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Let $A$ be a $\mathbb{K}$-algebra and $M$ be an $A$-module, $N \subset M$ be a submodule.

A module complement of $N$ is defined as a submodule $N'\subset M$ so that $M = N \oplus N'$.

Now in our lecture we defined a semisimple $A$-module as an $A$-module so that for every submodule exists a module complement.

But isn't that a trivial property that is true for all $A$-modules, because of this:

$M \cong N \oplus M / N$, where $M / N$ is an $A$-module by the operation $a.(m+N)=a.m+N$ for $a \in A$, $m+N \in M/N$, so $M/N$ is always a module complement of $N$?

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    $\begingroup$ Is $\mathbb{Z}\simeq 2\mathbb{Z}\oplus\mathbb{Z/2Z}$ ? Hint : the latter has a non zero element $x$ such that $x+x=0$. $\endgroup$ – Roland Sep 1 '17 at 13:41
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  • The module $M/N$ is not a submodule of $M$, and therefore cannot be a complement of $N$.
  • It is not necessarily true that $M \cong N \oplus (M/N)$. Take for example $A = k[x]$, $M = A$ and $N = Ax = (x)$. Then $$ N \oplus (M/N) = (x) \oplus (k[x]/(x)) \ncong k[x] = M $$ since $k[x]$ is torsion-free, but $k[x]/(x)$ is not.
  • In the above example the submodule $N$ has no direct complement: Otherwise they would exist some ideal $(f) \subseteq k[x]$ with $k[x] = (x) \oplus (f)$. Then $xf \in (x) \cap (f) = 0$ and therefore $f = 0$, which would result in $k[x] = (x)$.
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    $\begingroup$ Thank you very much! So can I say $M$ and $N \oplus (M/N)$ are isomorphic as vector spaces, but not as $A$-modules? In your example, this is because $(x)\oplus (k[x]/(x))\to k[x],(f,g)\mapsto f+g$ is a vector space isomorphism, but not a module isomorphism? $\endgroup$ – S. M. Roch Sep 1 '17 at 14:40
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    $\begingroup$ 1) Yes, $M$ and $N \oplus (M/N)$ are always isomorphic as vector spaces, but in the given example not as $A$-modules. 2) The mapping $(f,g) \mapsto f + g$ that you give is not well-defined! That $k[x]$ and $(x) \oplus (k[x]/(x))$ are not isomorphic as $A$-modules can instead be seen by considering the torsion elements of both modules: The element $m := (0,\overline{1}) \in (x) \oplus (k[x]/(x))$ is nonzero and satisfies $xm = 0$. But in $k[x]$ no such element exists, since $xf \neq 0$ for every nonzero $f \in k[x]$. $\endgroup$ – Jendrik Stelzner Sep 1 '17 at 18:39
  • $\begingroup$ Thanks, now I understand it better! $\endgroup$ – S. M. Roch Sep 1 '17 at 18:53
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No, for example, with $A=M=\mathbb R[x]/(x^2)$, and $N=(x+(x^2))$, there is no way to complement $N$.

Yes, you always have the short exact sequence $0\to N\to M\to M/N$, but $N$ is a summand of $M$ iff this sequence is split exact.

Added later: Actually I rediscovered a post I had written some time ago on when complementation can fail. It might come in useful too.

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