# faithfully flat ring extensions where primes extend to primes

I am interested in unital ring homomorphisms (and classes thereof) $R \rightarrow S$ of commutative rings that have the following pair of properties:

• $S$ is faithfully flat as an $R$-module, and
• For any prime ideal $P$ of $R$, $PS$ is a prime ideal of $S$.

I know of the following examples:

• $R \rightarrow R[X]$, where $X$ is an indeterminate over $R$ (or any set of algebraically independent indeterminates over $R$)
• $R \rightarrow R[[X]]$, where $X$ is an analytic indeterminate over $R$ (or any finite set of such)
• If $R$ is a field, take any (unital) ring map $R \rightarrow S$ such that $S$ is an integral domain.
• If $R$ is a Noetherian valuation domain, the map $R \rightarrow \hat{R}$ will work.
• Whenever $g: R \rightarrow S$ is an example of this and $W$ is a multiplicatively closed subset of $S$ such that $m S_W \neq S_W$ for any maximal ideal $m$ of $R$, then the induced map $R \rightarrow S_W$ works as well. (Some other localizations will be fine too.)

Then there are cases that sometimes work. For instance, one might expect $R \rightarrow \hat{R}$ to work, where $(R,m)$ is an arbitrary Noetherian local ring and $\hat{R}$ is its $m$-adic completion. (As noted above, this works for valuation rings.) However, it only works when $R/p$ is analytically irreducible for every prime ideal $p$ of $R$. It fails, for instance, when $R$ is the localization of $k[x,y]$ at $(x,y)$ when $k$ is a field of characteristic other than $2$. (This is because the element $x^2-y^2-y^3$ is prime in $R$ but not in $\hat{R}$.)

Do people know of any other important examples?

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This question might be easier to consider geometrically.

You want the morphism Spec $S \to$ Spec $R$ to be faithfully flat, and the fibre over each point $\mathfrak p \in$ Spec $R$ (which equals Spec $S\otimes_R \kappa(\mathfrak p)$, where $\kappa(\mathfrak p)$ is the residue field of $R/\mathfrak p$) should be integral. (Actually, the question asks that $S/\mathfrak p S$ be integral, but since $S$ is flat over $R$ by assumption, the embedding $R/\mathfrak p \hookrightarrow \kappa(\mathfrak p)$ induces an embedding $S/\mathfrak p S \hookrightarrow S\otimes_R \kappa(\mathfrak p).$ Thus if the target is a domain, so is the source, and conversely, since the target is a localization of the source.)

So for example, any smooth morphism of affine varieties all of whose fibres are connected will give an example. (Since a smooth connected variety is irreducible.)

This will give an enormous class of examlpes.

E.g. as a slight variation on the preceding remark, let $k$ be an algebraically closed field, and let $A \subset B$ an inclusion of f.g. $k$-algebras that are domains, such that, if $K$ denotes the algebraic closure of Frac$(A)$, then $K\otimes_A B$ is a domain. (In other words, the generic fibre of the map Spec $B \to$ Spec $A$ is geometrically irreducible.) Then over some non-empty open subset of Spec $A$ the restriction of Spec $A \to$ Spec $B$ will be faithfully flat and will have geometrically irreducible fibres, and hence we can find $a \neq 0$ such that $A[1/a] \to B[1/a]$ will have the desired property.

A concrete example is given by the inclusion $\mathbb C[t] \subset \mathbb C[x,y,t]/(y^2 - x(x-1)(x-t)),$ but there is nothing particularly special about this example; it is just my favourite example of a flat family of irreducible curves.

Added: This is an example taken from the discussion in comments below. It is intended to illustrate why the above example is not particularly special or limited in scope.

Let $A$ be any f.g. $k$-algebra, and let $f(x_1,\ldots,x_n)$ be a polynomial whose coefficients are elements of $A$, i.e. $f \in A[x_1,\ldots,x_n]$. Now consider the ring $B = A[x_1,\ldots,x_n]/(f)$.

Geometrically, $f$ is the equation for a family of hypersurfaces, parameterized by Spec $A$.

Again, suppose that $A$ is a domain (so Spec $A$ is a variety), and let $a \in A$ be the discriminant of $f$. Unless you are very unlucky in your choice, $a$ will be non-zero. (Geometrically, this is saying that the family of hypersurfaces $f= 0$ doesn't have every member being singular. Of course it could, if you chose it that way, but if you choose it generically, then it won't.) So if we pass to $A[1/a]$ and $B[1/a]$, we have a family of smooth hypersurfaces. Provided that $n \geq 2$, a smooth hypersurface is irreducible, so $A[1/a] \subset B[1/a]$ will satisfy the conditions of the problem.

[In the example with cubic curves, I didn't invert the discriminant, because I happened to choose a flat family of curves which remain irreducible even when they become singular.]

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I don't understand. Are you saying that if Spec $S/pS$ is connected, then $S/pS$ is an integral domain? I thought connectedness just meant that the ring isn't a product of nontrivial subrings. – neilme Mar 21 '13 at 3:05
Also, isn't the usual definition of "fiber over a point $p$" then "Spec $S_p / p S_p$"? If so, that would explain your linking of connectedness with integrality. – neilme Mar 21 '13 at 3:08
@neilme: Dear neilme, I wrote the first version of the answer in some haste, but have corrected/amplified it. If you still have questions, let me know. Regards, – Matt E Mar 21 '13 at 3:21
@neilme: Dear neilme, By assumption $B$ is f.g. over $k$, hence f.g. over $A$. Write $B = A[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$. Then $K \otimes_A B = K[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$, and we are asking if $f_1,\ldots,f_r$ cut out an irreducible variety over $K$. This can be a complicated question if the $f_i$ are complicated equations, but it needn't be. E.g. suppose that $r = 1$, so that we are just looking at a single hypersurface $f = 0$ in $\mathbb A^n$. Then you are asking if this hypersurface is irreducible, which is pretty easy: it is just a question as to whether or not $f$ ... – Matt E Mar 21 '13 at 11:24
@neilme: Dear neilme, It will be the resultant of $f$ and all its partial derivatives, and will vanish exactly at the singular points of the variety $f= 0$. Regards, – Matt E Mar 21 '13 at 14:35