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I'd like to have some clarification and help on this problem.

Suppose $M$ is a module and $K = \{(m, m) : m \in M\} \subset M \oplus M$. Show $K$ is a submodule of $M\oplus M$ which is a direct summand.

Showing $K$ to be a submodule in $M \times M$ should just follow from definition. But I'm a bit unsure of what a summand is so I'm not sure how to show this last bit that $M \oplus M$ is a summand (or is it asking whether $K$ is a summand?)

randomafk
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  • You need to show there exists another module $P$ such that $P \oplus K \cong M \oplus M$. To further clarify the terminology: this shows that $K$ is a summand of $M \oplus M$. –  Jan 02 '13 at 22:51
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    Actually, you should not merely show $P\oplus K\cong M\oplus M$, but rather $P\oplus K=M\oplus M$. Note that $\cong$ is trivial as $K\cong M$ (as you may have noted). We are talking of "internal direct sum" here. So you need to show that $P+K=M\oplus M$ and $P\cap K=0$. – Hagen von Eitzen Jan 02 '13 at 22:56
  • That's true. $P$ and $K$ aren't just abstract modules; they're concrete submodules of $M \oplus M$. –  Jan 02 '13 at 23:01
  • Then would we simply choose a $P$ such that $P \cap K = 0$? The only situation in which $P \cap K \ne 0$ is if $ x = (a, a) \in L$, but we can remove this from P since it is contained in $K$?. i.e. $K = M \oplus M - P$

    And so we can construct any $x \in M \oplus M$ from $P + K {p + k : p \in P, k \in K}$ ?

    It seems like everything sort of follows from construction or definition?

     – randomafk Jan 02 '13 at 23:13

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You can take $P=\{0\}\oplus M$ or $P=M\oplus \{0\}$. Both satisfy the two conditions: $P\cap K=\{(0,0)\}$ and $P+K=M\oplus M$.