φ(x) = (f1(x), …, fm(x))
andg(f1(x), …, fm(x)) = 0
whenever x∈ X and g ∈ I(Y).Mor(X, A1) ≈ k[x1, …, xn]/I(X) = A(X).
φ# : A(Y) → A(X)
defined byφ#(f) = f°φ.
In this way, we can view A as a contravariant functor from the category ALGSET/k (whose objects are all algebraic sets with the definition of morphism just presented) taking values in the category k-ALG (whose objects are all commutative k-algebras with the evident definition of morphism).A(Y) = k[y1, …, ym]/I(Y)
andA(X) = k[x1, …, xn]/I(X).
Define si = f(yi) ∈ A(X), and choose an arbitrary lifting s'i∈ k[x1, …, xn]. Then there is a morphisms = (s'1, …, s'm) : X → Y
that is independent of the choice of lifting, and clearly satisfies s# = f.Mor(X, Y) = Homk(A(Y), A(X))
for any pair of structured algebraic sets. It follows easily from the definitions that there is a fully faithful contravariant functorA : STRALGSETS/k → k-ALG
that identifies the category of structured algebraic sets with the category of finitely generated k-algebras. In particular, one has the following result.A(X× Y) = A(X)⊗k A(Y).
Y | ||
↓ | ||
X | → | An |
I(X∩ Y) = (I(X)∪ I(Y)).
k[x1,…,xn] | → | k[x1,…,xn]/I(X) |
↓ | ||
k[x1,…,xn]/I(Y) |
I(X∩ Y) = (y-x2, y) = (y, x2),
so the coordinate ring isA(X∩ Y) = k[x,y]/(y, x2) ≈ k[x]/(x2).
In particular, the intersection of algebraic sets can turn out to be a structured algebraic set; in the present case, this structured algebraic set conatins the extra geometric information that the two curves are tangent at the point of intersection.(X×Am)∩(An× Y),
where the intersection is computed inside An×Am = An+m.I(X) ⊂ k[x1,…,xn] ⊂ k[x1,…, xn, y1, …, ym]
andI(Y) ⊂ k[y1,…,ym] ⊂ k[x1,…, xn, y1, …, ym]
clearly allow us to identify the images withI1(X) = I(X×Am) and I2(Y) = I(An× Y).
By the previous lemma, we haveI((X×Am)∩(An× Y)) = (I1(X)∪ I2(Y)).
Now one can verify that multiplication defines an isomorphismA(X)⊗k A(Y) → k[x1,…,xn,y1,…,ym]/(I1(X)∪ I2(Y)),
by the "formula"f(x)⊗ g(y) → f(x,0)g(0,y).
Mor(X, *) = Homk(k, A(X)),
and it is the set of k-homomorphism that can easily be seen to consist of exactly one element.Mor(*, X) = { maximal ideals in A(X)} = { points in X} = X(k).
A(X) = k× k.
To see this, notice that we can takeA(X) = k[x]/(x2-x) ≈ k[x]/(x) ⊕ k[x]/(x-1).
More generally, if X and Y are structured algebraic sets, then their disjoint union satisfies A(X∐ Y) = A(X)× A(Y). In other words, A takes disjoint unions into Cartesian products of algebras.Δ:X→ X× X.
This is actually a morphism of structured algebraic sets; it corresponds to the algebra homomorphism A(X)⊗ A(X) → A(X) by multiplication.x3 + x2(1-m2) = 0.
Ignoring the obvious point at the origin, we get an additional pointx = m2-1, y = m3-m.
These equations define an algebra homomorphismk[x, y]/(y2-x3-x) → k[m],
since one can easily check that(m3-m)2 - (m2-1)3 - (m2-1)2 = 0.
Mor(X,H) = Homk(k[x,x-1], A(X)) = A(X)*,
the set of invertible elements in the ring A(X). It is interesting to note that, in this case, the set of morphisms always forms an abelian group. One interpretation of this remark is that there is a functorMor(-,H) : STRALGSET → AB.
Another interpretation says that H is a group-object in the category of structured algebraic sets. Concretely, there are morphismsμ : H× H → H | μ#(x) = x⊗ x |
ε:*→ H | ε#(x) = 1 |
ι:H→ H | ι#(x) = x-1. |
associativity: | μ°(μ×1) = μ°(1×μ) |
existence of identity: | μ(ε×1) = 1 |
existence of inverse: | μ(ι,1) = επ. |
A(T) = k[x]/(x2) = k[ε].
How should one interpret the set Mor(T,X)? By definition, every such morphism corresponds to an algebra homomorphism f:A(X)→ k[ε]. By composing with the canonical surjection k[ε]→ k (which takes ε to 0), one obtains a well-defined point x∈ X, corresponding to the maximal ideal M that is the kernel of the composite. Now we can decompose the set Mor(T,X) into subsetsMor(T,X; x) = {φ:T→ X | φ(*) = x}.
We now observe that f(M)⊂(ε) and f(M2) ⊂(ε2) = 0. Therefore,Mor(T,X; x) = Homk-VS(M/M2, k).
In a natural sense, this set of vector space homomorphism can be thought of as the tangent space to X at the point x. For example, let H be the hyperbola considered earlier, and look at the set Mor(T, H; (1,1)). Any such morphism is given byf(x) = 1+λε and f(y) = 1+με.
Moreover, since xy=1, we have0 = (1+λε)(1+με) - 1 = (λ+μ)ε.
Therefore, μ=-λ and the points x=1+λε, y = 1-λε all lie along the line x+y=2, the usual tangent line to the hyperbola at that point.Comments on this web site should be addressed to the author:
Kevin R. Coombes