Every non-zero module homomorphism $f:M\rightarrow N$ is injective. Prove that $M$ is simple.

$f$ is not the zero map so $M$ is not $\{0\}$. I'm guessing I should let $N$ be a proper submodule of $M$ and pick a non-zero module homomorphism for which $N$ is the kernel. Any ideas?



Consider the projection $M \twoheadrightarrow M/N$.

  • $\begingroup$ Ah, the beauty of simplicity. +1 $\endgroup$ – DonAntonio Dec 12 '13 at 14:50
  • $\begingroup$ just realised this implication is false unless they have the added assumption M is not {0}. If M={0}, then no non-zero homomorphism exist, so the hypothesis is vacuously satisfied, yet M would not be simple. $\endgroup$ – user108605 Dec 12 '13 at 16:13
  • $\begingroup$ Yes. This is often forgotten. $\endgroup$ – Martin Brandenburg Dec 12 '13 at 16:40

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