0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why is $a$ a unit? Since we divide by $a$, we need it to be a unit. But wouldn't this mean that $(a)=R$? $\endgroup$ – MEG Aug 17 '19 at 22:31
  • $\begingroup$ $a $ is only assumed non-zero; since $f (x) $ is divisible by $a $, $f (x)/a $ is defined. $\endgroup$ – Fabio Lucchini Aug 17 '19 at 22:48
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.