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Let $0\to L\stackrel{\alpha}\to M\stackrel{\beta}\to N\to 0$ be a splitting s.e.s. where $\alpha$ has a retraction $r$.

(a) Show that in this case $M=\alpha(L)\oplus\ker(r)$.

I am a little bit confused because we have defined $\oplus$ as "direct sum" that is $A\oplus B$ is all ordered pairs $(a,b)$ with $a\in A$ and $b\in B$. This cannot be equal to $M$ (it can at most be homomorphic to $M$). How can I otherwise interpret $\oplus$?

egreg
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Vinyl_cape_jawa
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  • There are actually two definitions of $\oplus$: the internal direct sum, and the external direct sum. In this case, it does seem like an external direct sum. The statement $M = \alpha(L) \oplus \text{ker}(r)$ simply means $\alpha(L)$ and $\text{ker}(r)$ are independent submodules of $M$ (their intersection is $0$), and they span $M$ (their sum is $M$). – Tunococ Oct 22 '14 at 11:05
  • Thanks. Now one more question. Why does the intersection have to be $0$? – Vinyl_cape_jawa Oct 22 '14 at 11:10
  • That is the definition of the notation $\oplus$. Otherwise the statement should have been written $M = \alpha(L) + \text{ker}(r)$. Knowing that $\alpha(L) \cap \text{ker}(r) = 0$ tells you that if $m \in \alpha(L) + \text{ker}(r)$, then there exist unique $\ell \in \alpha(L)$ and $k \in \text{ker}(r)$ such that $\ell + k = m$. – Tunococ Oct 22 '14 at 11:13
  • Oh, I think I did make a serious typo in my original comment. The direct sum in this case is internal, not external. – Tunococ Oct 22 '14 at 11:14

1 Answers1

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The notation $\oplus$ is used in two slightly different meanings. If $A$ and $B$ are submodules of $M$, then $A+B$ is written $A\oplus B$ when $A\cap B=\{0\}$. If $M$ and $N$ are modules, then $M\oplus N$ denotes the module whose elements are ordered pairs, with componentwise addition and multiplication by scalars.

The direct sum $M\oplus N$ (in the second sense) can be seen as $M'\oplus N'$ in the first sense, where $M'=\{(x,0):x\in M\}$ and $N=\{(0,y):y\in N\}$. Since $M$ and $M'$ are “canonically” isomorphic by $x\mapsto(x,0)$ and similarly for $N$, using the same notation is abusive, but handy. Often the two are distinguished by saying “internal direct sum” (first sense) and “external direct sum” (second sense).

What you have to prove is that the obvious morphism $\sigma$ from $\alpha(L)\oplus\ker r$ (external) to $M$ defined by $$ \sigma\colon (x,y)\mapsto x+y $$ is an isomorphism. Since $\ker\sigma=\{(x,y):y=-x\}$, this amounts to saying that

  1. $\alpha(L)\cap\ker r=\{0\}$ (injectivity of $\sigma$)
  2. $\alpha(L)+\ker r=M$ (surjectivity of $\sigma$)

which is the same as saying that $\alpha(L)\oplus\ker r$ (internal direct sum) equals $M$.

So, let $z\in\alpha(L)\cap\ker r$; then $z=\alpha(x)$ for some $x\in L$, so $$ 0=r(z)=r\alpha(x)=x $$ because $r$ is a retract of $\alpha$. Therefore $z=\alpha(0)=0$.

For the second condition, let $z\in M$; then $x=r(z)\in L$ and, setting $y=z-\alpha r(z)=z-\alpha(x)$ we have $$ r(y)=r(z-\alpha r(z))=r(z)-r\alpha r(z)=r(z)-r(z)=0 $$ so $y\in\ker r$. Therefore $z=\alpha(x)+y\in\alpha(L)+\ker r$.

egreg
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