I've mentioned several times in this 'blog my joint paper with Jean-Louis Colliot-Thélène on Del Pezzo surfaces over fields of cohomological dimension one, paper which I presented in Besançon. I would now like to try to explain something of what it is about for the benefit of non-experts. Unfortunately, I won't make my explanations accessible to all laymen, because that would either take too much space or require to remain at a level of generality such that it wouldn't be interesting to anyone; but I have plans for writing a general introduction to arithmetical geometry, at some point, that would be understandable by everyone and his dog. For the moment, I'll assume that the reader has undergraduate (math major) knowledge of algebra, including knowing what a field is (though I'll recall it in a minute), and also a polynomial in several indeterminates and a vector space. I guess that excludes a lot of people, but at least I'm not assuming that you know what an étale cohomology group (and, therefore, a field of cohomological dimension one) is. Anyway, here we go.
A (commutative) field (I recall just to make sure everyone agrees, but I assume the reader knows this already) is a domain E with two operations, addition and multiplication, so that addition puts an abelian group structure on E (in other words, addition is associative and commutative and has a neutral element written 0 (zero), and for every element x of E there exists an element written -x and called the negative of x, such that x+(-x) = (-x)+x = 0), multiplication distributes over addition, multiplication is associative and commutative and has a neutral element written 1 (one), and for every element x of E other than 0 there exists an element x-1 such that x·x-1 = x-1·x = 1. Examples of fields are the set R of real numbers or that, C, of complex numbers; or the set Q of rational numbers, or the set R(t) of rational functions in one indeterminate t with real coefficients; we also have the set Z/pZ of integer (residues) modulo a prime number p, or, for those who know what that is, the set of p-adics Qp.
If we take this definition of a (commutative) field and remove just the assumption that multiplication is commutative, leaving everything else as it is, we get the definition of a division algebra, or skew-field (some people just say “field”). The simplest example is probably the skew-field of quaternions, H: it is obtained by adjoining three symbols, i, j and k to the real numbers subject to the relations that i² = j² = k² = ijk = -1 (this implies, for example, that ij = k whereas ji = -k). It is a famous theorem of Wedderburn that every finite division algebra (in the sense that it has only finitely many elements) is, in fact, commutative (in other words, is a field). In any division algebra D, the set of elements x which commute with every other element y, that is, so that xy=yx for every y, forms a commutative field (a subfield of D) called the center of D (for example, the center of H is just R).
Recall that a field E is said to be algebraically closed iff every non-constant polynomial equation (with coefficients in E) in one indeterminate has a solution. For example, the equation x²-2=0 (in the indeterminate x) should have a solution (“the square root of two” up to a sign) in any algebraically closed field: this shows that the field Q of rational numbers is not algebraically closed, and neither is the field R of real numbers (because the equation x²+1=0 has no root in R, that is, -1 has no real square root); on the other hand, the field C of complex numbers is algebraically closed, a theorem due to Gauß (based on preliminary work by D'Alembert) and often referred to as the “fundamental theorem of algebra”.
A mere matter of notation: rather than considering polynomial equations in one indeterminate such as x²-2=0, algebraic geometers prefer to consider polynomial equations in two indeterminates which are homogeneous, meaning that the total degree (sum of the degree in each indeterminate) of each monomial es the same, such as X²-2T²=0: this can be arranged by the procedure (homogenizing the equation) consisting of adding a new indeterminate (here T) and multiplying each monomial by the appropriate power of it so that the total degree is the same everywhere. Now we look for non-trivial solutions of the homogeneous equation, that is, solutions in which all indeterminates are not simultaneously zero: it will readily be seen that (X,T) is a solution to the homogeneous equation X²-2T²=0 if and only if the ratio x=X/T is a solution of the initial equation x²-2=0. So it is the same to ask that every homogeneous polynomial equation in two indeterminates of (strictly) positive degree have a non-trivial solution as to ask that every non-constant polynomial equation in one indeterminate has a solution: both mean that the working field is algebraically closed. Actually it turns out that the restriction on the number of variables is irrelevant: over an algebraically closed field, any (one) homogeneous polynomial equation of (strictly) positive degree has a non-trivial solution so long as there are at least two indeterminates (and any (one) non-constant polynomial equation has a solution as long as there is at least one indeterminate). In fact, we can do better: over an algebraically closed field, if we consider at least m homogeneous equations of (strictly) positive degrees in n indeterminates, then so long as n-m is at least 1, they have a common non-trivial zero.
Homogeneous polynomials of degree two are called quadratic forms, and of degree three, cubic forms. When they have a non-trivial zero (in a given field) they are called isotropic, otherwise anisotropic. So the above paragraph tells us, in particular, that a form (quadratic, cubic or whatever) in at least two variables over an algebraically closed field is always isotropic. For quadratic forms, this is not true over the field of real numbers: in fact, there are quadratic forms with arbitrarily many variables over the real numbers that are not isotropic, for example X²+Y²+Z²+T² cannot be zero over the reals unless all of X, Y, Z and T are zero (the trivial solution), for reasons of positivity; on the other hand, any cubic form over the reals in at least two variables is always isotropic.
