`
φ(x) = (f _{1}(x), …, f_{m}(x))
`

`
g(f _{1}(x), …, f_{m}(x)) = 0
`

`
Mor(X, A ^{1}) ≈ k[x_{1}, …, x_{n}]/I(X) = A(X).
`

`
φ ^{#} : A(Y) → A(X)
`

`
φ ^{#}(f) = f°φ.
`

`
A(Y) = k[y _{1}, …, y_{m}]/I(Y)
`

`
A(X) = k[x _{1}, …, x_{n}]/I(X).
`

`
s = (s' _{1}, …, s'_{m}) : X → Y
`

`
Mor(X, Y) = Hom _{k}(A(Y), A(X))
`

`
A : STRALGSETS/k
→ k-ALG
`

`
A(X× Y) = A(X)⊗ _{k} A(Y).
`

Y | ||

↓ | ||

X | → | A^{n} |

`
I(X∩ Y) = (I(X)∪ I(Y)).
`

k[x_{1},…,x_{n}] | → |
k[x_{1},…,x_{n}]/I(X) |

↓ | ||

k[x_{1},…,x_{n}]/I(Y) |

`
I(X∩ Y) = (y-x ^{2}, y) = (y, x^{2}),
`

`
A(X∩ Y) = k[x,y]/(y, x ^{2}) ≈ k[x]/(x^{2}).
`

`
(X×A ^{m})∩(A^{n}× Y),
`

`
I(X) ⊂ k[x _{1},…,x_{n}] ⊂ k[x_{1},…, x_{n},
y_{1}, …, y_{m}]
`

`
I(Y) ⊂ k[y _{1},…,y_{m}] ⊂ k[x_{1},…, x_{n},
y_{1}, …, y_{m}]
`

`
I _{1}(X) = I(X×A^{m}) and I_{2}(Y) =
I(A^{n}× Y).
`

`
I((X×A ^{m})∩(A^{n}× Y)) = (I_{1}(X)∪ I_{2}(Y)).
`

`
A(X)⊗ _{k} A(Y) →
k[x_{1},…,x_{n},y_{1},…,y_{m}]/(I_{1}(X)∪ I_{2}(Y)),
`

`
f(x)⊗ g(y) → f(x,0)g(0,y).
`

- Let
`*`denote the algebraic set consisting of a single point. Then`A(*) = k`and`*`is a final object in the category of structured algebraic sets. In other words, given any structured algebraic set`X`, there is always a unique morphism`X→*`corresponding to the structure map`k→ A(X)`that includes`k`as the ring of constant functions on`X`. More precisely, one has

and it is the set of`Mor(X, *) = Hom`_{k}(k, A(X)),`k`-homomorphism that can easily be seen to consist of exactly one element. - Consider again the one point algebraic set
`*`. We have seen that`Mor(X, *)`always contains exactly one element. What is the set`Mor(*, X)`? Turning the question into one about algebras, we are asking about the set of`k`-algebra homomorphisms`Hom`. Since these must preserve the_{k}(A(X), k)`k`-structure, and`k`is a field, such homomorphisms are necessarily onto. Each surjective homomorphism to the field`k`corresponds to a maximal ideal in`A(X)`(obtained as the kernel). So, using the Weak Nullstellensatz, one has`Mor(*, X) = { maximal ideals in A(X)} = { points in X} = X(k).` - Let
`X`be the (disjoint) union of two points, viewed as an algebraic set over`k`. Then

To see this, notice that we can take`A(X) = k× k.`

More generally, if`A(X) = k[x]/(x`^{2}-x) ≈ k[x]/(x) ⊕ k[x]/(x-1).`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. - Let
`X`be a structured algebraic set, and consider the diagonal map

This is actually a morphism of structured algebraic sets; it corresponds to the algebra homomorphism`Δ:X→ X× X.``A(X)⊗ A(X) → A(X)`by multiplication. - Let
`X`be the affine plane curve defined by the equation`y`. So,^{2}=x^{3}+x^{2}`A(X) = k[x,y]/(y`. We can define a morphism from^{2}-x^{3}-x^{2})`A`as follows. Let^{1}→ X`y=mx`be a general line through the origin. This line intersects`X`at the points where

Ignoring the obvious point at the origin, we get an additional point`x`^{3}+ x^{2}(1-m^{2}) = 0.

These equations define an algebra homomorphism`x = m`^{2}-1, y = m^{3}-m.

since one can easily check that`k[x, y]/(y`^{2}-x^{3}-x) → k[m],`(m`^{3}-m)^{2}- (m^{2}-1)^{3}- (m^{2}-1)^{2}= 0. - Let
`H`be the hyperbola defined by the equation`xy=1`. So, the coordinate ring of`H`is`A(H) = k[x,y]/(xy-1)≈ k[x,x`. Now let^{-1}]`X`be any structured algebraic set. Then

the set of invertible elements in the ring`Mor(X,H) = Hom`_{k}(k[x,x^{-1}], A(X)) = A(X)^{*},`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 functor

Another interpretation says that`Mor(-,H) : STRALGSET → AB.``H`is a group-object in the category of structured algebraic sets. Concretely, there are morphismsμ : H× H → H μ ^{#}(x) = x⊗ xε:*→ H ε ^{#}(x) = 1ι: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. - Let
`T`be the structured algebraic set with coordinate ring

How should one interpret the set`A(T) = k[x]/(x`^{2}) = k[ε].`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 subsets

We now observe that`Mor(T,X; x) = {φ:T→ X | φ(*) = x}.``f(M)⊂(ε)`and`f(M`. Therefore,^{2}) ⊂(ε^{2}) = 0

In a natural sense, this set of vector space homomorphism can be thought of as the tangent space to`Mor(T,X; x) = Hom`_{k-VS}(M/M^{2}, k).`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 by

Moreover, since`f(x) = 1+λε and f(y) = 1+με.``xy=1`, we have

Therefore,`0 = (1+λε)(1+με) - 1 = (λ+μ)ε.``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. CoombesDepartment of Biomedical Informatics

The Ohio State University

Columbus, Ohio 43210