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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?

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I am confused: if $K$ is an algebraic number field and $f: K \rightarrow K$ is a function, what are your definitions of (i) analytic function and (ii) partial derivative in this context? – Pete L. Clark Dec 13 '10 at 8:06
K can be written with a basis $\{ 1, v, v^2, ..., v^{n-1} \}$, where $v$ solves $P(v)=0$. If we write both the dependent and independent variables as a linear combination in the span of this basis, then we can define partial derivatives by differentiating each of the dependent variable's components with respect to one of the independent variable's components. We can also define the whole derivative as per the usual limit definition - but in order for this limit to exist, there must be equations governing f's partial derivatives. This is the same exact situation as with the complex numbers. – chroma Dec 13 '10 at 17:39
@chroma: Except that you are working over $\mathbb{Q}$, not over $\mathbb{R}$, so how are you defining derivatives "as per usual"? That's the main difference between the algebraic number field and the complex number case. You don't have a pre-existing notion of derivatives over for functions $\mathbb{Q}\to\mathbb{Q}$. – Arturo Magidin Dec 13 '10 at 17:53
Hmm, you're right, I want R, not Q... What would I call fields constructed with the basis I mentioned and coefficients in R instead? I'll try editing the original post to reflect the terminology change. (And does an answer to my question exist then?) – chroma Dec 13 '10 at 18:03
@chroma: You would call them either $\mathbb{R}$ or $\mathbb{C}$, because those are the only two algebraic extensions of $\mathbb{R}$. You're not going to get anything new doing that, just either the trivial extension, or back to Cauchy-Riemann except writing complex numbers as $a + b\theta$ with $\theta$ some non-real complex number instead of as per usual. – Arturo Magidin Dec 13 '10 at 18:04

4 Answers 4

up vote 11 down vote accepted

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).

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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.)

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The link to the question/answer in question is… – Arturo Magidin Dec 13 '10 at 18:10

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.

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According to Wikipedia, the Jugendtraum wants to find what algebraic numbers are enough to construct all abelian extensions of a field. That wasn't what I was thinking about, if I understand you correctly. (See my reply to Clark above.) Also, you can't write the CR equ's as what you did, because that is a much weaker condition than them. If you define derivatives over quadratic fields the same as with complex numbers, and seek to find the appropriate 'CR' equations to guarantee their existence, the formula I gave is the one needed. (You can check it reduces correctly when $d=-1$.) – chroma Dec 13 '10 at 17:58
@chroma: you are being rude and dismissive to people who are only trying to help you and who are more knowledgeable than you. This is not an attitude conducive to receiving any further help. – Qiaochu Yuan Dec 13 '10 at 23:07
@chroma: asking for a Laplacian is the first minimum requirement for a working theory, since this exists in many situations where CR equations do not or require extra structure (such as graph theory). David is telling you that the (or "a", or "one") naive generalization gives up ellipticity so that the theory has no resemblance to complex analysis or harmonic function theory. There are known derivative-like structures used in number theory and commutative algebra, such as Kaehler differentials and the different but there is no obvious analogue of analysis related to these. – T.. Dec 14 '10 at 0:54
@Qiaochu Yuan: I don't understand. I didn't mean to communicate any sort of emotion or attitude. I definitely see that people here are more knowledgeable than I, and I understand that the main problem here is my own ignorance and subsequent lack of ability to express my mathematical ideas in this discussion. That I chose not to reference any of these metafacts, but instead (1) said I didn't think the Jug = my problem, (2) pointed out a factual inaccuracy, and (3) tried to illustrate what it is I'm working on exactly, is all because I wanted to keep on-topic, not to condescend. Apologies. – chroma Dec 14 '10 at 2:49
@T: I see. My original motivation wasn't so grandiose as to build a new kind of theory or analysis on number fields analogous to that of known calculus, but it's nice to know from others here that it has been looked into and found relatively impossible. (I also didn't realize David was trying to naively generalize CR; I am prone to taking people too literally, especially online.) – chroma Dec 14 '10 at 2:55

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Since links can expire it is always helpful to summarize the contents of the links in a short way. – Julian Kuelshammer Feb 16 '13 at 10:47

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