Prime ideals of a localization

Remarks

The preimage of a prime ideal under a ring homomorphism is prime, and $\lambda^{-1}(\mathfrak q)$ misses $S$ because elements of $S$ become units. Conversely, if $\mathfrak p \cap S = \emptyset$, then $a/s \in S^{-1}\mathfrak p$ forces $ua \in \mathfrak p$ for some $u \in S$, hence $a \in \mathfrak p$ since $u \notin \mathfrak p$. This shows $\lambda^{-1}(S^{-1}\mathfrak p) = \mathfrak p$ and that $S^{-1}\mathfrak p$ is a proper prime ideal. Finally every ideal $J$ of $S^{-1}R$ is the extension of its contraction, which gives $S^{-1}\lambda^{-1}(\mathfrak q) = \mathfrak q$.

Example: for $R = \Z$ and $S = \set{1, 2, 4, \ldots}$, the prime ideals of $\Z[1/2]$ are $(0)$ and $p\Z[1/2]$ for the odd primes $p$; the prime $2\Z$ meets $S$ and disappears.

For $S = R \setminus \mathfrak p$ with $\mathfrak p$ prime, the primes of $R_{\mathfrak p}$ correspond to the primes of $R$ contained in $\mathfrak p$. So $R_{\mathfrak p}$ has the unique maximal ideal $\mathfrak p R_{\mathfrak p}$, that is, it is a local ring.

The hypothesis that $\mathfrak p$ be prime matters: for a non-prime ideal $I$ with $I \cap S = \emptyset$ one can have $\lambda^{-1}(S^{-1}I) \ne I$. For instance, with $R = \Z$, $I = 6\Z$ and $S = \set{1, 2, 4, \ldots}$, one gets $\lambda^{-1}(S^{-1}I) = 3\Z$.

Going deeper

For a student

This result is the reason why localization is a "local" operation on the spectrum. For $S = \set{1, f, f^2, \ldots}$ it identifies $\operatorname{Spec} R_f$ with the basic open set $D(f)$. For $S = R \setminus \mathfrak p$ it identifies $\operatorname{Spec} R_{\mathfrak p}$ with the primes inside $\mathfrak p$, so the height of $\mathfrak p$ equals the Krull dimension of $R_{\mathfrak p}$. Combined with the description of the primes of a quotient $R/I$, it describes the primes of rings such as $R_{\mathfrak p}/I R_{\mathfrak p}$ for an ideal $I$ of $R$.