decomposition of a module

**hollywood** · September 10th 2012, 06:25 AM

Here is the problem I'm working on:

Let $R$ be an associative ring with $1$ , $M$ a right $R$ -module, $F:M\rightarrow{R}$ a homomorphism of $R$ -modules with $f(M)=R$ . Prove that there is a decomposition $M=K\oplus{L}$ with $f(K)=0$ and $f|_L:L\rightarrow{R}$ is an isomorphism.

I think that $K$ should be the kernel of $F$ . Then $L$ should be something like $M/K$ , but it seems like I need something that's actually in $M$ instead. What should I use for $L$ , and how do I show that $M=K\oplus{L}$ and $f|_L$ is an isomorphism?

**Vlasev** · September 10th 2012, 09:13 PM

First off, if $x \in K$ , then $f(x) = 0$ , so $x \in ker(f)$ . So $K$ , if it exists, is a subset of $ker(f)$ . So, it definitely feels like $K = ker(f)$ is the right choice. As for the rest, there is an isomorphism theorem for modules that is much like the ring and group isomorphism theorems.

It basically says that $im(f) \cong M/ker(f)$ . Since $im(f) = R$ you have that $im(f) = R \cong M/ker(f)$ . Now, you are right that $M/ker(f)$ does not reside within $M$ so maybe you can try to find a submodule $L$ of $M$ that is isomorphic to $M/ker(f)$ and then show that $M = K \oplus L$ . From there it'll follow that $f|_L : L \to R$ is indeed an isomorphism.

**hollywood** · September 11th 2012, 12:31 PM

So I need $g:M\rightarrow{M}$ with kernel $K=ker{f}$ . Then the image will be a submodule of $M$ isomorphic to $M/K$ .

I think I can fix $m_0$ in $f^{-1}(1)$ and make $g(m)=f(m)m_0$ for all $m\in{M}$ . The kernel will be the same as the kernel of $f$ , so that works.

So my answer is:
Fix $m_0\in{f^{-1}(1)}$ , which is not empty since $f(M)=R$ . Let $g:M\rightarrow{M}$ be given by $g(m)=f(m)m_0$ for all $m\in{M}$ . This map is a module homomorphism since:
$g(rm)=f(rm)m_0=rf(m)m_0=rg(m)$ and
$g(m+n)=f(m+n)m_0=(f(m)+f(n))m_0$ $=f(m)m_0+f(n)m_0=g(m)+g(n)$ .
Let $L=g(M)$ be the image of this map and $K=ker(g)$ be the kernel. Then I claim that $M=K\oplus{L}$ .

Since both $K$ and $L$ are submodules of $M$ , $K\oplus{L}\subseteq{M}$ . If $m\in{M}$ , then it can be represented as $m=(m-g(m))+g(m)$ , where $m-g(m)\in{K}$ since $f(m-g(m))=f(m)-f(g(m))=f(m)-f(f(m)m_0)$ $=f(m)-f(m)f(m_0)=f(m)-f(m)=0$ , and of course $g(m)\in{L}=g(M)$ . So $M\subseteq{K}\oplus{L}$ and by double containment $M={K}\oplus{L}$ .

Given $M={K}\oplus{L}$ as described above, $f(K)=0$ since $K$ is the kernel of $f$ . If $f(m)=0$ and $m\in{L}$ , $m$ is of the form $rm_0$ for some $r\in{R}$ , so $0=f(m)=f(rm_0)=rf(m_0)=r$ , giving $m=rm_0=0$ , so $f|_L$ is injective. And every $r\in{R}$ has a preimage (namely $rm_0$ ) in $L$ , so $f|_L$ is surjective. Therefore $f|_L$ is an isomorphism.

Thank you for giving me the push I needed.

- Hollywood

**johnsomeone** · September 11th 2012, 02:03 PM

What follows is basically identical to your work, except that it avoids your function g. Notice that your $L = g(M)$ is the same as $L = m_0R$ , since $f(M) = R$ by assumption.
(One other minor point - the problem said that M is a right R module, but your work treated it as if ti were a left R module. It doesn't make a difference here - you just flip the order in which you need to write those things down, and nothing is really impacted.)

Let $m_0 \in f^{-1}(1)$ (possible since $f$ is surjective). Let $L = m_0R$ . Let $K = ker(f)$ . Both are submodules of $M$ . Let $M' = L + K$ , also a submodule.

Let $m \in K \cap L.$ Then $m \in L$ so $\exists r \in R \ni m = m_0r$ . Also, $m \in K$ , so $f(m) = 0$ , so $f(m_0r) = 0$ , so $f(m_0)r = 0$ , so $1r = 0$ , so $r = 0$ .

Thus $m \in K \cap L \implies m = m_0(0) = 0$ . Thus $M' = K \oplus L$ .

To show $M' = M$ , let $m \in M$ . Note that $m = (m - m_0f(m)) + m_0f(m)$ .

Since $f(m - m_0f(m)) = f(m) - f(m_0f(m)) = f(m) - f(m_0)f(m)$ ,

have $f(m - m_0f(m)) = f(m) - (1)f(m) = 0$ , so it follows that $(m - m_0f(m)) \in ker(f) = K$ .

Obviously $m_0f(m) \in L$ . Thus $m = (m - m_0f(m)) + m_0f(m) \in K + L = M'$ . Thus $M \subset M'$ . Since $M' \subset M$ , have $M' = M$ .

Since $m \in L$ means $\exists r \in R \ni m = m_0r$ , have that $f|L: L \rightarrow R$ by $f(m) = f(m_0r) = f(m_0)r = 1r = r$ .

Thus $ker(f|L) = \{m_0r \in L | f(m_0r) = 0 \} = \{m_0r \in L | r = 0 \} = \{ 0 \}$ . Also, $Im( f|L ) = R$ , since if $r \in R$ , then $(f|L)(m_0r) = r$ .

Thus $f|L$ is an isomorphism.

Have shown that $\exists K, L$ submodules of $M \ni M = K \oplus L$ , and $f|K \equiv 0$ , and $f|L$ is an isomorphism onto $R$ .

Thread: decomposition of a module

LinkBack

Thread Tools

Display

decomposition of a module

Re: decomposition of a module

Re: decomposition of a module

Re: decomposition of a module

Similar Math Help Forum Discussions

a singular module is the quotient of a module and its essential submodule

Module Help

k[x]-module

Jordan Decomposition to Schur Decomposition

Module 77

Search Tags