## Rational Varieties — An Introduction Through Quadrics

Projective spaces are the most basic algebraic varieties we know (at least in some sense) and rational varieties are those that are as close as possible to being projective spaces.  Informally, a rational variety is one admitting a parametrization by projective space.  The project of determining which varieties are rational (or not) has led to the development of a large amount of rich and beautiful theory – and many (seemingly straightforward) questions are still open.  On a historical note, the rationality probelm for quadric and cubic surfaces was settled “classically”, that is, over one hundred years ago.  The rationality of quadric hypersurfaces in general is also relatively easy to deal with, and occupies the bulk of this post.  It wasn’t until the 1970s that Clemens and Griffiths proved that smooth cubic threefolds are not rational by identifying a cohomological obstruction – a method involving the intermediate Jacobian.  Showing that a given variety is not rational is usually quite difficult; as an example, the problem of determining whether all smooth cubic fourfolds are rational or not is (I believe) still open.  There has also been much recent work in identifying other classes of varieties which are “close” to being rational – but that discussion will have to wait for another day.  The goal of this post will be more modest, I’ll discuss the definitions and some more basic examples of rational varieties.

Fix a ground field $k$ and then $X$ will always denote a variety defined over $k$.  It will be important to remember that this ground field need not be algebraically closed.  Even if one is only interested in studying varieties over algebraically closed fields (or even just over the complex numbers), non-closed fields quickly enter the picture – the main example being function fields.

Definition: A variety $X$ is rational if it is birationally equivalent to projective space.  To be more explict, $X$ is rational if there is a birational mormphism over $k$$\mathbb{P}^n_k \rightarrow X$.  Note : All rational maps should be marked with a dotted arrow, but I couldn’t get them to come up in the post, so please keep this in mind in the following!

Remark:  There are examples of varieties which are not rational but become rational after a field extension.  Thus it’s important to keep track of the field over which $X$ lives, and over which fields $X$ is rational.  To see an example of this…

Example:  Fix a prime number $p$ congruent to 3 mod 4.  Let $C$ be the curve defined over $k = \mathbb{Q}$ by the homogeneous equation $x^2 + y^2 = pz^2$ in $\mathbb{P}^2_k$.  Suppose that $C$ has a rational point over $k$, that is, a set of solutions in the rational numbers.  By clearing denominators we may assume that $(x:y:z)$ is a solution in the integers not all divisible by $p$.  If neither x nor y are divisible by p, considering the equation mod p, we see that $x^2 = -y^2 (\text{mod } p)$.  Over $\mathbb{F}_p$ then, this would give a solution to $u^2 = -1 (\text{mod } p)$ which is impossible because $p$ is congruent to 3 mod 4.  If x and y are divisible by p, then we quickly see that z must also be divisible by p, a contradiction.  Thus $C$ has no rational points.  If $C$ were rational, then the map $\mathbb{P}^1_k \rightarrow C$ would give plenty of rational points on $C$.  Thus $C$ is not rational over $\mathbb{Q}$.  However, if we consider $C$ defined over the bigger field $K = \mathbb{Q}( \sqrt{p})$, then $C$ is rational.  To see this, consider the explicit parametrization $t \rightarrow (t^2 - 1: 2t: (1 / \sqrt{p}) (t^2 + 1) )$.  In the language of algebraic geometry, we say that $C_K = C \times_{Spec(k)} Spec(K)$ is rational over $K$.  If a variety $X$ becomes rational after a base change to the algebraic closure of the field, then we say that $X$ is geometrically rational.

Over any field, a rational map from a curve to a smooth projective variety is already a morphism.  Thus rational curves must actually be isomorphic to $\mathbb{P}^1$.  It’s well known that any rational curve over an algebraically closed field can be embedded in $\mathbb{P}^2$ as a smooth conic, and indeed over any field we have:

Proposition:  If $C$ is a smooth, geometrically rational curve, then $C$ can be embedded in $\mathbb{P}^2$ as a smooth conic.

