Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Note. I have edited my question to make it more transparent, following some very good comments that I received. I am sorry if it is a bit long.

A local homomorphism of local rings $(A,\mathfrak{m})\stackrel{\varphi}{\longrightarrow}(B,\mathfrak{n})$ is called a scalar extension (terminology due to Hans Schoutens) if:

  • $\varphi(\mathfrak{m})B=\mathfrak{n}$, and
  • $\varphi$ is a flat extension.

Let's fix a field $K$ (algebraically closed, if you wish) and let $\mathscr{C}_K$ be the category of Noetherian local rings whose residue field is a subfield of $K$, with morphisms being local homomorphisms.

Question A. Is there a functorial way of producing scalar extensions with a prescribed residue field? More precisely, is it possible to define a functor $F_K:\mathscr{C}_K\rightarrow\mathscr{C}_K$ in such a way that for every $A\in\mathscr{C}_K$ the local ring $F_K(A)$ is a scalar extension of $A$ with residue field $K$?

Here are some things that I know about this question:

(1) Grothendieck proved that scalar extensions with prescribed residue field always exist:

Theorem. (EGA III, Proposition 10.3.1, page 20). Let $(A,\mathfrak{m})$ be Noetherian local ring with residue field $k$, and let $K$ be a field extension of $k$. Then there exists a scalar extension $(A,\mathfrak{m})\stackrel{\varphi}{\longrightarrow}(B,\mathfrak{n})$ from $A$ to a Noetherian local ring $B$, with the property that $B/\mathfrak{n}$ is $k$-isomorphic to $K$.

Grothendieck's construction of the desired scalar extension depends on various 'choices' that he makes in his proof, and hence, does not produce a unique answer. For this reason I think it is hopeless to get a functor there.

(2) Various mathematicians have used a different method to construct scalar extensions with prescribed residue field, which seems 'more hopeful' to be functorial. In [b] (pp. 776-777) Kunz calls a special case of this construction the constant field extension. A version of this construction in the equicharacteristic case appears in [a] (pp. 18-19, 6.3). A more detailed description of this method can be found in [c] (pp. 4-7) and in [d] (pp. 36-38). I describe it in the equicharacteristic case: Given a local ring $(A,\mathfrak{m},k)$ and a field extension $K$ of $k$, take a coefficient field $k\hookrightarrow\hat{A}$ and complete $\hat{A}\otimes_kK$ with respect to the ideal $\mathfrak{m}(\hat{A}\otimes_kK)$. This is your $F_K(A)$. It is easy to see that this $F_K(A)$ is an scalar extension of $A$ with residue field $K$. ($F_K(A)$ depends on the choice of a coefficient field of $\hat{A}$, but is unique up to isomorphism).

Question B. Is the $F_K(\:\cdot\:)$ that was just described a functor from $\mathscr{C}_K$ to $\mathscr{C}_K$? To clarify the question, if $A_1\stackrel{\psi}{\longrightarrow} A_2$ is a local homomorphism of Noetherian local rings in $\mathscr{C}_K$, then does $\psi$ extend to a local homomorphism $B_1:=F_K(A_1)\rightarrow B_2:=F_K(A_2)$?

I can see how the method described in [c] provides an affirmative answer in equicharacteristic $0$ to Question B (it comes down to the fact that in equicharacteristic $0$ every maximal subfield of a complete local ring is a coefficient field) but I don't see how the method of [c] would still work in equicharacteristic $p>0$. I haven't checked the mixed characteristic case, yet, because I thought the equicharacteristic case is easier and if it cannot be settled positively, then there is even less hope for the mixed characteristic.

