vector space as an affine variety

Overview

Every vector space has a canonical affine algebraic variety structure. This makes sense intuitively since (informally) each vector space is somehow linear… and thus affine.

Affine algebraic variety structure

Let $V$ be a finite-dimensional vector space. Given an ordered basis $\beta$ of $V$, consider the vector space isomorphism

$$\begin{array}{rl}[-{]}_{\beta }:V& ⟶{\mathbb{C}}^{\dim V}\\ v& ⟼[v{]}_{\beta }.\end{array}$$

The set $V$ has a unique affine algebraic variety structure that makes $[-{]}_{\beta }$ an isomorphism of affine algebraic varieties for all ordered bases $\beta$ of $V$.

Proof

Throughout this proof, to make things easier to write, assume $\dim V=n$ for some $n\in {\mathbb{Z}}_{\ge 1}$.

In order to prove the statement above, we can show that we can put an affine algebraic variety structure on V, then we must show that it is unique.

Existence of affine variety structure

To give V an affine algebraic variety structure, first we need to make sure $[-{]}_{\beta }:V\to {\mathbb{C}}^n$ is a homeomorphism. Therefore, for every closed set $U\in {\mathbb{C}}^n$, let $[-{]}_{\beta }^{-1}(U)$ be a closed subset of $V$.

Now, we need to define $\mathbb{C}[V]$. Since we have a bijection between ${\mathbb{C}}^n$ and $V$, we can use that to build $\mathbb{C}[V]$.

For $\varphi \in \mathbb{C}[x_1,\dots ,x_n]$ let $\tilde{\varphi }\in \mathbb{C}[V]$ be defined as the map $\varphi \circ [-{]}_{\beta }$.

Therefore, we have the pair $(V,\mathbb{C}[V])$ such that $\mathbb{C}[V]\subseteq {\mathbb{C}}^0(V,\mathbb{C})$ and

$$[-{]}_{\beta }:V\to V(\{0\})={\mathbb{C}}^n$$

where

$[-{]}_{\beta }^{\ast }(\mathbb{C}[V(I)])=[-{]}_{\beta }^{\ast }(\mathbb{C}[x_1,\dots ,x_n])=\mathbb{C}[V].$

This means $V$ has an affine algebraic variety structure.

Uniqueness

Using this lemma, given two ordered basis $\beta$ and $\gamma$ for V, it suffices to show that $id:V_{\beta }\to V_{\gamma }$ is an isomorphism where $V_{\beta }$ denotes the affine algebraic variety structure inherited from $\beta$.

We can illustrate this with the commutative diagram below:

We know that the diagram commutes, therefore

$id=[-{]}_{\gamma }^{-1}\circ ([-{]}_{\gamma }\circ [-{]}_{\beta }^{-1})\circ [-{]}_{\beta }.$

We know by construction that $[-{]}_{\beta }$ and $[-{]}_{\gamma }$ are isomorphisms. However, we need to show that the middle term $[-{]}_{\gamma }\circ [-{]}_{\beta }^{-1}$ is an isomorphism. It is apparent that it is a vector space isomorphism, but not that it is a morphism of affine algebraic varieties.

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Lemma For $A\in Ma{t}_{m\times n}(\mathbb{C})$ the map

$$\begin{array}{rl}\phi :{\mathbb{C}}^m& \to {\mathbb{C}}^n\\ z& \mapsto Az\end{array}$$

is a morphism of affine algebraic varieties.

Proof Fix $i\in \{1,\dots ,m\}$.

$$\begin{array}{rl}({\phi }^{\ast }(x_i))(z_1,\dots ,z_n)& =x_i(\phi (z_1,\dots ,z_n))\\ & =x_i\left(A\left[\begin{array}{c}{z}_1\\ ⋮\\ z_n\end{array}\right]\right)\\ & =\sum _{k=1}^n{a}_{ik}{z}_k\text{ the ith entry of }A\vec{z}\\ & =a_{i1}{z}_1+\cdots +a_{in}{z}_n\\ & =(a_{i1}{x}_1+\cdots +a_{in}{x}_n)(z_1,\dots ,z_n)\end{array}$$