Proof:  The first thing to note is that over any field, it makes sense to speak of a canonical divisor $K_C$ on a smooth curve.  The second thing to note is that the dimension of the complete linear system associated to any divisor does not depend on the field over which the curve is defined.  Indeed for any coherent sheaf $F$ on $C$, and for any extension $K$ of $k$, we have $H^0(C_K, F_K) \cong H^0(X, F) \otimes K$.  This identity can be seen, for example, by computing using a Cech complex.  (Note that so far, all this would hold for any variety, not just a curve).  With these tools at hand then, the proof is pretty straightforward.  Over the algebraic closure of $k$, the sheaf $\mathcal{O}(-K_C) \cong \mathcal{O}(2)$ and so has three global sections.  So over $k$ also the sheaf $\mathcal{O}(-K_C)$ has three global sections and then one checks that the complete linear system associated to $-K_C$ embeds $C$ in $\mathbb{P}^2$ as a smooth conic.

Then rationality questions for curves are reduced to the study of plane conics.  To tell when a smooth conic in $\mathbb{P}^2$ is rational, we have…

Proposition:  For any geometrically rational curve $C$ the following conditions are equivalent:

1.  The curve $C$ is rational, that is, isomorphic to $\mathbb{P}^1$.

2.  The curve admits a $k$ point.

3.  The curve has a point defined over some odd degree extension of $k$.

4.  There is an odd degree line bundle on the curve defined over $k$.

Proof:  (1) clearly implies (2) and (2) clearly implies (3).  To see that (3) implies (4), suppose that $P$ is a point defined over $k'$ for an odd degree extension of $k$.  If $k'$ is separable over $k$ of degree d, then the point $P$ has d conjugates $P=P_1, \ldots, P_d$ under the action corresponding to the Galois group of $k \subset k'$.  Since the divisor $D = P_1 + \ldots + P_n$ is invariant under the Galois action, it must actually be defined over $k$.  The proof of that fact can be outlined as follows:  If $f$ is a function vanishing on $D$, then $\sigma \cdot f$ also vanishes on $D$ for any $\sigma$ in the Galois group (which acts on the coefficients of the polynomial).  By considering the elementary symmetric polynomials in $\sigma \cdot f$ (taken over all elements of the Galois group), one gets polynomials defined over $k$ which vanish exactly on $D$.  That fact we’ll leave as an exercise.  Then $\mathcal{O}(D)$ is a degree d line bundle defined over $k$.  In the inseparable case, we’ll leave as another exercise that if the degree of the extension in $d$, then the divisor $d \cdot P$ can be defined over $k$.  To see that (4) implies (1), let $L$ be a line bundle of degree 2r + 1 defined over $k$.  Then the line bundle $L \otimes \mathcal{O}(rK_C)$ has degree $1$, is defined over $k$, and its 2 global sections give the required isomorphism.

In fact, the case of degree 2 hypersurfaces in any projective space is similar:

Proposition: If $Q$ is a smooth quadric hypersurface in projective space, then the following conditions are equivalent:

