The constant field of a plane curve #
Let F be a field generated over k by a transcendental element x and an element y with
φ(x, y) = 0, for an irreducible φ ∈ k[X, Y]. This file proves the absolute-irreducibility
criterion for the constant field: if φ stays irreducible over an algebraically closed field
K ⊇ k, then k is algebraically closed in F, that is, k is the exact field of constants of
F. This is Stichtenoth's Corollary 3.6.8, stated in his Section 3.6, where k is assumed
perfect; the proof here needs no hypothesis on k. It supplies the exactness of constants for
function fields given by an absolutely irreducible plane equation.
The exact-constants criterion passes through the tensor product F ⊗[k] K and reuses the
domain criterion after converting the bivariate equation to two-variable polynomial form.
- The relations between
xandyare exactly the multiples ofφ: the kernel of evaluationk[X, Y] → Fat(x, y)is the principal ideal(φ). Indeedφ(x, Y)has positive degree inY, sincexis transcendental, so by Gauss's lemma it is irreducible overk(x). It is therefore a multiple of the minimal polynomial ofyoverk(x)and divides everyψ(x, Y)withψ(x, y) = 0, first overk(x)and then, again by Gauss's lemma, overk[x]. - So the ring
k[x, y]isk[X, Y] / (φ), and sinceKis flat overk, the ringk[x, y] ⊗[k] KisK[X, Y] / (φ). This is a domain whenφis irreducible, hence prime, inK[X, Y]. Its localizationF ⊗[k] Kis then a domain as well (TauCeti.isDomain_tensorProduct_of_isLocalization), becauseFis the fraction field ofk[x, y]. - When
Kis algebraically closed,F ⊗[k] Kbeing a domain forceskto be algebraically closed inF(TauCeti.algebraicClosure_eq_bot_of_isDomain_tensorProduct).
Main results #
TauCeti.ker_aeval_eq_span_singleton_of_irreducible: the ideal of relations of(x, y)overkis generated byφ.TauCeti.isDomain_tensorProduct_of_irreducible_map: ifφis irreducible over a fieldKcontainingk, thenF ⊗[k] Kis a domain.
References #
- H. Stichtenoth, Algebraic Function Fields and Codes, 2nd ed., GTM 254, Springer, 2009, Corollary 3.6.8.
The relations of a point on a plane curve. If x is transcendental over k and
φ(x, y) = 0 for an irreducible φ ∈ k[X, Y], then the polynomials vanishing at (x, y) are
exactly the multiples of φ.
Plane curves stay integral under extension of scalars. Let F be generated over k by a
transcendental x and an element y with φ(x, y) = 0. If φ ∈ k[X, Y] is irreducible over a
field K containing k, then F ⊗[k] K is a domain.
The absolute-irreducibility criterion #
An absolutely irreducible plane equation has exact constants (Stichtenoth,
Corollary 3.6.8). Let F = k⟮x, y⟯ with x transcendental over k, and let φ(x, y) = 0 for a
polynomial φ ∈ k[X][Y] which stays irreducible over an algebraically closed field L
containing k. Then k is the exact constant field of F.
No perfectness of k is needed. Absolute irreducibility is only sufficient: over an imperfect
field the constants of F can be exact although φ factors over L, as for
y ^ p = t x ^ p + u over 𝔽_p(t, u).