# abstract algebraic variety.

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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You can forget about inverse matrices and adjoints: just write out the $n^2$ polynomial equations (in the entries $x_{ij}$ of $M$) translating the matrix equality $M^T\cdot M=I_n$ –  Georges Elencwajg Apr 16 '12 at 12:08
@Georges Elencwajg Good Idea! it generates the same system!. But Why this algebra works? How can I define the bijection <.<? I have no idea! –  Arkj Apr 16 '12 at 12:12
Dear @Arkj: you should ask your teacher or mathematicians who advocate that varieties be defined as functors in elementary courses. –  Georges Elencwajg Apr 16 '12 at 12:17
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. –  Georges Elencwajg Apr 16 '12 at 12:50

## 1 Answer

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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Wow! And How can I prove that O.o? –  Arkj Apr 15 '12 at 18:06
Sorry for my stupid question –  Arkj Apr 15 '12 at 18:35
Sorry for ask, could you tell me what is the bijection map? <.<! –  Arkj Apr 16 '12 at 1:35
A functional of $K[ \dots ]$ is pluggin some values for $K$ in $x_{i,j}$. I will not give any further hints, since this seems like a homerwork problem. –  plusepsilon.de Apr 16 '12 at 8:19
@Arkj: What is O.o ? –  Georges Elencwajg Apr 16 '12 at 12:11