1.  The quadric $Q$ is rational.

2.  The quadric $Q$ has a rational point over $k$.

3.  The quadric has a rational point over some odd degree field extension of $k$.

Proof:  That (1) implies (2) seems easy, because one would like to say that the rational map $\mathbb{P}^{n - 1} \rightarrow Q$ is defined on an open set $U$ which must have lots of $k$ points (the image of which give $k$ points on $Q$).  This is true when $k$ is an infinite field because the set of $k$ points is Zariski dense in an open set.  However over a finite field, there are open sets of projective space without any $k$ points at all!  There is a way around this pitfall though which we can at least outline.  By blowing up a $k$ point where the rational map $\phi: \mathbb{P}^{n-1} \rightarrow Q$ is not defined, we get a rational map from the blown up variety $\phi' : Bl_p(\mathbb{P}^{n-1}) \rightarrow Q$.  Since $Q$ is smooth and projective, the map $\phi'$ is defined outside of codimension 2.  As the exceptional divisor of the blowup  is a projective space (with lots of rational points), we get a rational map from this projective space (which is one dimension smaller) to $Q$.  By induction on the dimension then, it follows that we can find a $k$ point on $Q$.  To see that (2) implies (1), let $P$ be a $k$ point on $Q$, and consider the projection from $P$ to a hyperplane in $\mathbb{P}^n$.  The projection is generically one to one because a generic line through $P$ hits $Q$ in only one other point.  That condition (2) implies (3) is clear, so we must finally show that (3) implies (2).  If $k'$ is a degree d (= odd) extension of $k$, we may assume that it is generated by one element $z$, which is immediate if the extension is separable (and in the general case we can successively add elements to reduce to this one).  Pick then $P$ to be a point over $k'$.  We may write it’s coordinates as $(f_0(z) : \ldots : f_n(z))$ where $f_i(z)$ is a degree $d -1$ polynomial in z.  We can think of this as a degree $d' \leq (d-1)$ map from a rational curve $\mathbb{P}^1$ to $\mathbb{P}^n$.  If the image, $C$ lies on $Q$, then we are finished, there are lots of rational points on $Q$.  Otherwise, pull the equation of $Q$ back to $\mathbb{P}^1$ and dehomogenize to get a polynomial $f(t)$ of degree 2d’.  Since $z$ is a root of $f$, the minimal polynomial $g$ of z divides $f$.  The quotient polynomial has odd degree, so one of it’s irreducible factors $h$ has odd degree $d'' < d$.  For a root of $h$, say $z'$, the image of the point $z' \in C$ is a point of $Q$ defined over a field extension of degree d”.  By induction on the degree of the field extension then, we are finished.

Remark:  In fact, this proposition holds even if the quadric is singular as long as it’s not the union of two planes.  To see this, note that a singular quadric $Q$ is a cone over a smooth one $Q'$.  As a not too hard exercise, show that $Q$ has a smooth point over $k$ if and only if $Q'$ has a point over $k$

There are many more interesting theorems in the long history of trying to classify rational varieties.  For example:  1.  By the Hasse-Minkowski theorem, a smooth quadric hypersurface defined over the rational numbers is rational over $\mathbb{Q}$ if and only if it has a point defined over $\mathbb{R}$.  2.  Quadric hypersurfaces (which are not the union of two planes) are always rational over finite fields.   3.  Cubic surfaces over algebraically closed fields are rational.  Etc.

Perhaps more can be said in the future!

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### 4 Responses to Rational Varieties — An Introduction Through Quadrics

1. You want the command dashrightarrow. See? $\dashrightarrow$.

2. Perhaps you can talk about Unirational varieties that aren’t rational? Do you know a short way to prove the theorem of Clemens and Griffiths about Cubic Threefolds?

3. Aron Simis says:

Actually, I would like to see an explicit rational map from projective 3-space to a cubic threefold. What is a reference for that?

4. Matt DeLand says:

Hi Aron,

I’ve never thought about how to write down an explicit map from projective space to a smooth cubic threefold. But I can explain how to prove it’s unirational:

Let V be a smooth cubic hypersurface in projective 4-space. Pick a line L on V. Consider all the planes containing L, this can be identified with a $\mathbb{P}^2$ (inside the Grassmannian). None of these planes are contained inside V. So each plane P, when intersected with V, gives a cubic curve, which contains L. That is, each plane gives rise to a conic (the residual to L in the intersection – that is, the other component). The total space, Q, of this conic bundle over $\mathbb{P}^2$ maps birationally to V (and should be identified with the blowup of V along L). Call E the exceptional divisor (it’s rational) and the fibers of E to L correspond to lines in the family of planes. The map from E to the base of the conic bundle is two to one, because each conic intersects L in two points.

In any case, we can prove the following general statement: Suppose $W \rightarrow S = \mathbb{P}^2$ is a conic bundle and $E \subset W$ is a rational variety dominating the base, then W is unirational. To prove this form the base change $W' = W \times_S E \rightarrow E$. This is again a conic bundle, which has a section! The section allows us to conclude (see the proposition above) that the conic over the generic point is $\mathbb{P}^1_K$ where K is the function field of E (which is rational), and so the function field of W’ is K(t), that is, purely transcendental. So W’ is rational and maps finitely to W (which is birational to our original V).