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An abstract affine algebraic $k$-variety is the correspondence which assigns to each $k$-algebra $K$ a set $X(K)$. This assignment must satisfy the following:

(i). For each homomorphism of $k$-algebras $φ : K → K_0$ there is a map $X(φ): X(K) → X(K_0)$

(ii). $X(\operatorname{id}_K) = \operatorname{id}_X(K)$

(iii). For any $φ_0 : K → K_0$ and $φ_1 : K_0 → K_1$ we have $X(φ_1 \circ φ_0) = X(φ_1) \circ X(φ_0)$.

(iv). There exists a finitely generated $k$-algebra $A$ such that for each $K$ there is a bijection $X(K) → \operatorname{Hom}_k(A, K)$ and the maps $X(φ)$ correspond to the composition maps $\operatorname{Hom}(A, K) → \operatorname{Hom}_k(A, K_0)$, where the composition maps are the maps induced by homomorphism of $k$-algebras $\beta :K \to K_0$ in the obvious way (composing), and where $\operatorname{Hom}_k(A,B)$ denotes the set of all homomorphisms of $k$-algebras $f:A \to B$ (i.e ringhomomorphism that fix the field $k$).

Problem

Let the correspondence taking a $k$-algebra $K$ $ K \to O(n,K) = \left\{n\times n \text{ matrices with entries in }K\text{ such that }M^T = T^{-1}\right\} $ (i.e orthogonal matrix over the $K$-algebra $K$).

Prove that it's an abstract algebraic variety. I proved everything, except that has (iv). I don't know what $K$-algebra $A$ to consider :/.

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    What I mean is that the link between the elementary definition of algebraic variety and the functorial definition is a long story if you want to tell it well.2012-04-16

1 Answers 1

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Write $M^{-1} = \frac{1}{\det M} M^{ad},$ then you get a bunch of Polynomial equations from $M^{T} = M^{-1},$ i.e. for every entry $P_{i,j}(x_{1,1}, \cdots, x_{n,n}) = 0.$

Your $K$-algebra is then the quotient A = K[x_{1,1}, \cdots ,x_{n,n}] / < P_{i,j} : 1 \leq i,j \leq n>.

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    @Arkj: Wha$t$ is O.o ?2012-04-16