References.

a. M. Hochster and C. Huneke, $F$-regularity, test elements, and smooth base change, Trans. Amer. Math. Soc., 346 (1994).

b. E. Kunz, Characterizations of regular local rings of characteristic $p$, Amer. Jour. of Math., 41 (1969).

c. H. Schoutens, Classifying singularities up to analytic extensions of scalars, Ann. of Pure and Applied Logic, 162, (2011) (also available on the Arxiv, here).

d. H. Schoutens, The use of ultraproducts in commutative algebra, Lecture Notes in Mathematics, 1999, Speringer (2010).

share|improve this question
2  
The answer to the detailed question is no, take $A_1=A_2=B_2=\mathbf Z_{(p)}$, $B_1=\mathbf Z_p$, $k_1=k_2=K=\mathbf F_p$. –  user2035 Aug 9 '11 at 8:47
1  
The existence of a local homomorphism $\hat A_1^K\to\hat A_2^K$ is a special case of the weak universal property of Schoutens's completion of $A_1$ along $K$. Regarding the functor question: which category should be the domain? –  user2035 Aug 9 '11 at 14:29
1  
I don't see how you can restrict attention to a particular constructino of scalar extensions. E.g. suppose given $A_1$, and let $A_2$ be a randomly constructed scalar extension with residue field $k_2 := K$. Then surely $B_2 = A_2$, and so your are looking for a map $B_1 \to A_2$, i.e. a map from your (or Schouten's) particular scalar extension to my randomly constructed one. This more or less means (I think) that you are asking for Cohen rings to be functorial, which they are not (as far as I know). –  Emerton Aug 9 '11 at 15:31
1  
In what follows, I am referring to the preprint version (v2): You are right, there seems to be a problem with this part of the paper. As in Eisenbud Exercise 7.17b take a complete local noetherian ring $S$ containing a maximal subfield $\kappa$ which is not a coefficient field. Let $\lambda$ be the residue field of $S$ and take $R=\kappa$ for simplicity. Then, $\kappa\to S$ cannot be extended to $\lambda$ since $\kappa$ was assumed maximal. –  user2035 Aug 9 '11 at 18:53
1  
Another problem: Take an arbitrary field $\kappa$ and let $R=\kappa$. The author's construction yields $\hat R^\lambda=\lambda$, but in the case $S=\lambda$ there are several non-isomorphic $\kappa$-homomorphisms $\lambda\to\lambda$ in general. –  user2035 Aug 9 '11 at 18:53
show 15 more comments

1 Answer

up vote 4 down vote accepted

I post this as an answer since it is too long, actually answers Question B and sheds some light on Question A. The example is taken from Eisenbud, Commutative Algebra, Exercise 7.17b.

Let $A_1=\mathbf F_p(t)$, $A_2=\mathbf F_p(u)[[x]]$, $\psi\colon A_1\to A_2$, $t\mapsto u^p+x$. On the residue fields, $\psi$ induces an isomorphism $\mathbf F_p(t)\cong\mathbf F_p(u^p)$. If $K$ is any extension field of $\mathbf F_p(u)$, the $F_K$ in Question B has $F_K(A_1)=K$ and $F_K(A_2)=K[[x]]$ with the obvious homomorphisms $A\to F_K(A)$. However, the diagram $$\begin{array}{ccc} \mathbf F_p(t) & \xrightarrow{\psi} & \mathbf F_p(u)[[x]] \\ \downarrow && \downarrow \\ K && K[[x]]\end{array}$$ cannot be completed since $t$ becomes the $p$-th power $u^p$ in $K$, whereas it is mapped to $u^p+x$ in $K[[x]]$ which is not a $p$-th power.

As for Question A, depending on the exact interpretation, any $F_K$ should satisfy $F_K(k)=K$ for subfields $k\subset K$, so this example shows that $k[[x]]\to F_K(k[[x]])$ cannot just be the canonical homomorphism $k[[x]]\to K[[x]]$ in general.

EDIT: The above assumes, contrary to what you said in the comments, that the embedding of the residue field into $K$ is part of the data in $\mathscr C_K$. Otherwise, you can still apply the same argument if you assume $K$ algebraically closed (so that the image of $t$ is still a $p$-th power). However, I think that in this case even the simpler problem of choosing a natural embedding into $K$ for all fields in $\mathscr C_K$ is already impossible except in trivial cases.

share|improve this answer
    
Nice (counter-)example! I am convinced that (as noted by a-fortiori) even if we allow choice of embedding of the residue field into $K$, it may still be impossible to extend the homomorphism. Big thank you to a-fortiori and Emerton for taking their time to discuss this problem with me. –  Mahdi Majidi-Zolbanin Aug 11 '11 at 18:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.