0

Let $R$ a commutative ring. I'm trying to show that $$\mathrm{Spec}( R_{\mathfrak p})=\{\mathfrak qR_{\mathfrak p}\mid \mathfrak q\subset \mathfrak p, \mathfrak q\in Spec(R)\}.$$ I recall that $$R_{\mathfrak p}=\left\{\frac{x}{s}\mid s\notin \mathfrak p\right\}.$$

For the inclusion $\supset$ I did as follow. Suppose by contradiction that it's not prime, i.e., there is $\frac{x}{s},\frac{x}{s'}\in \mathfrak qR_{\mathfrak p}$ s.t. neither $\frac{x}{s}$ nor $\frac{x'}{s'}$ are in $\mathfrak q R_{\mathfrak p}$.

I can't arrive to a contradiction. For the other inclusion, I have no idea.

Any help is welcome.

user26857
  • 53,190
user386627
  • 1,908
  • $R_{\mathfrak p}=\left{\frac{x}{s}\mid s\notin \mathfrak p\right}$ is not quite right. Especially if the complement of $\mathfrak p$ contains zero divisor it's important to keep in mind that the elements in the localisation are equivalence classes of the above set, signifying that you're allowed to simplify and expand any fraction. – Arthur Nov 12 '16 at 20:09
  • Presumably, you mean $\frac{x}{s}\frac{x'}{s'}\in\mathfrak(q)R_{\mathfrak p}$. The comma is in error, and you are missing the $'$ character. – Thomas Andrews Nov 12 '16 at 20:10
  • Related: https://math.stackexchange.com/questions/1280885 – Watson Nov 12 '16 at 20:10
  • @Arthur The implicit assumption is that $\mathfrak p$ is prime, so $\mathfrak p$ doesn't have (relevant) zero-divisors, and the definition is good enough. – Thomas Andrews Nov 12 '16 at 20:12
  • @ThomasAnsrews ${(n,0)\mid n\in \Bbb Z}$ is a prime ideal of $\Bbb Z\times \Bbb Z$, even though there are zero divisors everywhere, both inside and outside the ideal, and you're allowed to localise at that ideal. – Arthur Nov 12 '16 at 20:14
  • How is the notation $\frac{x}{s}$ problematic in that case? @Arthur – Thomas Andrews Nov 12 '16 at 20:20
  • @ThomasAndrews That depends. If you see $\frac xs$ as a true fraction that may be simplified and expanded to give the same value, then it's not problematic, but if you see it as an ordered pair, then it is problematic. I guess I was a bit quick on the trigger. – Arthur Nov 12 '16 at 20:22
  • @Arthur $\frac{x}{s}$ is always an equivalence class of ordered pairs. The real risk is if you think of $R_{\mathfrak{p}}$ as somehow a subset of some "larger" ring of fractions, like a field of fractions. – Thomas Andrews Nov 12 '16 at 21:46

2 Answers2

1

Ok so I'm going to sketch the main steps of the proof. First let's set ourselves straight:

$$\mathfrak{q}R_{\mathfrak p}=\left\{\dfrac{x}{s}\bigg| x\in \mathfrak{q},\ s\notin \mathfrak{p}\right\}$$

is the ideal generated in $R_{\mathfrak p}$ by the image of $\mathfrak q \subset R$ by $\phi: R\rightarrow R_{\mathfrak p},\ x\mapsto \dfrac{x}{1}$. You actually need to prove that it is an ideal but it should be easy.

$\supseteq$: First let's prove it is prime. So you fix $\dfrac{x}{s},\ \dfrac{x'}{s'}\in R_{\mathfrak{p}}$ such that $\dfrac{xx'}{ss'}=\dfrac{y}{t}\in \mathfrak{q}R_{\mathfrak{p}}$, where $y\in \mathfrak{q}$ and $t\notin\mathfrak{p}$. Thus by the définition of localisation you get $\sigma\notin\mathfrak{p}$ such that: $$\sigma(yss' -txx')= 0$$ Thus you get $(\sigma t)xx'=\sigma yss'\in\mathfrak{q}$ but $\sigma t\notin \mathfrak{p}$ so $\notin\mathfrak{q}$ thus $xx'\in\mathfrak{q}$ thus $x\in \mathfrak{q}$ or $x'\in \mathfrak{q}$.

$\subseteq$: You take $\mathfrak Q$ prime in $R_{\mathfrak p}$ and consider $$\mathfrak{q}=\phi^{-1}(\mathfrak{Q})$$ By construction you know that it is prime.

In fact what we need to do, is prove that these constructions are reciprocal one to another.

Thus you need to prove $(1)$: $$\phi^{-1}(\mathfrak{q} R_{\mathfrak p})=\mathfrak q$$

and $(2)$: $$\phi^{-1}(\mathfrak{Q})R_{\mathfrak p}=\mathfrak Q$$

You should be able to handle this easily, much like what I have done in $\supseteq$. Tell me if you need more help.

Andrei.B
  • 1,035
0

There is a canonical homomorphism $\varphi:R\rightarrow R_{\mathfrak{p}}$ by sending $x\mapsto x/1$, prime ideals $\mathfrak{q}$ in $R_{\mathfrak{p}}$ go to prime ideals in $R$ under the preimage of $\varphi$. In symbols $\varphi^{-1}(\mathfrak{q})$ is a prime ideal in $R$. If $\mathfrak{q}':=\varphi(\mathfrak{q})$ is not contained in $\mathfrak{p}$ then there are elements in $\mathfrak{q}'$ inside $R-\mathfrak{p}$ these elements are units in $\mathfrak{q}$ so this is a contradiction. We must have $\mathfrak{q'}\subset \mathfrak{p}$.

mathma
  • 2,323