1 Recap of last time Def. An alg. subset of k^n or a subset which is algebraically closed is a set of the form V(I) = {a | f(a) = 0 ∀ f ∈ I} where I ⊆ k[x] is an ideal. Thm. 1 There is a natural bijection between {points in V(I)} ↔ {max. ideals in k[x]/I} I and V(I) define the same algebraic set. (If we consider k[x]/I and k[x]/J the maximal ideals are the same; the only difference is the possible presence of some nilpotent elements. Thm. 2. (There is a natural bijection) 1 1. {alg. subsets of k^n} ↔ {ideals I ⊆ k[x] with I = √I} In the forward direction, this is Σ ↦ I(Σ) = {f ∈ k[x₁, ..., xₙ] | f(Σ) = 0} and in the reverse direction this is V: A ↦ I(A) = {f ∈ k[x₁, ..., xₙ] | f(A) = 0}. Proof. It should be clear that V(I(Σ)) ⊇ Σ. If a ∉ Σ then there is an f ∈ A such that f(a) ≠ 0. Note: I(Σ) = √I(Σ). By definition I(Σ) is the set of functions whose some power of that function vanishes on Σ, but this is over a field, so the function itself vanishes on Σ. Thm 2.1 (sometimes called Nullstellensatz.) Let A ⊆ k^n, and let g ∈ k such that if f₁(a) = ... = fₘ(a) = 0 then g(a) = 0. Then we must show that g ∈ √(I(A)), where √I(A) is the radical ideal of I(A). Consider k[x₁, ..., xₙ₊₁] and let B = A + (1-gxₙ₊₁)k[x₁, ..., xₙ₊₁] Either B is a proper ideal or B is the whole ring. If B is a proper ideal it lies in a maximal ideal m ⊆ k[x₁, ..., xₙ₊₁] since any maximal ideal is of this form. Thus, (a, aₙ₊₁) = m ⊆ k[x₁, ..., xₙ₊₁] so g(a) = 0 and aₙ₊₁ = 1 so g(a) = 0; contradiction. If B is the whole ring k[x₁, ..., xₙ₊₁] then 1 ∈ B. So, 1 = h₁ + h₂(1-gxₙ₊₁) for some h₁ ∈ I(A), h₂ ∈ k[x₁, ..., xₙ₊₁]. So, we get 1 = h₁(x₁, ..., xₙ) + h₂(x₁, ..., xₙ)(1 - gxₙ₊₁) to clear the xₙ₊₁ terms from all these monomials since we get in k[x₁, ..., xₙ] sums of monomials of this problem, and we get inject into this ring. We choose I large enough since now we can rewrite h₁ as some sum of monomials in x₁, ..., xₙ multiplied by g, which is also a polynomial in x₁, ..., xₙ. This proves theorem 2.1 which proves theorem 2.