Localization and ultrapowers I'm finishing up my senior thesis on ultraproducts and am trying to replace some of my longer proofs with nicer ones.  
Let $P$ be a prime ideal of $R$, and $F$ an ultrafilter on a set $X$.  If $R_F$ is the ultrapower of $R$ with respect to $F$, it is straightforward to prove that $(R_P)_F \cong (R_F)_{P_F}$ as rings; this is done by constructing an explicit map and showing it is an isomorphism.  
Can anyone think of a way to give a shorter proof, perhaps analagous to the following module-theoretic argument?
If $I$ is an ideal of $R$, we have a SES of $R$-modules $$0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$$
which then induces another SES $$0 \rightarrow I_F \rightarrow R_F \rightarrow (R/I)_F \rightarrow 0$$ so we can conclude that $R_F/I_F \cong (R/I)_F$ (I know there are a few more details to this proof, I just wanted to give the basic idea).
Any help would be greatly appreciated.
 A: For a more conceptual proof, you might try showing that $(R_P)_F$ is a localization of $R_F$ at $P_F$, that is, that every homomorphism $f: R_F\to B$ such that $P_F$ is mapped to units, that is, $f(P_F)\subset B^\times$, factors through the natural map $i: R_F\to (R_P)_F$. Any two localizations, as usual in the case of universal properties, are isomorphic by a unique isomorphism. This still requires talking about $i$, but may let you avoid quite as much explicit fiddling with indices.
A: When I see ultrapowers, I tend to think along the lines of non-standard analysis; e.g. the transfer principle.
That is, the statement "$P$ is a prime ideal of $R$, and $S$ is the localization of $R$ at $P$" is true if and only if its transfer is true. So all that remains is to check that for internal nonstadnard sets $P$ and $R$, phrases like "$P$ is a *-prime ideal of $R$" (for nonstandard rings $P$ and $R$) are equivalent to "$P$ is a prime ideal of $R$".
As an example, the predicate "$R$ is a local ring" is equivalent to "$\forall x \in R: \exists y \in R: xy = 1_R \vee (1-x)y = 1_R$", which clearly remains unchanged after transfering, so for nonstandard rings, the internal and external notions of being local are identical.
For comparison, an example of something that isn't preserved by transfer is the property of being Noetherian. While every internal ascending chain of ideals in a *-Noetherian ring does eventually stabilize, they are indexed by ${}^\star \mathbb{N}$ rather than $\mathbb{N}$: the point at which they stabilize may be an (externally) infinite hyperinteger, and such a chain would violate the external ascending chain condition. Furthermore, a non-standard ring is likely to have ideals that are not internal, and 
the *-Noetherian property has nothing to say about chains of ideals involving those!
A: I doubt that there is an abstract nonsense proof. Although the ultraproduct has a universal property, this involves infinite products, which don't commute with localizations in general. Also, when I try to prove the isomorphism, I need additional assumptions, which I cannot get rid of (but perhaps someone else):

Let $F$ be a $\kappa$-complete ultrafilter on an infinite set $X$, $R$ a commutative ring, $M$ an $R$-module and $S \subseteq R$ a multiplicative subset with $|S|<\kappa$. Then $(S^{-1} M)_F \cong S^{-1} (M_F)$.

Proof: The universal map $M \to S^{-1} M$ induces $M_F \to (S^{-1} M)_F$. Since the right hand side is an $S^{-1} R$-module, it factors as $S^{-1} (M_F) \to (S^{-1} M)_F$. We will show that it is injective and surjective, hence an isomorphism:


*

*Let $[m]/t$ be in the kernel, where $[m]$ denotes the class of $m \in \prod_x M$ in the ultraproduct. Then $F$ contains $\{x \in X : \exists s \in S : s m_x = 0\} = \cup_{s \in S} \{x \in X : s m_x = 0\}$, so by $\kappa$-completeness we get that $F$ contains $\{x \in X : s m_x = 0\}$ for some $s \in S$, which means precisely that $s [m] = 0$ in $M_F$, and therefore $[m]/t = 0$ in $S^{-1} (M_F)$ as desired.

*Let $[r] \in (S^{-1} M)_F$ be given. Let $M'$ be the image of $M \to S^{-1} M$. Then $X = \cup_{s \in S} \{x \in X : s r_x \in M'\}$, hence by $\kappa$-completeness we have that $\{x \in X : s r_x \in M'\} \in F$ for some $s \in S$. For all these $x$ write $r_x = m_x/s$ for some $m_x \in M$. For all the other ones let $m_x=0$ or anything else. Then $[m]/s$ is a preimage.
