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?
[Math] Why doesn’t stuff hold in characteristic non-zero
abstract-algebrafield-theory
Related Solutions
Let's suppose you care about finite groups, one way or another. Then you probably care about the classification of finite simple groups. The bulk of this classification is the groups of Lie type, which were discovered by finding analogues of Lie groups over finite fields.
Finite fields are also (as one might guess) very important in computer science. I'm certainly no expert, but here are some applications I know of:
- Cryptographic protocols like Diffie-Hellman have as their basis the simple fact that it is difficult to invert exponentiation in finite fields.
- The standard way that one factors polynomials over the integers is to apply something like Berlekamp's algorithm to factor them over several finite fields first, then combine the factorizations.
- The classic theorem that IP=PSPACE requires some work over finite fields.
- Elliptic curves over finite fields are used for elliptic curve cryptography.
- I have also been told that vector spaces and varieties over finite fields can be used to construct error-correcting codes. I don't know anything about this, but here is a book on the subject. For the special case of linear codes this leads to a beautiful analogy between lattice sphere packings and error-correcting codes which is described, for example, by Noam Elkies here.
Finally, even if you are only interested in varieties over $\mathbb{C}$ (say), if your variety happens to also be nice and defined over $\mathbb{Z}$ then it can be nice and defined over $\mathbb{F}_p$ for all but finitely many $p$ and you can use the Weil conjectures to compute its Betti numbers by counting. This is particularly easy to do for varieties with nice moduli interpretations like flag varieties.
Edit: You might also be interested in reading Serre's expository article How to use finite fields for problems concerning infinite fields, as well as Manin's Reflections on arithmetical physics. I got the latter link from an excellent answer to an MO question on mirror symmetry over finite fields.
Claim: Let $n \geq 2$ be a natural number and $F$ be a field. Then $F \to F, a \mapsto a^n$ is additive iff $F$ has positive characteristic $p$ and
- either $F$ is finite, say $|F|=p^e$, and $p^k \equiv n \bmod p^e-1$ for some $k$,
- or $F$ is infinite and $n$ is a power of $p$.
Proof. First we note that if $a \mapsto a^n$ is additive, it is actually a field homomorphism (since it is multiplicative anyway). It follows that if $F=\mathbb{F}_p$ for a prime $p$ the map is additive iff it is the identity iff $a^{n-1} = 1$ for all $a \in \mathbb{F}_p^*$ iff $p-1\mid n-1$, using that $\mathbb{F}_p^*$ is cyclic.
If more generally $F=\mathbb{F}_{p^e}$ is a finite field, the map is additive iff it equals one of the elements of $\mathrm{Gal}(\mathbb{F}_{p^e}/ \mathbb{F}_p)$. This group is cyclic, generated by the Frobenius. Hence, the condition becomes $\forall a \in F : a^n = a^{p^k}$ for some $k$. We may assume $a \in F^*$ here. Using that $F^*$ is cyclic, this reduces to $p^k \equiv n \bmod p^e -1$.
Now let $F$ be an infinite field, and consider the equation $(a+b)^n=a^n + b^n$ in $F$. It is trivial for $b=0$. If $b \neq 0$, we may as well assume $b=1$ by dividing $b^n$ on both sides. Thus, we reduce to $(a+1)^n = a^n+1$, which becomes $P|_F=0$ for $P(x)=\sum_{k=1}^{n-1} \binom{n}{k} x^k$. Since $F$ is infinite, this implies $P=0$, i.e. $\binom{n}{k}=0$ for all $0 < k < n$. In particular, $n=\binom{n}{1}=0$ in $F$, and we see that $F$ has positive characteristic $p$. But then these equations already hold in $\mathbb{F}_p$, and we get $\binom{n}{k} \equiv 0 \bmod p$ for $0<k<n$. Now it is well-known that $n$ is a power of $p$: Write $n = p^v m$ with $p \nmid m$. Since $a \mapsto a^{p^v}$ is injective and additive, it follows that $(a+b)^m=a^m+b^m$. By the same reasoning as above, it follows $\binom{m}{k} \equiv 0 \bmod p$ for all $0<k<m$. But since $\binom{m}{1}=m \not\equiv 0 \bmod p$, it follows that $m=1$. $\square$
The condition is in both cases that $n \bmod \exp(F^*)$ is a power of $p$. Here, we define the exponent $\exp(G)$ of a group to be the nonnegative generator of the ideal $\{z \in \mathbb{Z} : \forall g \in G : g^z=1\}$. If $F$ is a finite field, $\exp(F^*)=|F|-1$, and if $F$ is an infinite field, then $\exp(F^*)=0$.
Best Answer
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.