Definition An algebraic variety X is called rational if k(X) is isomorphic to the field of rational functions k(x1,,xn) = k(An).


  1. Let X=Z(xy-1)A2. Then the function field of A(X)=k[x,x-1] is clearly isomorphic to k(X). In fact, projection on the x-coordinate defines a morphism XA1 that induces an isomorphism of function fields.
  2. Let X=Z(y2-x3-x2)A2. In the previous chapter, we constructed a morphism A1 X given by the equations

    x = m2-1


    y = m3-m.

    This morphism defines an injection

    A(X) k[m] k(m),

    and hence induces a map k(X) k(m). The inverse of this map on function fields is given by m = x/y.
  3. Suppose the characteristic of the field k differs from 3. Then the curve Z(x3+y3-1) is not a rational curve. To see this, suppose that it is rational. Then we could write

    x = p(t)/r(t), and y = q(t)/r(t),

    where p, q, r k[t] are relatively prime polynomials. Now the equations
    p3 + q3 -r3 = 0
    3p2p' + 3q2q' -3r2r' =0

    can be viewed as a linear system
    p q -r
    p' q' -r'
    Since p0, this system is equivalent to
    p q -r
    0 pq'-qp' rp'-pr'
    whose general solution is
    = C
    By symmetry, we can assume that

    deg(r) deg(q) deg(p).

    Since p2, q2 and r2 are relatively prime, we conclude that

    p2 divides rq'-qr'.


    2deg(p) deg(r) + deg(q) -1.

    This contradicts the assumption we made about the degrees, and so no such polynomials can exist. Note that this proof immediately generalizes to show that Z(xn+yn-1) is not a rational curve if n>2 and the characteristic of k is relatively prime to n.
  4. The surface Z(x3+y3+z3-1) in A3 is rational. To see this, consider the lines
    L = Z(z-1, x+y)
    M = Z(z-ω, x+ω y),

    where ω is a primitive cube root of unity. Then L and M X, but L M=. Now choose a plane EA3 that contains neither L nor M. Every point xA3\ L\ M uniquely determines a line N(x, L, M) that contains x and has a nonempty intersection with both L and M. Thus, one gets a morphism

    A3\ L \ M A2


    x |→ N(x, L, M) E.

    This map restricts to a well-defined morphism everywhere on X. Since N(x, L, M) X generically determines a unique point of X\ L \ M, the map induces an isomorphism of function fields k(X) k(A2).

Remark A variety X is called unirational if k(X) k(t1, , tn). For a long time, it was an open question whether unirational implies rational. For n=1 or n=2, it does. Among other examples, Clemens and Griffiths showed that the threefold Z(x3+y3+z3+w3-1) is unirational but not rational. It is still unknown if Z(x3 + y3 + z3 + u3 + v3-1) is rational, unirational, or neither.

Definition Two varieties X and Y are called birationally isomorphic if their function fields are isomorphic.

Notice that a variety is rational if and only if it is birationally isomorphic to An.

Lemma Let U X be a nonempty open subset of an algebraic variety X. Then U is birationally isomorphic to X.

Proof: Shrink U further and assume that it is an affine open subset. Then A(U) = A(X)[1/f] clearly has the same field of fractions that A(X) has.

Definition A rational mapping φ:X--→ Y between algebraic varieties is defined as a collection of rational functions φi k(X) such that

g(φ1(x), , φm(x)) = 0

whenever g I(Y) and all the φi are regular at the point x X.

Since each φi is regular on a nonempty open set, so is φ. A rational map is called dominant if φ(U) is dense in Y for any nonempty open subset U contained in the domain of regularity of φ.

Theorem There is a natural equivalence between the category of algebraic varieties with dominant rational maps and the category of finitely generated field extensions of k.

Proof: Let φ:X--→ Y be a dominant rational map, and let U X be a nonempty open set where φ is regular. Then

φ# : A(Y) A(U) k(X).

If φ#(g) = 0, then g=0, since φ(U) is dense in Y. So, φ# is an embedding of an integral domain into a field. By the universal property of the field of fractions, φ# extends uniquely to a homomorphism of fields

φ* : k(Y) k(X).

Conversely, let ψ:k(Y) k(X) be any homomorphism of fields. Restrict to the subring A(Y) = k[y1,,ym]/I(Y). Then

(ψ(y1), , ψ(ym))

is a rational map from X to Y. It is easy to check that these two procedures are inverses.

Proposition Two varieties X and Y are birationally isomorphic if and only if they contain isomorphic open subsets.

Proof: One direction is immediate from the lemma before the theorem. The other follows by interpreting maps of function fields as morphisms defined on open subsets and applying the previous theorem.

Theorem Every algebraic variety is birationally isomorphic to a hypersurface.

Proof: Since k(X)/k is a separable extension of fields, it has a separating transcendence basis t1, , tn k(X). In other words, k(X) is a finite separable algebraic field extension of the rational function field k(t1, , tn). By the primitive element theorem, we can write

k(X) = k(t1, , tn)[s]/( minimal polynomial of s).

By clearing denominators in the coefficients of the minimal polynomial, we can assume the coefficients live in the ring of polynomials. Now define

A(Y) = k[t1, , tn, s]/( minimal polynomial of s).

Then Y is a hypersurface and k(Y)=k(X).

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210