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There are a bunch of theorems in algebra that require the underlying field to be characteristic 0. I seem to remember that these all stemmed from one basic fundamental theorem that only holds in characteristic 0 fields, but I can't remember what it was. Is there a basic fact that doesn't hold in fields of positive characteristic that is the base reason that more advanced theorems require characteristic 0, and if so, what is it?

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By the way: the standard term to refer to "characteristic nonzero" is positive characteristic. –  Arturo Magidin Jun 21 '11 at 6:29
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Here are three phenomena specific to the positive characteristic case which I have found to lead to many of the differences from characteristic zero.

1: Let $k$ be a field, let $k[t]$ be the univariate polynomial ring over that field, and consider the derivative $\frac{d}{dt}$, i.e., the unique $k$-derivation on $k[t]$ which maps $t$ to $1$. The derivative is in particular a $k$-linear map, so it has a kernel. In characteristic zero the kernel is just the subring $k$ of constant polynomials, a very small and well-understood subspace. But in characteristic $p > 0$ the kernel consists of $k[t^p]$, i.e., all polynomials of the form $f(t^p)$, which is a big, infinite dimensional subring abstractly isomorphic to the original ring. Therefore having zero differential means something subtly, but profoundly, different in positive characteristic. Both separability of field extensions and (more generally) etaleness of maps can be characterized in terms of differentials, so this leads to a lot of the well-known differences. But it comes up in other places too: for instance, the theory of Weierstrass points on algebraic curves in positive characteristic is totally different and apparently less satisfactory: for a curve in characteristic zero of genus $g \geq 2$, the Weierstrass points form a set of $g^3-g$ points. However in characteristic $p$ it is possible for every point to be a Weierstrass point! The problem it seems is that early on in the theory of Weierstrass points one needs to write down a Wronskian, i.e., some matrix of derivatives. This can go haywire in positive characteristic.

1bis: Similarly in positive characteristic there are nonlinear polynomials whose derivative is everywhere nonzero: e.g. the Artin-Schreier map $x^p - x$. Thus in characteristic $p$ the affine line is not simply connected. In general there are many more unramified coverings of affine varieties in characteristic $p$ than in characteristic zero.

2: In any ring of characteristic $p > 0$, the map $x \mapsto x^p$ is an endomorphism, the Frobenius map. This gives everything in positive characteristic an extra structure that is not present in characteristic zero: for instance it gives rise to "new" isogenies of abelian varieties, and so forth. Here is a very low brow exploitation of the Frobenius map on $R = \mathbb{Z}/p\mathbb{Z}$: clearly $1^p = 1$, and by the homomorphic property of Frobenius, $2^p = (1+1)^p = 1^p + 1^p = 2$. In general, if $x \in R$ is such that $x^p = x$, then $(x+1)^p = x^p + 1^p = x+1$. Thus $x^p = x$ for all $x$ in $R$: we have proved Fermat's Little theorem by induction on $x$!

3: There exist finite fields of characteristic $p$, whereas any field of characteristic $0$ must contain $\mathbb{Q}$ and thus be infinite. Working in a finite field can be strange, because any given nonzero univariate polynomial over a field vanishes at only finitely many points. When the field is infinite, this is a very small set, but over a finite field there are nonzero polynomials which vanish at every point. Another way to say this is that the map which associates to $f \in k[t]$ its function $x \mapsto f(x)$ is injective iff $k$ is infinite (it is also surjective iff $k$ is finite!). When $k$ is finite of cardinality $p$, the kernel is the ideal generated by $t^p-t$. There's that Artin-Schreier map again! These considerations actually lead swiftly to a proof of the Chevalley-Warning Theorem.

3bis: a more geometric way of expressing the above is that a field $k$ is infinite iff the set of $k$-rational points of affine $n$-space over $k$ is Zariski dense in the whole space. When this does not hold (i.e., when $k$ is finite), "general position" arguments can fail dramatically, so that often in big theorems finite fields need to be treated separately from infinite fields. Sometimes the simple structure of finite fields leads to alternate, easier proofs -- e.g. in the Primitive Element Theorem or the existence of separable splitting fields for central simple algebras -- but sometimes not: e.g. the proof of Noether Normalization is much nastier over a finite field, to the extent that some texts retreat to the case of an algebraically closed field (when what the standard proof needs is precisely that the field be infinite). Finally, Bjorn Poonen rather recently came up with a useful reparation for Bertini type theorems over finite fields: one cannot intersect with a hyperplane because there are not "enough" $k$-rational hyperplanes. Rather one needs to intersect with a hypersurface of relatively low degree.

