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Suppose we have $f : M\rightarrow N$ a homomorphism of $R$-modules. If $M'\leq M$, then $f$ induces a homomorphism $f' : M/M'\rightarrow N/f(M')$.

What is the kernel of $f'$? Is it $(\ker f + M')/M'$?

Certainly $(\ker f + M')/M' \leq \ker f'$, but I can't show the other inclusion. If $m + M'\in \ker f'$, then $f(m) \in f(M')$, but this doesn't necessarily imply that $m\in M'$, does it? So can I split $m$ up into a sum $m = m' + m''$ where $m'\in M'$ and $m''\in \ker f$?

IAlreadyHaveAKey
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  • I figured it out. Since $f(m)\in f(M')$, there exists $m'\in M'$ such that $f(m) = f(m')$, so then $m = m - m' + m'$, so $m'\in M'$ and $f(m-m') = f(m) - f(m') = 0$, so $m-m'\in \ker f$. Therefore $m \in \ker f + M'$. – IAlreadyHaveAKey Sep 22 '17 at 00:19

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It's going to be $f^{-1}(f(M'))/M'$. Indeed, this is certainly contained in the kernel, and if $f'(m + M') = f(m) + f(M') = 0$, then $f(m) \in f(M')$, whence $m \in f^{-1}(f(M')) \supseteq M'$.

By the argument in your comment, this can alternatively be written as $(\ker f + M')/M'$.

Mr. Chip
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