Bryden Cais.  Algebraic Geometry, HW 1.  January 29, 2003

1. Let X = Spec A with P ∈ X and let U be any open neighborhood of P. We show that there is an affine open neighborhood V of P contained in U. Since U is open, we have U = X \ V (I) for some ideal I ⊂ A. Pick any f ∈ I such that f ∉ P. This is possible since if I ⊂ P then P ∈ V (I) = {Q : Q ⊃ I} and hence P ∉ U. We know that D(f) = X \V (f) is an open affine neighborhood of P since P ∉ V (f) = {Q : Q ⊃ (f)} and f ∉ P, and therefore P ∈ X \ V (f) = D(f). Moreover, since f ∈ I, and prime ideal Q containing I also contains (f) so that

V (f) = {Q : q ⊃ (f)} ⊃ {Q : Q ⊃ I} = V (I).

It follows that X \ V (f) ⊂ X \ V (I) and hence D(f) ⊂ U.

2. If X = 𝔸_{k}^{2} − {(0, 0)} were an open affine subvariety of 𝔸_{k}^{2}, we would have a surjection k[x, y] ↠ O(X) given by restriction. We claim that this surjection is also an injection. For let f, g ∈ k[x, y] and suppose that f = g on X. Then in particular, f − g is a polynomial vanishing on the line x = 0 without the point (0, 0). But when restricted to this line, f − g is a polynomial in one variable with infinitely many roots, so that f − g is identically 0. Hence, we have an isomorphism k[x, y] ≃ O(X). But this is impossible, as X ≄ 𝔸_{k}^{2}, so that X is not affine.

3. Let A = k[t]/(t^{2}). Observe that A is a local ring with unique maximal ideal (t). Moreover, it is not hard to see that Spec A = {(t)} since prime ideals of the quotient k[t]/(t^{2}) correspond to prime ideals of k[t] containing (t^{2}), and it is clear that the only such ideal is (t). Since A is not an integral domain, (0) is not a prime ideal. It follows that the closed sets of Y = Spec A are just {∅, Y } and that these are also the open sets.

Let X be any variety over k. Then the associated scheme is (Spec O(X), O_{X}) where O_{X}(U) = O(U) for any open set U ⊂ X. To define a morphism of schemes f : Y → X we must give a continuous map of topological spaces f : Y → X together with a morphism of sheaves f^{#} : O_{X}(U) → O_{Y} (f^{−1}(U)) for any open u ⊂ X. Observe that to define f : Y → X (as a map of topological spaces) it is enough to pick a single point x ∈ X as the image of (t) ∈ Y. Clearly f will be continuous no matter the choice of x, since Y has the discrete topology. We show that to give a map of sheaves f^{#} corresponds to giving a tangent vector ϕ ∈ T_{x}.

Let U ⊂ X be open. If x ∉ U then f^{−1}(U) = ∅, and hence, by definition, O_{Y} (f^{−1}(U)) = O_{Y} (∅) = 0 so that for any U not containing x, we define f^{#}(O_{X})(U) = 0. On the other hand, if x ∈ U, then f^{−1}(U) = Y so for such U we must define f^{#} : O_{X}(U) → O_{Y} (Y) = A. Now giving such a map f^{#} for every open U containing x is equivalent to giving a map of stalks

f̃: O_{X,x} → A,

with f̃ a local homomorphism of local rings. However, giving such a map f̃ is equivalent to giving a map f̄ : 𝔪_{X,x} → (t) (since for φ ∈ O_{X,x} we then define f̃(φ) = f̄(φ − φ(x)) + φ(x), which is then a local homomorphism of local rings), and since (t) = {at : a ∈ k}, we see that this is equivalent to giving a map ϕ : 𝔪_{X,x} → k and then defining f̄(φ) = ϕ(φ)t. However, observe that since f̃ must be a local homomorphism of local rings, we must have f̄(φ^{2}) = f̄(φ)^{2} = (ϕ(φ)t)^{2} = 0 in A. Thus, giving a local homomorphism of local rings f̃: O_{X,x} → A is equivalent to giving a homomorphism ϕ : 𝔪_{X,x} → k that kills 𝔪_{X,x}^{2}, or in other words, a homomorphism ϕ : 𝔪_{X,x}/𝔪_{X,x}^{2} → k. But this is precisely an element of T_{x} = (𝔪_{X,x}/𝔪_{X,x}^{2})^{∗}. Thus we see that a morphism of schemes f : Y → X is given by a point x ∈ X (to be the image of (t)) and a tangent vector ϕ ∈ T_{x} (which gives the map f^{#} of sheaves when f^{−1}(U) ≠ ∅). Conversely, from our discussion it is clear that any morphism f of schemes gives a point f((t)) of X and a tangent vector induced by the homomorphism of local rings given by f^{#}.

4. Let 𝔪 be a maximal ideal of A, i.e. 𝔪 is a closed point of X. We show that the maximal ideals of A′ containing 𝔪 lie in the same Z/2Z orbit on X′. Let 𝔪_{1}, 𝔪_{2} be distinct maximal ideals of A′ containing 𝔪 and suppose that σ is a generator for Z/2Z. Suppose that m_{1} ≠ σ𝔪_{2}. Then 𝔪_{1} ⊈ σ𝔪_{2} since 𝔪_{1} is maximal. Then there exists x ∈ 𝔪_{1} with x ∉ σ𝔪_{2}. Put y = xσ(x). By construction, y ∈ 𝔪_{1} since 𝔪_{1} is an ideal. Clearly, y lies in the field of fractions of A since it is Z/2Z invariant, and since A is integrally closed and x ∈ A′ is integral over A, we have y ∈ A ∩ 𝔪_{1} = 𝔪 (since A ∩ 𝔪_{1} is an ideal of A containing 𝔪 and 𝔪 is maximal). But 𝔪 is contained in 𝔪_{2} so that y ∈ 𝔪_{2}. Since 𝔪_{2} is maximal, and hence prime, and x ∉ 𝔪_{2}, we have σ(x) ∈ 𝔪_{2}, so that, since σ has order 2, x ∈ σ(𝔪_{2}). Since x ∈ 𝔪_{1} was arbitrary, it follows that 𝔪_{1} ⊆ σ𝔪_{2} so that, since 𝔪_{1} is maximal, 𝔪_{1} = σ𝔪_{2}, and this is a contradiction. It follows that 𝔪_{1} = σ𝔪_{2}. Conversely, if 𝔪_{1} = σ𝔪_{2} then A ∩ 𝔪_{1} = A ∩ 𝔪_{2} is a maximal ideal of A so that the closed points of X correspond to the Z/2Z orbits on the closed points of X′.