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**/`p`**Z** of integer (residues) modulo a prime
number `p`, or, for those who know what that is, the set of
`p`-adics **Q**_{p}.

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
`i`² = `j`² = `k`² =
`i``j``k` = -1 (this implies, for example,
that `i``j` = `k` whereas
`j``i` = -`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
`x``y`=`y``x` 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 `x`²-2=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 `x`²+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 `x`²-2=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 `X`²-2`T`²=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 `X`²-2`T`²=0 if and only if
the ratio `x`=`X`/`T` is a solution of the
initial equation `x`²-2=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
`X`²+`Y`²+`Z`²+`T`² 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,
`x`_{1} through `x`_{n}
of `D` such that every element of `D` can be written
as `t`_{1}`x`_{1} + … +
`t`_{n}`x`_{n}
for certain elements `t`_{1} through
`t`_{n} 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 C_{1} fields: a field
`E` is said to be C_{1} 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, C_{r} fields can be defined by
replacing the inequality `d`<`n` by
`d`^{r}<`n`.) For example,
over a C_{1} 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 C_{1} 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 C_{1} fields. And it can
be shown that all C_{1} fields are of cohomological dimension
at most one (the C_{0} fields are precisely the algebraically
closed fields, and quite generally a C_{r} field is
of cohomological dimension at most `r`).

Over a field `E` of cohomological dimension one (just as
over any C_{1} 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
`a``X`²+`b``Y`²-`Z`²=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
`i`²=a, `j`²=b, and
`k`²=`i``j``k`=-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 C
_{1}fields are of cohomological dimension at most one. - Over a C
_{1}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.