of generality assume $N \neq \{0\}$. And as $R$ is principal, $\varphi(N) = (a_\varphi)$ for all $\phi \colon M \rightarrow R$. So we set $\sum = \{(a_\varphi) \colon \varphi \in Hom(M,R) \}$. This is non-empty and has a maximal element, which we denote $(a_\nu) = (a_1)$. Let's say $a_1 = \nu(y)$.
We can show that $a_1$ is not zero, and that $a_1$ divides $\varphi(y)$ for every $\varphi \in Hom(M,R)$. Letting $\pi$ be the natural projection onto the $i$-th coordinate of the basis, we get $\pi_i(y) = a_1 b_i$. We define $y_1 = \sum_{i=1}^n b_i x_i$ where $x_1, \ldots, x_n$ is the basis of $M$. We can show that $a_1 y_1 = y$. Now, we want to show the below.
I'm having trouble understanding the proof for $b)$. It write for arbitrary $n \in N$, $n = \nu(n) y_1 + (n – \nu(n) y_1)$. What I'm trouble having see is
-Why is $n-\nu(n) y_1 \in N$?
-Why do we need $N \cap \text{ker } \nu$, instead of just $\text{ker } \nu$?
Thanks for your help!
EDIT: For my second point, my proof says to (to show that the sum is direct. We do know that N is the sum of the two modules at this point) suppose $x \in Ra_1 y_1 \cap (N \cap \text{Ker } \nu)$. Then we write $x = ra_1y_1 \in N$, and $\nu(x) = 0$.
Then $0 = \nu(x) = \nu(r a_1 y_1) = ra_1 \nu(y_1) = ra_1$, so we get $r = 0$ as $a_1 \neq 0$. My problem is that in this line, it seems $x \in N$ seems unnecessary.
EDIT EDIT: well $x$ must be in $N$ since we want that whole thing to be $N$. It's all good now I think.
Best Answer
We have found $\nu\colon N\to R$ with image $(a_1)$, and taken $y\in N$ with $\nu(y)=a_1$. Given $n\in N$ we have $\nu(n)\in(a_1)$, say $\nu(n)=ta_1$, so that $$ \nu(n)y_1 = ta_1y_1 = ty \in N. $$ Thus both $n$ and $\nu(n)y_1$ lie in $N$, so their difference $n-\nu(n)y_1$ does too.
In general a decomposition $M=Rm\oplus K$ will not pass to the submodule $N$, so we will not in general have $N=(Rm\cap N)\oplus(K\cap N)$. The statement (b) says that we do have it in our special situation, with $m=y_1$ and $K=\mathrm{Ker}\,\nu$.
Once we have such a compatible decomposition, we can pass to the case of the free module $M\cap K$, of rank one less that $M$, and its submodule $N\cap K$, and apply induction.