1

Is the localization of a direct product of two rings at a maximal (or prime) ideal identified with a localization of one of them? I would appreciate for any detailed answer.

karparvar
  • 5,730
  • 1
  • 14
  • 36

1 Answers1

5

First prove the following: if $S_i\subset R_i$ is a multiplicative set, then $$(S_1\times S_2)^{-1}(R_1\times R_2)\simeq S_1^{-1}R_1\times S_2^{-1}R_2.$$

A maximal ideal of $R_1\times R_2$ is of the form $M_1\times R_2$ or $R_1\times M_2$ with $M_i$ maximal in $R_i$. To localize $R_1\times R_2$ at $M_1\times R_2$, consider the multiplicative set $$(R_1\times R_2)\setminus (M_1\times R_2)=(R_1\setminus M_1)\times R_2,$$ use the previous isomorphism and recall that $S^{-1}R=0$ if $0\in S$.

user26857
  • 52,094
  • 2
    can you give more detail why the isomorphism is true? – annimal Dec 09 '14 at 15:26
  • @annimal There is natural map $R_1\times R_2\to S_1^{-1}R_1\times S_2^{-1}R_2$. It is easy to verify that this map satisfies Corollary 3.2 of Atiyah, MadDonald, Introduction to Commutative Algebra. This shows the isomorphism. – Elías Guisado Villalgordo Jun 18 '23 at 16:05