# Thread: decomposition of a module

1. ## decomposition of a module

Here is the problem I'm working on:

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

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

2. ## Re: decomposition of a module

First off, if $\displaystyle x \in K$, then $\displaystyle f(x) = 0$, so $\displaystyle x \in ker(f)$. So $\displaystyle K$, if it exists, is a subset of $\displaystyle ker(f)$. So, it definitely feels like $\displaystyle K = ker(f)$ 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 $\displaystyle im(f) \cong M/ker(f)$. Since $\displaystyle im(f) = R$ you have that $\displaystyle im(f) = R \cong M/ker(f)$. Now, you are right that $\displaystyle M/ker(f)$ does not reside within $\displaystyle M$ so maybe you can try to find a submodule $\displaystyle L$ of $\displaystyle M$ that is isomorphic to $\displaystyle M/ker(f)$ and then show that $\displaystyle M = K \oplus L$. From there it'll follow that $\displaystyle f|_L : L \to R$ is indeed an isomorphism.

3. ## Re: decomposition of a module

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

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

Fix $\displaystyle m_0\in{f^{-1}(1)}$, which is not empty since $\displaystyle f(M)=R$. Let $\displaystyle g:M\rightarrow{M}$ be given by $\displaystyle g(m)=f(m)m_0$ for all $\displaystyle m\in{M}$. This map is a module homomorphism since:
$\displaystyle g(rm)=f(rm)m_0=rf(m)m_0=rg(m)$ and
$\displaystyle g(m+n)=f(m+n)m_0=(f(m)+f(n))m_0$$\displaystyle =f(m)m_0+f(n)m_0=g(m)+g(n). Let \displaystyle L=g(M) be the image of this map and \displaystyle K=ker(g) be the kernel. Then I claim that \displaystyle M=K\oplus{L}. Since both \displaystyle K and \displaystyle L are submodules of \displaystyle M, \displaystyle K\oplus{L}\subseteq{M}. If \displaystyle m\in{M}, then it can be represented as \displaystyle m=(m-g(m))+g(m), where \displaystyle m-g(m)\in{K} since \displaystyle f(m-g(m))=f(m)-f(g(m))=f(m)-f(f(m)m_0)$$\displaystyle =f(m)-f(m)f(m_0)=f(m)-f(m)=0$, and of course $\displaystyle g(m)\in{L}=g(M)$. So $\displaystyle M\subseteq{K}\oplus{L}$ and by double containment $\displaystyle M={K}\oplus{L}$.

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

Thank you for giving me the push I needed.

- Hollywood

4. ## Re: decomposition of a module

What follows is basically identical to your work, except that it avoids your function g. Notice that your $\displaystyle L = g(M)$ is the same as $\displaystyle L = m_0R$, since $\displaystyle f(M) = R$ 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 $\displaystyle m_0 \in f^{-1}(1)$ (possible since $\displaystyle f$ is surjective). Let $\displaystyle L = m_0R$. Let $\displaystyle K = ker(f)$. Both are submodules of $\displaystyle M$. Let $\displaystyle M' = L + K$, also a submodule.

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

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

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

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

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

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

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

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

Thus $\displaystyle f|L$ is an isomorphism.

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