Localization preserves prime ideals

A prime ideal disjoint from the multiplicative set extends to a prime ideal in the localized ring.
Localization preserves prime ideals

Prime ideals behave well under : if a prime ideal does not meet the elements being inverted, then it stays prime after localization. This is one half of the .

Theorem

Let RR be a , let SS be a in RR, and let p\mathfrak p be a prime ideal of RR such that pS=\mathfrak p\cap S=\varnothing.

Then the extended ideal

S1p  =  {xsS1R:xp, sS} S^{-1}\mathfrak p \;=\; \left\{ \frac{x}{s}\in S^{-1}R : x\in\mathfrak p,\ s\in S \right\}

is a prime ideal of S1RS^{-1}R.

Moreover, S1pS^{-1}\mathfrak p is precisely the kernel of the canonical map

S1RS1(R/p), S^{-1}R \longrightarrow S^{-1}(R/\mathfrak p),

and contracting back recovers p\mathfrak p:

(S1p)R=p. (S^{-1}\mathfrak p)\cap R=\mathfrak p.

If pS\mathfrak p\cap S\neq\varnothing, then S1p=S1RS^{-1}\mathfrak p=S^{-1}R (so there is no corresponding prime in the localization), which explains the disjointness hypothesis.

Examples

  1. Localizing Z\mathbb Z at a prime.
    Let R=ZR=\mathbb Z and S=Z(p)S=\mathbb Z\setminus(p), so S1R=Z(p)S^{-1}R=\mathbb Z_{(p)}. The prime ideals of Z\mathbb Z are (0)(0) and (q)(q) for primes qq. One checks:

    • (p)S=(p)\cap S=\varnothing, so (p)(p) extends to the (unique) maximal ideal pZ(p)p\mathbb Z_{(p)}, which is prime.
    • If qpq\neq p, then (q)S(q)\cap S\neq\varnothing (since qSq\in S), so (q)(q) becomes the unit ideal after localization.
    • Also (0)S=(0)\cap S=\varnothing, so (0)(0) extends to (0)(0) in Z(p)\mathbb Z_{(p)}.
  2. Inverting a variable in a polynomial ring.
    Let R=k[x,y]R=k[x,y] and S={1,y,y2,}S=\{1,y,y^2,\dots\}, so S1R=k[x,y,y1]S^{-1}R=k[x,y,y^{-1}]. The ideal p=(x)\mathfrak p=(x) is prime and disjoint from SS (since no power of yy lies in (x)(x)), hence S1(x)S^{-1}(x) is prime in k[x,y,y1]k[x,y,y^{-1}].
    By contrast, the prime ideal (y)(y) meets SS (it contains yy), so S1(y)=S1RS^{-1}(y)=S^{-1}R.

  3. Localization at a prime and the resulting local ring.
    If p\mathfrak p is a prime ideal of RR and S=RpS=R\setminus\mathfrak p, then S1RS^{-1}R is the , which is a . The extension S1pS^{-1}\mathfrak p becomes the unique maximal ideal of RpR_{\mathfrak p}, and it is prime by the theorem.

For the full bijective correspondence between primes of S1RS^{-1}R and primes of RR disjoint from SS, see .