1
$\begingroup$

Let $R$ be a commutative ring, and let $S \subseteq R$ be a multiplicatively closed subset (not containing $0$). Then we construct the localised ring $R [ S^{-1} ]$. I understand that prime ideals in $R [ S^{-1} ]$ correspond to prime ideals in $R$ which do not intersect $S$. Is it always true that if $\mathfrak{q}$ is prime in $R$, then $\mathfrak{q}R[S^{-1}]$ is prime in $R[S^{-1}]$?

1 Answers 1

3

Either $\mathfrak q$ does not intersect $S$, and then you know that the extension to $S^{-1}R$ is prime, or it intersects $S$ in which case $\mathfrak q(S^{-1}R) = S^{-1}R$ is a very specific non-prime.

Here's an elementary way to prove the first assertion: show that $a/s$ is in $\mathfrak q(S^{-1}R)$ if and only if $a \in \mathfrak q$. You'll need to use that $\mathfrak q$ is prime and that $S \cap \mathfrak q=\varnothing$.

  • 0
    Great. I am a bit of a scatterbrain and frequently make mistakes so I always like to check what I've done is right2012-05-23