# Example of affine variety with infinite dimension

It is a typical question in Algebraic Geometry to find a Noetherian topological space with infinite dimension: we can take $X=[0,1]$ and check that.

However, it is not so obvious for me to:

Find an affine variety of infinite dimension

Some remarks (because definitions diversify):

1. An affine variety is an irreducible algebraic set in $\mathbb{A}^n$

2. An algebraic set is a subset of $\mathbb{A}^n$ of the form $Z(S)$ (locus of $S \subseteq \mathbb{K}[X_1,\ldots,X_n]$)

3. A topological space $X$ is called reducible if $X=X_1 \cup X_2$ where $X_1, X_2$ are non-empty, proper subsets of $X$ and closed. In the context of the problem, closed means ''is an algebraic set'' (the Zariski topology)

• How could a closed subset of $\mathbf A^n$ have dimension bigger than $n$? – user64687 Mar 5 '15 at 23:41
• I don't assume that $\mathbb{A}^n$ has dimension $n$. – Leafar Mar 5 '15 at 23:43
• But it does, @Leafar, and the quotient of a ring of dimension $n$ has dimension at most $n$. This is problematic for your question. – user98602 Mar 5 '15 at 23:45
• Leafar, perhaps you meant affine scheme. In this case, you can take something like $\mathbb{A}^\infty$, which is $\mathrm{Spec}(k[x_1,x_2,\ldots])$. – Alex Youcis Mar 6 '15 at 1:01
• Also, $[0, 1]$ is not a noetherian topological space. (It has a non-compact subset, which is impossible in a noetherian topological space.) – Zhen Lin Mar 6 '15 at 8:45

## 1 Answer

Lemma: Let $X \neq \varnothing$ be an irreducible topological space and $A \subseteq X$ non-empty, closed and irreducible. Then, $dim(A) \leq dim(X)$.

Proof: By definition, the dimension of $X$ is the largest integer $n$ such that there exists a chain $$\varnothing \neq X_0 \subsetneq X_1 \subsetneq \cdots \subsetneq X_n = X$$ of closed and irreducible subsets of $X$. Since $A$ is closed, irreducible and non-empty, let $m$ be the dimension of $A$. Let $$\varnothing \neq Y_0 \subsetneq Y_1 \subsetneq \cdots \subsetneq Y_m = A$$ be a maximal chain of closed and irreducible subsets $A$. If $A=X$, then $$\varnothing \neq Y_0 \subsetneq Y_1 \subsetneq \cdots \subsetneq Y_m = A = X,$$ and then $m = dim(A) = dim(X)$. If $A \neq X$, then $$\varnothing \neq Y_0 \subsetneq Y_1 \subsetneq \cdots \subsetneq Y_m = A \subsetneq X,$$ and then $m = dim(A) \lneq dim(X)$. Therefore $dim(A) \leq dim(X)$. $\clubsuit$

We know that $dim(\mathbb{A}^n)=n$.

An affine variety is by definition always contained in $\mathbb{A}^n$, which has dimension $n$.

Since $\mathbb{A}^n$ is an affine variety, $\mathbb{A}^n$ is irreducible, then we can apply the previous lemma and conclude that every affine variety is of finite dimension.