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Thanks for the extensive reply. I think the first item is what I was looking for. This at least seems to be the reason that we can't look at all characteristics equally in the study of differential algebra, or even computer algebra (e.g., the squarefree factorization algorithm for characteristic 0 polynomials breaks in characteristic nonzero for exactly this reason). –  asmeurer Jun 22 '11 at 1:30
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"Is there a basic fact that doesn't hold in fields of positive characteristic that is the base reason that more advanced theorems require characteristic 0, and if so, what is it?"

In some sense, there can't be such a basic fact, at least if you interpret "basic" from the point of view of mathematical logic. Specifically, we have Robinson's principle, which says the following:

If a first-order sentence holds in every field of characteristic zero, then there exists a constant p such that the sentence holds for every field of characteristic larger than p

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There certainly can be such a basic fact because every finite extension of fields in char. 0 is separable and there are counterexamples in characteristic p for every prime number p. The question asked is not about mathematical logic. –  KCd Jun 21 '11 at 20:59
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So there is no single sentence in the language of fields which holds in a field iff it is perfect, or iff it has characterstic zero. Nevertheless the class of fields of given characteristic (or the class of perfect fields, or imperfect fields) is elementary: you just need the sentence to depend on the field. So I agree with K: the logical perspective does not seem so relevant here. –  Pete L. Clark Jun 21 '11 at 22:25
    
I wasn't referring to some logic definition of "basic". I guess it's impossible to ask the question without using some term that means something rigorously (and rather different from what I want) in some branch of mathematics. –  asmeurer Jun 22 '11 at 1:25
    
@asmeurer: to the best of my knowledge (and I know a bit about model theory, although some others here know much more) "basic" is not a technical term in mathematical logic: "elementary" is. Thus the above comments of myself and @KCd that this answer seems a bit out of left field... –  Pete L. Clark Jun 22 '11 at 7:13
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I'm not sure if there are really that many results that only hold in characteristic 0; usually assuming characteristic 0 is a sufficient, but not necessary, condition that replaces a more technical one. The more technical condition is usually about separability; any extension of characteristic 0 fields is separable, while in positive characteristic it is possible for an extension to be inseparable (the simplest example is $\mathbb{F}_p(\sqrt[p]{\,\,\,T\,\,})/\mathbb{F}_p(\,T\,\,)$).

In fact, I have a hunch as to the "basic fundamental theorem" you are thinking of: the primitive element theorem. Many times, I have seen this presented as a theorem about finite extensions of characteristic 0 fields, when in fact it holds for any finite and separable field extension.

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I don't know. I think I might be looking for something more basic. Why is it, fundamentally, that a positive characteristic extension can be inseparable? –  asmeurer Jun 21 '11 at 5:55
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@asmeurer: The reason is that the derivative $f'$ of an irreducible polynomial $f\in K[x]$ can have a common factor with $f$ when $K$ is of positive characteristic. This seems counterintuitive (if $f$ is irreducible, how can it have any common factors with a polynomial of lesser degree?), but it occurs when $f'=0$. For example, $x^p-T$ is an irreducible polynomial in $\mathbb{F}_p(T)[x]$, but its derivative is $px^{p-1}=0$ (because $p=0$ in characteristic $p$). –  Zev Chonoles Jun 21 '11 at 6:02
    
Ah, that's why the squarefree algorithm doesn't work in positive characteristic. –  asmeurer Jun 21 '11 at 6:20
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Zev: I would say the more basic theorem involved here is that all finite extensions of fields in characteristic 0 are separable, because separable extensions lead to Galois extensions. A field extension which is primitive may or may not lie in a Galois extension. –  KCd Jun 21 '11 at 21:01
    
@KCd: Aren't all extensions of characteristic 0 fields separable? Also I didn't really go into Galois theory (at least, I don't think the proof of the primitive element theorem requires it). But I agree the more basic fact is the separability, from which the primitive element theorem is a consequence. –  Zev Chonoles Jun 22 '11 at 2:42
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