# When is the radical of the extension of a prime ideal prime?

(All rings assumed to be commutative and unital)

Given a ring $R$, a prime ideal $\mathfrak{p}$ of $R$, and an extension ring $S$ (the algebra map $R\to S$ is injective), are there any nontrivial sufficient conditions for $rad(\mathfrak{p}S)$ to be prime? How about when $S$ is integral over $R$? How about when it is finitely presented as an $R$-module?

EDIT: Qing Liu's comment means that I should be even more specific.

Assume the third question above ($S$ is finitely presented as an $R$-module.).

Suppose further that there exists an element $f \in R$ such that $R_f$=$S_f$. Then even further, suppose $f \in \mathfrak{p}$.

EDIT 2: It is now open season for adding any extra conditions on $S$ that you want. Please don't add stupid conditions like S=R. Note that the condition about $f$ is actually a condition on $S$.

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Think about the case $R=\mathbb Z$ and $S$ equal to the ring of integers in a finite extension of $\mathbb Q$. Hard to give a smart answer I think. – Qing Liu Feb 4 '10 at 22:12
Isn't this about (non-)splitting of a prime? So for $\mathbb{Z}$ and suggested $S$, isn't this non-trivial? – Dror Speiser Feb 4 '10 at 22:35
I voted up the previous two "answers", which didn't answer the question before I changed it. Had they answered the question sufficiently, I would have accepted one and started a new question. That is not to say that the answers are not good information, just that they didn't answer the question pre-or-post edit. – Harry Gindi Feb 4 '10 at 22:58
I voted to close, since I figured out another proof of the fact I was looking for that doesn't depend on this information. Just a remark as I go. This question didn't look nearly as unanswerable when it was a commutative algebra problem, but when it became an algebraic geometry problem, it suddenly started to look much harder. (Not because it's geometry, just because the translated problem just looks much harder). – Harry Gindi Feb 4 '10 at 23:37

Probably Harry has already thought of it like this, but rephrased in terms of algebraic geometry, this question is basically "under what conditions is the preimage of an irreducible variety under a map be irreducible."

When you phrase it like that, it sounds like you might be able to get a statement like "when all maximal prime ideals have prime (up to taking radical) extension" (I think this really does follow over an algebraically closed field by the Nullstellensatz, and some easy topology) but I don't think you'll really do any better than that. This at least gives you some geometric intuition (lots of maps, like smooth one with connected fibers satisfy this condition).

I have no idea how the condition that $f$ exists with $R_f=S_f$ is supposed to help. This says that the map is an isomorphism away from the 0 set of f, but $V(\mathfrak{p})$ is in that zero set, so you haven't gained any control.

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If you look at the situation geometrically, you have a morphism of affine schemes $Spec(S)\to Spec(R)$ and you are asking if the inverse image of the integral subscheme $V(\mathcal P) \subset Spec(R)$ is irreducible. My intuition is that this happens only rarely.

For example if $k$ is a field, consider the diagonal embedding $k \to k\times k$. You get an algebra which has all the properties you require:injectivity of structure map, integrality, finite presentation (after all it is a 2-dimensional vector space...) Yet the zero ideal of $k$ extends to the zero ideal in $k \times k$ , which is reduced but not prime.

Still, in order to end on an optimistic note, here is a positive result. If you take for $S$ the polynomial ring $R[X_1,\ldots ,X_n]$ over $R$, then the extension of any prime ideal of $R$ will remain prime. This corresponds geometrically to the product of the subscheme $V(\mathcal P)$ with affine space $\mathbb A^n_R$.

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Also, the obvious in between answer (between rare and always): $\mathbb{Z} -> \mathbb{Z} [\sqrt{D}]$ has half the prime ideals satisfy the condition. – Dror Speiser Feb 4 '10 at 22:45
Ah yes,interesting observation, Dror. – Georges Elencwajg Feb 4 '10 at 22:57

You might be looking for Zariski's Main Theorem (Hartshorne, III.11.4):

If $f: X \to Y$ is a birational projective morphism of noetherian integral schemes, and $Y$ is normal, then, for every $y \in Y$, the inverse image $f^{-1}(y)$ is connected.

This isn't what you asked for in two ways. (1) $f^{-1}(y)$ connected doesn't mean that the corresponding ideal is prime, it just rules out a specific way for that ideal to be nonprime. (2) There are almost no examples where $X$ is affine. Furthermore, if $U$ is an affine open in $X$, then $U \cap f^{-1}(y)$ need not be connected.

But this is a theorem which says that, under birational maps, and certain conditions, preimages of irreducible sets are something-like-irreducible.

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There is an algebraic form of ZMT, and precisely what I'm trying to prove is a limited converse. – Harry Gindi Feb 4 '10 at 23:13