Let $R$ be an infinite comutative ring with unity, $M,N$ be $R$-modules, $f:M \to N$ be a surjective module homomorphism; then $|M|=|N ||\ker f|$?

Question

Let $R$ be an infinite commutative ring with unity, $M,N$ be modules over $R$, let $f:M \to N$ be a surjective module homomorphism; then is it true that $|M|=|N || \ker f|$ ($M,N$ are not necessarily finite ) ?

Answer 1

I'm on a cell phone so let the kernel be $A$. Using the axiom of choice, construct a set $N'$ consisting of one element from each coset of $A$. There is an obvious bijection of $N'$ with $N$. Then every element of $M$ can be represented uniquely as the sum of an element of $N'$ and an element of $A$, hence there is a bijection between $N'\times A$ and $M$ defined by $(n,a)\mapsto n+a$. Since $|N'\times A|=|N||A|$, the result follows.