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Let $M$ be a finitely presented $A$-module and $S \subset A$ be a multiplicative set. Hence have a surjective map $f:A^{\oplus n} \rightarrow M$

Let $S^{-1}M$ be a free $S^{-1}A$-module. Then we have an isomorphism $S^{-1}f:(S^{-1}A)^{\oplus n} \rightarrow S^{-1}M$.

Is it true that $S^{-1}$ker($f$)=Ker($S^{-1}f$)?

I think $"\subseteq"$ holds true but I am not sure about the other inclusion.

mathStudent
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  • Is $A$ a commutative ring? If so, both inclusions hold even without the finite presentation assumption. Given $x/s \in \ker(S^{-1}f)$, then $\frac{0}{1} = S^{-1}f(x/s) = f(x)/s$. Then there exists $t \in S$ such that $t f(x) = 0$. Then $\frac{x}{s} = \frac{tx}{ts} \in S^{-1} \ker(f)$. – Viktor Vaughn May 23 '20 at 07:45

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