Math 631, Lecture 1

Notes by Bryden Cais

Sept. 6, 2002

Fix a field k and denote by k[T] the polynomial ring in n variables k[T_{1} … T_{n}]. The ring k[T] has the natural structure of a k-algebra, since it is a ring, k injects (via inclusion) into k[T] and this injection realizes k[T] as a k-vector space.

Definition 1. A system of algebraic equations is any set S ⊂ k[T].

Now let K be any k-algebra (in particular, this includes the cases when K/k is a field extension), and let α = (α_{1} … α_{n}) ∈ K^{n}. Then we say that α is a K-solution of S if F (α) = 0 for all F ∈ S. We denote the set of all K-solutions of S Sol(S; K).

Example 1. If n = 2, k = ℚ, and S = {F (T_{1}, T_{2})}, then Sol(S, k) = {(T_{1}, T_{2}) ∈ ℚ^{2} : F (T_{1}, T_{2}) = 0}. Similarly, we see that Sol(S; ℝ) ⊂ ℝ^{2} is a real algebraic curve and that Sol(S; ℂ) ⊂ ℂ^{2} has the structure or a Riemann surface.

Example 2. We have K^{n} = Sol({0}; K) for any K.

Given any k-algebra homomorphism (that is, a ring homomorphism that is the identity on k)

φ : K ⟶ K′,

we have an induced homomorphism

⨁^{n} φ : Sol(S; K) ⟶ Sol(S, K′)

given by the obvious homomorphism from K^{n} to K′^{n} (since after all, Sol(S; K) ⊂ K^{n}).

Proposition 1. Let (S) be the k[T] ideal generated by S. Then for any K, S, Sol(S; K) = Sol((S); K).

Proof. Suppose α ∈ K^{n} is in S. Then for every F ∈ S we have F (α) = 0, and therefore α ∈ Sol((S); K) since every G ∈ (S) has the form

G = ∑_{F_{s}∈S} A_{s}F_{s},

where A_{s} ∈ k[T]. The converse is even more obvious since S ⊂ (S).

Theorem 1. We have Sol(S; K) = Sol(S′, K) for all K if and only if (S) = (S′).

Proof. The if direction is trivial. For the reverse, observe that by Proposition 1 it suffices to show that Sol((S); K) = Sol((S′), K) for all K implies (S) = (S′). Let α = (α_{1} … α_{n}) ∈ Sol(S; K) and define

φ_{α} : k[T] ⟶ K

by φ(T_{i}) = α_{i}. It is clear that φ_{α} is a homomorphism of k-algebras. Moreover, we see that φ_{α}((S)) = 0 so that (S) ⊂ ker φ_{α}. We therefore obtain the commutative diagram

k[T]  k[T]
↓^{q}  ^{φ_{α}}↓
k[T]/(S) ⟶^{φ̄_{α}} K

where q is the natural quotient map. Observe that this diagram tells us two things:

1. Given any α ∈ Sol((S); K) we obtain φ̄_{α}, an element of Hom_{k} (k[T]/S, K).

2. Given any element φ ∈ Hom_{k} (k[T]/S, K), we obtain the composite homomorphism

φ̃ = φ ∘ q : k[T] ⟶ K

satisfying φ̃((S)) = 0. This then gives us α = (φ̃(T_{1}), … φ̃(T_{n})) ∈ Sol((S); K).

The upshot is that we have a bijection

Hom_{k} (k[T]/(S), K) ↔ Sol(S; K).

Therefore, if Sol((S); K) = Sol((S′); K) for all K, we have a bijection

Hom_{k} (k[T]/(S), K) ↔ Hom_{k} (k[T]/(S′), K)

for all K. Now let K be the k-algebra k[T]/(S). Observe that the identity map is in

Hom_{k} (k[T]/(S), k[T]/(S)) ,

so that we have some homomorphism φ : k[T]/(S′) ⟶ k[T]/(S). In particular, this tells us that if we lift φ to a map φ̃ : k[T] ⟶ k[T] in the natural way, we have φ̃(S) ⊂ (S′). Interchanging ALMOST THERE!!! S and S′ then completes the proof.

Observe that the hypothesis that K be any k-algebra is essential: If S = x^{2}+1, then Sol(S; ℝ) = Sol(1, ℝ) but we clearly do not have (1) = (S). Notice also that the above proof uses this variability of field in a crucial way.