x = m2-1
andy = m3-m.
This morphism defines an injectionA(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.x = p(t)/r(t), and y = q(t)/r(t),
where p, q, r∈ k[t] are relatively prime polynomials. Now the equationsp3 + q3 -r3 | = 0 |
3p2p' + 3q2q' -3r2r' | =0 |
|
| = |
| . |
|
| = |
| , |
| = | C |
| . |
deg(r)≤ deg(q) ≤ deg(p).
Since p2, q2 and r2 are relatively prime, we conclude thatp2 divides rq'-qr'.
Therefore,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.L = | Z(z-1, x+y) |
M = | Z(z-ω, x+ω y), |
A3\ L \ M → A2
byx |→ 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).g(φ1(x), …, φm(x)) = 0
whenever g∈ I(Y) and all the φi are regular at the point x∈ X.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 fieldsφ* : 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.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 defineA(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