**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](http://arxiv.org/abs/math/0508236)).


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



On the functoriality of scalar extensions of local rings (edited)
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.


