# Module over PID is isomorphic to the direct sum of a submodule and the kernel of a module homomorphism

I'm reviewing some module theory before I take an algebra prelim at my university, and I came across a question that I do not know how to start.

"Let $$R$$ be a PID and let $$M$$ be an $$R$$-module. If $$f:M\rightarrow R$$ is a module homomorphism, show there exists a submodule $$N$$ of $$M$$ such that $$M=N\oplus \ker(f)$$."

My background with modules is limited. How can I approach this proof?

First note that the image of $$M$$ under $$f$$ is an ideal of $$R$$; since $$R$$ is PID, this ideal is principal generated by an (assume non-zero) element $$a\in R$$. Thus we get a surjective $$R$$-module homomorphism $$f:M\to aR$$. Since $$aR\cong R$$ is a (free, hence) projective $$R$$-module, the exact sequence $$\{0\}\to\operatorname{Ker}f\hookrightarrow M\xrightarrow f aR\to\{0\}$$ splits. Consequently, $$\operatorname{Ker}f$$ is a direct summand of $$M$$.

More explicity, let $$x_0\in M$$ such that $$f(x_0)=a$$ and $$N=x_0R\subseteq M$$. Then clearly $$N\cap\operatorname{Ker}f=\{0\}$$ and for every $$x\in M$$ we have $$x=\underbrace{x_0\frac{f(x)}a}_{\in N}+\underbrace{\left(x-x_0\frac{f(x)}a\right)}_{\in\operatorname{Ker}f}$$ thus proving $$M=N+\operatorname{Ker}f$$ and hence $$M=N\oplus\operatorname{Ker}f$$ as internal direct sum.

Alternatively, \begin{align} N\times\operatorname{Ker}f&\to M& (x,y)&\mapsto x+y\\ M&\to N\times\operatorname{Ker}f& x&\mapsto\left(x_0\frac{f(x)}a,x-x_0\frac{f(x)}a\right) \end{align} are inverse each other isomorphisms.

• Why is $a$ a unit? Since we divide by $a$, we need it to be a unit. But wouldn't this mean that $(a)=R$? – MEG Aug 17 '19 at 22:31
• $a$ is only assumed non-zero; since $f (x)$ is divisible by $a$, $f (x)/a$ is defined. – Fabio Lucchini Aug 17 '19 at 22:48

If $$\ker(f)=M$$, then we can let $$N=0$$, and we're done.

So assume $$\ker(f)\ne M$$.

Then $$f(M)\ne 0$$, hence $$f(M)=(y)$$, for some nonzero $$y\in R$$.

Let $$x\in M$$ be such that $$f(x)=y$$, and let $$N=\langle{x}\rangle$$.

Claim:$$\;M=N\oplus \ker(f)$$.

Proof:

First we show $$N\cap \ker(f)=0$$.

Let $$z\in N\cap \ker(f)$$.

Since $$z\in N$$, we have $$z=rx$$ for some $$r\in R$$. \begin{align*} \text{Then}\;\;&z\in N\cap \ker(f)\\[4pt] \implies\;&z\in \ker(f)\\[4pt] \implies\;&f(z)=0\\[4pt] \implies\;&f(rx)=0\\[4pt] \implies\;&rf(x)=0\\[4pt] \implies\;&ry=0\\[4pt] \implies\;&r=0\;\;\;\text{[since y\ne 0]}\\[4pt] \implies\;&rx=0\\[4pt] \implies\;&z=0\\[4pt] \end{align*} hence $$N\cap \ker(f)=0$$.

It remains to show $$N+\ker(f)=M$$.

Let $$m\in M$$. Then we have \begin{align*} &f(m)\in f(M)\\[4pt] \implies\;&f(m)\in (y)\\[4pt] \implies\;&f(m)=ry\;\text{for some}\;r\in R\\[4pt] \end{align*} Let $$n=rx$$ and let $$z=m-n$$. Then we have $$n\in N$$, and \begin{align*} f(z)&=f(m-n)\\[4pt] &=f(m)-f(n)\\[4pt] &=ry-f(rx)\\[4pt] &=ry-rf(x)\\[4pt] &=ry-ry\\[4pt] &=0\\[4pt] \end{align*} hence $$z\in\ker(f)$$, so $$m=n+(m-n)=n+z\in N+\ker(f)$$.

Thus we have $$M=N+\ker(f)$$, which completes the proof.