Algebraically closed fields of positive characteristic - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T09:27:14Z http://mathoverflow.net/feeds/question/3551 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic Algebraically closed fields of positive characteristic Harrison Brown 2009-10-31T15:01:36Z 2010-03-31T02:05:12Z <p>I'm taking introductory algebraic geometry this term, so a lot of the theorems we see in class start with "Let k be an algebraically closed field." One of the things that's annoyed me is that as far as intuition goes, I might as well add "...of characteristic 0" at the end of that.</p> <p>I know the complex numbers from kindergarten algebra, so I have a fairly good idea of how at least one algebraically closed field of characteristic 0 looks and feels. And while I don't have nearly the same handle on the field of algebraic numbers, I can pretty much do arithmetic in it, so that's two examples.</p> <p>But I've never (<em>really</em>) seen an algebraically closed field of characteristic p > 0! I can build one just fine, and if you put a gun to my head I could probably even do some arithmetic in it, but there's no intuition of the sort that you get with the complex numbers. So: does anyone know of an intuitive description of such a field, that it's possible to get a real sense of in the same way as C?</p> http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/3555#3555 Answer by John Goodrick for Algebraically closed fields of positive characteristic John Goodrick 2009-10-31T15:44:59Z 2009-10-31T15:44:59Z <p>I doubt I can give an "intuitive" description of such a field, but I hope this is useful anyhow:</p> <p><em>Any two algebraically closed fields of the same characteristic and same transcedence degree (over their prime subfields) are isomorphic.</em></p> <p>So, in some sense, there really aren't too many examples of the kind of fields you're looking for.</p> <p>(This is one of those facts that model-theorists seem to like more than geometers, even though it has a very nice proof, IMHO. It can be proved in essentially the same way that you prove any two vector spaces of the same dimension over the same field are isomorphic, replacing spans by algebraic closures and "linearly independent" by "algebraically independent." The key point is that after making this translation, the Steinitz Exchange Lemma works just fine inside of any algebracially closed field, of any characteristic.)</p> http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/3564#3564 Answer by Ben Webster for Algebraically closed fields of positive characteristic Ben Webster 2009-10-31T16:44:48Z 2009-10-31T16:44:48Z <p>So first of all, there's only one algebraically closed field of characteristic p that you should think about for the sake of intuition, the algebraic closure K of F_p. And that's very easy, it's just the limit of the finite fields of order p^n as n goes to infinity.</p> <p>In fact, almost any construction involving K can be done in a sufficiently large finite field. For example, a statement like "every variety over <code>F_p</code> has a point over K" is equivalent to "every variety over <code>F_p</code> has a point over a sufficiently large finite field."</p> <p>Also, the multiplicative group of K is easy to understand: it's just the direct limit over all numbers n coprime to p of Z/nZ, where Z/nZ -> Z/mZ if n|m by 1 -> m/n. Which is another way of saying that K is gotten from F_p by just adding all roots unity of order coprime to p.</p> http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/3568#3568 Answer by Sam Lichtenstein for Algebraically closed fields of positive characteristic Sam Lichtenstein 2009-10-31T17:01:28Z 2009-10-31T17:01:28Z <p>It's certainly not too hard to understand everything there is to understand about the algebraic closure of Fp. Perhaps the reason this is unsatisfying as an example for founding intuition is because it doesn't really have a nice topological structure; it lacks anything like a natural metric. So here's an attempt to explain why what is in some sense the next simplest example puts you in a better situation, intuition-wise.</p> <p>If you have some intuition about the p-adic numbers look and feel (for example, topologically), then you secretly have intuition for the t-adic topology on the complete local field K=Fp((1/t)). Now, as far as characteristic p fields go, this sort of puts you in the position of (in your parlance) a "preschooler" who knows about R but hasn't yet gotten to kindergarten to learn about C. Why is K like R? First, it is locally compact. Second, it is at least analogous to completing Fp(t), which is very much like Q with Fp[t] as the analogue of Z, at an "infinite" valuation, namely the degree or (1/t)-adic valuation, rather than a "finite" place like a prime polynomial in Fp[t]. (The (1/t)-adic valuation corresponds to the point at infinity on the projective line over F_p. Likewise, number theorists love to say, perhaps partly to annoy John Conway, that the real and complex absolute values on Q correspond to "archimedean primes" or "primes dividing infinity". This is actually a pretty lame analogy, though, since K=Fp((1/t)) looks a lot more like Fp((t)), say, than R or C looks like Q2.)</p> <p>Unfortunately there are two extra difficulties in the characteristic p case. First, upon passing to the algebraic closure L of K we lose completeness. Second, we make an infinite field extension, unlike the degree 2 extension C/R. Thus, while L is an algebraically closed field of characteristic p, it bears little resemblance to R. In fact, it's a lot more like an algebraically closed field of characteristic 0 that is a bit scarier (at least to me) than C, namely Cp, or what you get when you complete the algebraic closure of Qp with respect to the topology coming from the unique extension of the p-adic valuation. While this may seem bad, I think it's actually good, because one can really get a handle on some of the properties of Cp. [Note that as another answerer pointed out, Cp = C as a field, but not as a topological or valued field, which is really a more interesting structure to consider from the viewpoint of intuition anyway.]