Now let us come to the definition of fields of cohomological dimension one: we say that E (a (commutative) field) is of cohomological dimension at most one iff every division algebra D which contains E in its center (in the sense that its addition and multiplication extend those of E and that any element of D commutes with any element of E) and is of finite dimension as a vector space over E, is actually commutative (is a field). The requirement that D be of finite dimension over E means that there exist a finite number of elements, x1 through xn of D such that every element of D can be written as t1x1 + … + tnxn for certain elements t1 through tn of E. For example, the division algebra H of quaternions contains the field R of real numbers in its center (it is its center), and H is a vector space over R of dimension four (with basis 1, i, j, k): but since H is not commutative, we know that R is not of cohomological dimension at most one (actually, it is of cohomological dimension infinity, if you want to know). On the other hand, any algebraically closed field is of cohomological dimension at most one; the proof, with some details omitted, goes like this: if D is a division algebra whose center contains a field E and is of finite dimension over it, then any element x of D must satisfy a polynomial equation over E (otherwise the powers of x would span an infinite-dimensional vector space over E in D and this cannot be) — but if E is algebraically closed this means that x is in E, and since this is true for every x we conclude that D is E, therefore D is, indeed, a commutative field (namely E). Furthermore, the result of Wedderburn which has already been mentioned (viz., all finite division algebras are commutative) means essentially that finite fields are of cohomological dimension at most one (because any finite dimensional division algebra over a finite field is itself finite, so Frobenius's theorem states that it is a (commutative) field).
We shall say that the fields of cohomological dimension zero are precisely the algebraically closed fields. The fields of cohomological dimension at most one (as defined in the previous paragraph) which are not of cohomological dimension zero will be said to have cohomological dimension (exactly) one. So the finite fields are of cohomological dimension (exactly) one.
Following Lang, we can define the C1 fields: a field E is said to be C1 iff every homogeneous polynomial equation over E of degree d (with d>0) in n indeterminates, such that d<n, has a non-trivial solution. (More generally, Cr fields can be defined by replacing the inequality d<n by dr<n.) For example, over a C1 field, a quadratic form in at least three variables is always isotropic, and a cubic form in at least four variables is always isotropic, and so on; furthermore, it can be proved that if we take two quadratic forms (note that 2+2=4) over a C1 field in at least five variables then they have a common non-trivial zero. A theorem by Warning and Chevalley states (essentially) that finite fields are C1 fields. And it can be shown that all C1 fields are of cohomological dimension at most one (the C0 fields are precisely the algebraically closed fields, and quite generally a Cr field is of cohomological dimension at most r).
Over a field E of cohomological dimension one (just as over any C1 field by the very definition), any quadratic form in three variables has a non-trivial zero. I won't give the full proof of that, but the basic idea is to write the quadratic form in the form aX²+bY²-Z²=0 and to note that if this has no zero then we can construct a quaternion(-like) algebra over E by adding three symbols i, j and k with the relations that i²=a, j²=b, and k²=ijk=-1: this defines a division algebra so long as the quadratic form is anisotropic. (At least this proof works when the characteristic of the field is not two, that is, when 2 is not equal to 0 in E. Otherwise the equation for the quadratic form is a bit different.)
To summarize:
- All C1 fields are of cohomological dimension at most one.
- Over a C1 field, a quadratic form in three variables has a non-trivial zero; or a cubic form in four variables; or two quadratic forms in five variables (have a common zero).
- Over a field of cohomological dimension at most one, a quadratic form in three variables has a non-trivial zero.
The natural question to ask is then whether a cubic form in four variables over a field of cohomological dimension (at most) one is necessarily isotropic, or whether two quadratic forms in five variables necessarily have a common non-trivial zero over such a field. And the answer is no: this is essentially what the joint paper proves.
A Del Pezzo surface of degree three is exactly a cubic
surface, that is, the surface defined in projective space by a cubic
form in four variable—if you don't know what projective space
is, just take a polynomial equation in three variables with degree at
most three and you'll miss only the points at infinity. A Del
Pezzo surface of degree four is the complete intersection of two
quadrics in projective four-space, that is, the locus defined by the
cancellation of two non-proportional quadratic forms. This is why the
result is phrased by geometers as: there exist (smooth) Del Pezzo
surfaces of degrees three and four over fields of cohomological
dimension one having no rational points
. I wish I could give a
general definition of Del Pezzo surfaces of all degrees that is more
or less understandable, but I wasn't able to find one. This note is
already long enough as it is, anyway.