All rings are assumed to be commutative with unity. Usually denoted $R$
Intro: I understand that there are two definitions of localization (or possibly more, but these two are somewhat canonical for me):
1 Defining ring of fractions with elements from multiplicative $S$ as denominators.
2 For any set $S$ by universal property of localization: $\alpha \in Hom(R, S^{-1}R)$ such that a)$\alpha(s)$ is invertible and b)If $\beta \in Hom(R, T)$ and $\beta(s)$ invertible in $T$ then there is $\gamma \in Hom(S^{-1}R, T)$ such that these ring homomorphisms commute.
I understand the "fractions definition" and from this, I built some understanding of Categorical definition. We also had a proposition in my Alg. Geometry class showing that rings of fractions from the first definition (together with homomorphism $r->r/1$) are basically the same thing as in the categorical definition.
Question However, the categorical definition needs not the assumption of $S$ being multiplicatively closed. How does a (categorical) localization with respect to $S$ not multiplicatively closed looks like?
My thoughts Of course the definition does not say that such localization exists. But what looks more natural to me is take $S'\supset S$ the smallest multiplicatively closed set containing $S$ and consider localization with respect to $S'$. I suppose that all elements from $S'$ would be invertible also in $S^{-1}R$ so maybe using universal property of localization should work?
Best Answer
That's correct: the localization with respect to $S$ is just the same as the localization with respect to the multiplicatively closed set $S'$ that $S$ generates. This is because if $\alpha:R\to T$ is any ring homomorphism, then the set of $s$ such that $\alpha(s)$ is invertible is always multiplicatively closed, so it contains $S$ iff it contains $S'$. So, the universal properties of localization with respect to $S$ and localization with respect to $S'$ are equivalent.