Generalizing Cauchy-Riemann Equations to Arbitrary Algebraic Fields Can it be done?
For an arbitrary quadratic field $Q[\sqrt{d}]$, it's easy to show the equations are simply $ f_x = -\sqrt{d} f_y $, where $ f : Q[\sqrt{d}] \to Q[\sqrt{d}]$. I'm working on the case of $Q[\theta]$, when $\theta$ is a root of $\theta^3 - a\theta - b$, but I'm not sure if it's even possible. Has there been any mathematical research done on this topic? What do you think about it?
 A: To answer the three literal questions asked:
A wise man said once, "define PDE over a number field and you will be a rich man".  This was meant not as research advice but to indicate that everyone would love to have such a thing, many have tried to create it, and obvious approaches don't work.
(p.s.  if you search this site for "arithmetic Kodaira-Spencer class" you will find an earlier discussion with a pointer from Matt E to a paper by Faltings on the nonexistence of (any linear version of) such an object over number fields.  In other words, the elusiveness of arithmetic differentiation is not just a sociological observation.)
A: I don't know if this will help, but I thought about something like this when I was an undergrad. I was thinking about the Jugendtraum: the fact that abelian extensions of imaginary quadratic fields can be described by values of analytic functions on $\mathbb{C}$. My thought was the following: Let $K=\mathbb{Q}(\sqrt{-D})$ be a quadratic imaginary field. Then $\mathbb{C} \cong K \otimes \mathbb{R}$ and we can write the Cauchy-Riemmann equations as $(D \partial_x^2 + \partial_y^2) f=0$, which seems to be built from the norm form $K \to \mathbb{Q}$.
Therefore (I thought), if we want to generalize the Jugendtraum to a real quadratic field $\mathbb{Q}(\sqrt{D})$, we should consider functions on $K \otimes \mathbb{R}$ which obey $(D \partial_x^2 - \partial_y^2) f=0$. 
Well, this didn't get anywhere. But I did show that a function $f: \mathbb{R}^2 \to \mathbb{R}$ obeying $(D \partial_x^2 - \partial_y^2) f=0$ is of the form $g(x+\sqrt{D}y) + h(x-\sqrt{D}y)$. So, if that's the road you're gong down, I can tell you where it ends.
A: If you want to take derivatives in a rather general context, you can.  For instance, let 
$K$ be any topological field.  Then for any function $f: K \rightarrow K$ and any 
$x \in K$, we say the derivative of $f$ exists at $x$ if the usual limit 
$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ 
exists.  So if your topology on $\mathbb{Q}$ is the usual Archimedean one coming from the restricting the Euclidean metric on $\mathbb{R}$, you can speak of continuous and differentiable functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$.  
However these functions lack most of the nice properties of the corresponding functions on $\mathbb{R}$ or $\mathbb{C}$, due to the lack of completeness of $\mathbb{Q}$.  That is, a continuous (and even differentiable) function on a closed 
interval in $\mathbb{Q}$ need not satisfy the intermediate value property, need not be bounded, if it is bounded need not assume a maximum or minimum value, and need not be uniformly continuous, need not satisfy the Mean Value Theorem or Taylor's Theorem, and so forth.  So it is fair to ask why one would want to study differentiable functions on $\mathbb{Q}$.
(I should say that it's not completely clear that there is no good answer to this.  For instance, in the case of the $p$-adic field $\mathbb{Q}_p$, it is not so common to speak of or study differentiable functions.  However, there is a nontrivial theory here, as I learned from Alain Robert's book on $p$-adic analysis.  While it is not as essential as in the real or complex case, it has definitely been studied and written about.)
The issue of defining partial derivatives over number fields is a bit more subtle, and here I think there are problems that the OP has yet to appreciate.  Think about the Cauchy-Riemann equations on $\mathbb{C}$: Step 0 here is identifying $\mathbb{C}$ with $\mathbb{R}^2$ and a function $f: \mathbb{C} \rightarrow \mathbb{C}$ as a function 
$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, i.e., a real function in several variables.
More generally, let $(K,| \ |)$ be a normed field, i.e., $| \ |: K \rightarrow \mathbb{R}^{\geq 0}$ is such that $\rho(x,y) := |x-y|$ is gives a metric on $K$ with the additional property that $|xy| = |x||y|$ for all $x,y \in K$.  And let $V$ be a finite-dimensional normed $K$-vector space, say of dimension $n$  Then a basic but perhaps underappreciated result is that the completeness of $K$ is essential to make an identification of normed $K$-linear spaces $V \cong K^n$.  So, for example, the Euclidean norm $| \ |$ on $\mathbb{C}$ is equivalent to any product metric on $\mathbb{R}^2$.
This property does not hold in general when we extend the Archimedean norm $| \ |$ on $\mathbb{Q}$ to an arbitrary number field $K$.  In fact, things work out okay exactly if $K = \mathbb{Q}$ or $K$ is an imaginary quadratic field.  So let's look at the next 
simplest case, that of a real quadratic field $K = \mathbb{Q}(\sqrt{D})$.  In this case 
there are two norms on $K$ extending the Euclidean norm, say $| \ |_1$ and 
$| \ |_2$ corresponding to the two different embeddings of $K$ into $\mathbb{R}$.  (So, for instance, if $|\sqrt{D}|_1$ is the positive square root of $D$, $|\sqrt{D}|_2$ is 
the negative square root of $D$.)  Neither $(K,| \ |_1)$ nor $(K,| \ |_2)$ is equivalent, as a normed $\mathbb{Q}$-vector space, to $\mathbb{Q}^2$ with the product norm.  Indeed, here is an even stronger statement: consider the (unique) embedding $\iota$ of $\mathbb{Q}$ into $K$.  Then, with respect to the topology induced by either $| \ |_1$ or $|\ |_2$ $\iota(\mathbb{Q})$ is dense, since indeed both are dense in their completions, which are isomorphic to $\mathbb{R}$.  On the other hand, the embedding of $\mathbb{Q}$ into $\mathbb{Q}^2$ via the diagonal, $x \mapsto (x,x)$, has closed image.  
So the very idea of partial derivatives here makes me nervous.  An upshot of the above discussion is that choosing the basis $\{1,\sqrt{D} \}$ for $\mathbb{Q}(\sqrt{D})$ over $\mathbb{Q}$, the ``directions'' $1$ and $\sqrt{D}$ are not metrically/topologically independent, even though they are independent in the sense of linear algebra.
A final remark to make is that, to a number theorist like myself, it is very unnatural to choose a particular Archimedean norm $| \ |$ on a number field $K$.  Rather, there is a finite set of equivalence classes of such norms ("Archimedean places") which can be determined by looking at the factorization of any polynomial $P(t) \in \mathbb{Q}[t]$ 
such that $K \cong \mathbb{Q}[t]/(P)$ over $\mathbb{R}$: if $P$ has $r$ real roots and 
$s$ complex-conjugate pairs of complex roots, then there are $r + s$ Archimedean places of $K$, and one needs to work with all of them at once in order to do topologically useful things.  In particular, the natural embedding here is really from $K$ into $K \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{R}^r \oplus \mathbb{C}^s$.  Note that this latter object is a field in exactly two cases: when $(r,s) = (1,0)$ (i.e., $K = \mathbb{Q})$ or when $(r,s) = (0,1)$ (i.e., $K$ is an imaginary quadratic field).