Therefore

$${\phi }^{\ast }(x_i)=a_{i1}{x}_1+\cdots +a_{in}{x}_n\in \mathbb{C}[x_1,\dots ,x_n]$$

Now for an arbitrary $f\in \mathbb{C}[x_1,\dots ,x_n]$,

$$f=\sum _{I=(i_1,\dots ,i_n)}{a}_I{x}_1^{i_1}\cdots x_n^{i_n}$$

Since we know that ${\phi }^{\ast }$ is an algebra homomorphism,

$${\phi }^{\ast }(f)=\sum _{I=(i_1,\dots ,i_n)}{a}_I{\phi }^{\ast }(x_1^{i_1})\cdots {\phi }^{\ast }(x_n^{i_n})$$

which is a product of polynomials, and is thus a polynomials. Stated another way, ${\phi }^{\ast }(f)\in \mathbb{C}[x_1,\dots ,x_n]$.

Now the only thing that remains is to prove that $\phi$ is continuous.

Let $Z\subseteq {\mathbb{C}}^n$ be closed. Thus

$$\begin{array}{rl}Z& =V(f_1,\dots ,f_k)\text{ for }{f}_i\in \mathbb{C}[x_1,\dots ,x_n]\\ & =\{z\in {\mathbb{C}}^n|f_1(z)=\cdots =f_k(z)=0\}\end{array}$$

Consider the preimage of the set $Z$:

$$\begin{array}{rl}{\phi }^{-1}\left(Z\right)={\phi }^{-1}(V(f_1,\dots ,f_k))& =\{z\in {\mathbb{C}}^n|\phi (z)\in V(f_1,\dots ,f_k)\}\\ & =\{z\in {\mathbb{C}}^n|f_1(\phi (z))=\cdots =f_n(\phi (z))=0\}\\ & =\{z\in {\mathbb{C}}^n|({\phi }^{\ast }(f_1))(z)=\cdots =({\phi }^{\ast }(f_n))(z)=0\}\\ & =V({\phi }^{\ast }(f_1),\dots ,{\phi }^{\ast }(f_k))\\ & =V(⟨{\phi }^{\ast }(f_1),\dots ,{\phi }^{\ast }(f_k)⟩)\end{array}$$

Thus ${\phi }^{-1}(Z)$ is closed in ${\mathbb{C}}^n$ and $\phi$ is continuous.

With both of these conditions, the map $\phi$ which is left multiplication by a matrix is a morphism.

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By this lemma, $[-{]}_{\gamma }\circ [-{]}_{\beta }^{-1}$ is a morphism because it can be written as the matrix representing the change of basis. Thus, $id:V_{\beta }\to V_{\gamma }$ is a composition of morphisms so it must be a morphism. Note this same process can be done with $id:V_{\gamma }\to V_{\beta }$, so $id$ is a morphism with an inverse morphism so it is an isomorphism. By the first lemma, for any choice of ordered basis on $V$ the two affine algebraic variety structures are the same.

Linear maps are morphisms

Suppose that $W$ is another finite-dimensional vector space. Equip the sets $V$ and $W$ with the affine algebraic variety structures resulting from Part (i), and let $\phi :V⟶W$ be a linear map. Prove that $\phi$ is a morphism of affine algebraic varieties.

For simplicity let $\dim W=m$.

Finite dimensional vector spaces are isomorphic to ${\mathbb{C}}^n$. Thus, we know we can find a matrix representation of $\phi$ such that the following diagram commutes for a given choice of ordered basis $\gamma$ for W.

Therefore, we see that the function can be written as

$$\phi =[-{]}_{\gamma }^{-1}\circ [\phi ]\circ [-{]}_{\beta }.$$

Similarly to above, we know that $[-{]}_{\beta }$ and $[-{]}_{\gamma }$ are isomorphism by construction. By the lemma above, we know that the map for left multplication by a matrix (in this case $[\phi ]$) is a morphism, thus $\phi$ is a composition of morphism and thus is a morphism.