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Let $R$ be a (commutative with identity) ring, and $S$ be a multiplicatively closed subset of $R$. We define the localization of $R$ at $S$ a ring homomorphism $\phi:R\to S^{-1}R$ into a ring $S^{-1}R$ such that each element of $\phi(S)$ is a unit in $S^{-1}R$, and such that whenever we have a ring homomorphism $f:R\to T$ such that $f(S)$ is contained in the group of units of $T$, we have a unqiue ring homomorphism $\bar f:S^{-1}R\to T$ satisfying $\bar f\circ \phi=f$.

I want to show that $\ker(\phi)=\{x\in R:\ sx=0 \text{ for some } s\in S\}$ using the above definition.

The usual proof goes by explicitly constructing $S^{-1}R$ by defining an equivalence relation on $(S\times R)$. But why should we need any explicit construction. When dealing with tensor products, we prove everything from universal property alone. So there must be a way here too. But I cannot see it.

caffeinemachine
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  • Suggestion: One direction of inclusion is direct. If $sx=0$, then since $s$ is invertible, $x=s^{-1}sx=s^{-1}0=0$. The other inclusion is from the universal property. Although one might need the explicit construction to show that a ring exists where the elements of $S$ are invertible and the elements not in the kernel are not zero. – Michael Burr Feb 19 '17 at 14:25
  • It is not clear how you are deducing the reverse inclusion. Is this your approach: Assume $sx\neq 0$ for all $s\in S$. We want to show $x\notin \ker(\phi)$. Then are we constructing a ring homomorphism $R\to T$ such that $S$ maps into $T^*$ and $x$ does not maps to $0$? Of course, one such ring $T$ is obtained by the usual construction of $S^{-1}R$. – caffeinemachine Feb 19 '17 at 15:28
  • That approach is exactly what I am hinting at. – Michael Burr Feb 19 '17 at 16:10
  • But that is tantamount to explicit construction. But I suppose that can't be avoided. – caffeinemachine Feb 19 '17 at 18:09
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    @caffeinemachine You can do it very efficiently using universal properties of polynomial and quotient rings, e.g. see here. – Bill Dubuque Feb 20 '17 at 01:53

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