</p> <p>For example of some similarities, miraculously Cp turns out to still be algebraically closed, and I believe the same proof goes through for L above. Another property L and Cp share is that in addition to "geometric" field extensions K'/K obtained by considering function fields of plane curves over Fp, there are also "stupider" extensions coming from extending the coefficient field. This is like passing to unramified extensions of p-adic fields, where one ramps up the residue field. (In fact, it's exactly the same thing.) Both L and Cp are complete valued fields with residue field the algebraic closure of Fp. (But the valuation is NOT discrete; it takes values in Q.) There are some dangerous bends to watch out for topologically, however. Some cursory googling tells me that Cp is not locally compact, although it is topologically separable. </p> <p>In addition, positive characteristic inevitably brings along the problem of inseparable field extensions sitting side L. This is, of course, an aspect where L/K is unlike Cp/Qp. Notwithstanding such annoyances, I would argue that the picture sketched above actually does give an example of an algebraically closed field of characteristic p for which it is possible to have some real intuition. </p> http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/3584#3584 Answer by Qiaochu Yuan for Algebraically closed fields of positive characteristic Qiaochu Yuan 2009-10-31T19:46:14Z 2009-10-31T20:03:00Z <p>The algebraic closure of F<sub>p</sub> has one really important property, which is that its Galois group over F<sub>p</sub> is <strike>isomorphic to</strike> the profinite completion of Z and is topologically generated by the Frobenius map x -&gt; x^p. The fixed field of the subgroup nZ is precisely F<sub>p^n</sub>; in other words, there is exactly one copy of each finite field of characteristic p and they are nested according to divisibility. Any statement you can make about a finite number of elements in this field takes place, as others have mentioned, in a sufficiently large finite subfield.</p> <p>One interesting way to get a grip on the elements of the algebraic closure is to think of them as <strong>aperiodic necklaces.</strong> As explained in <a href="http://mathoverflow.net/questions/769/exhibit-an-explicit-bijection-between-irreducible-polynomials-over-finite-fields" rel="nofollow">this question I asked</a>, there is a (non-canonical) bijection between irreducible polynomials of degree n over F<sub>p</sub> and Lyndon words of length n on an alphabet with p letters. One should think of conjugating as analogous to cyclic rotation, and the choice of a particular representative of a necklace up to rotation as the choice of a particular representative of a point up to conjugation. This is the categorification of the fact that as dynamical systems, words on an alphabet of length p and the algebraic closure of F<sub>p</sub> have the same zeta function.</p> <p>(The upshot of all of this is that some people think there should be a proof of the rationality of the zeta function of a variety over F<sub>p</sub> using formal language theory, by finding a regular language that describes those points. Unfortunately, all known translations give non-regular languages.)</p> http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/8325#8325 Answer by S. Carnahan for Algebraically closed fields of positive characteristic S. Carnahan 2009-12-09T07:18:07Z 2009-12-09T07:18:07Z <p>The spherical completion (aka maximal completion) of $\overline{\mathbb{F}_p((t))}$ is an example of an algebraically closed field of characteristic p that hasn't been mentioned here yet. The "spherically complete" property means that any sequence of nested nonempty balls has nonempty intersection (if the radii converge to zero, this is just completeness). One nice property of this field is that you can describe the elements "explicitly": take any power series with rational powers of t and coefficients in $\overline{\mathbb{F}_p}$, such that the set of exponents with nonzero coefficients is a well-ordered subset of Q. (It is rather tricky to verify that this set is closed under multiplication)</p> <p>If you like nonarchimedean analysis, you might have a reason to use this field instead of Cp or an algebraic closure of Fp((t)).</p> http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/16550#16550 Answer by L Spice for Algebraically closed fields of positive characteristic L Spice 2010-02-26T19:46:34Z 2010-03-31T02:05:12Z <p>It is a little unfair to pass this off as an answer to your question, but it also seems too interesting to be buried as a comment. (I can say that because I am citing other people's work!)</p> <p>The comments following <a href="http://mathoverflow.net/questions/3551/algebraically-closed-fields-of-positive-characteristic/3564" rel="nofollow">Ben's answer</a> point out some nice senses in which ‘any’ calculation over the algebraic closure of $\mathbb F_p$ is really a calculation over a suitably large finite extension of $\mathbb F_p$, but I didn't see a precise statement in the comments. I think it is very much worthwhile to note that, not just algebraically-closed-positive-characteristic computations, but even algebraically-closed-characteristic-$0$ calculations can be viewed this way. For example, this is one way to prove that an injective, polynomial self-map of $\mathbb C^n$ is bijective.</p> <p>See <a href="http://arxiv.org/abs/arxiv:0903.0517" rel="nofollow">Serre's lovely article</a> and <a href="http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem" rel="nofollow">Tao's lovely summary of it</a> for more details on this point of view.</p>