# How to fix this proof that isomorphic varieties have the same dimension? Is it possible?

I am trying to prove the following:

Show that affine algebraic varieties that are isomorphic have the same dimension.

For completeness let's state the definitions:

Let $V,W$ be varieties. Then they are isomorphic if there is a bijective polynomial map $F: V \to W$.

The dimension of a variety $V$ is the lenght of the longest chain of irreducible subvarieties: $V = V_d \supsetneq V_{d-1} \supsetneq \dots \supsetneq V_0$.

My proof idea was this:

Let $n$ be the dimension of $V$ and assume by contradiction the dimension of $W$ was $>n$. Let $W_i$ denote the chain of irreducible subvarieties in $W$. Then $F^{-1}W_i$ is a chain of irreducible subvarieties of $V$ longer than $n$.

The problem with this proof is that $F^{-1}W_i$ is not an irreducible subvariety it's merely an algebraic set.

Can this proof be fixed? And if not, what would be the correct way to show this?

This is one of the exercises in chapter 1 in Karen Smith's Inivation to algebric geometry.

• Isn't $F$ a homeomorphism? Being irreducible is a topological property (as is dimension, I guess). – Hoot May 24 '16 at 3:29
• A bijective polynomial map is not the same thing as an isomorphism... a counter-example is $t \to (t^2,t^3)$, some $\mathbb{A}^1 \to C$, where $C$ is cut out by $x^3 = y^2$ in $\mathbb{A}^2$. (The problem is that this map is not injective on tangent vectors...) – Lorenzo Najt May 24 '16 at 6:35
• Hello. Corollary 3 (in Chapter 9) in Cox's Ideals, Varieties, and Algorithms is the same theorem. However, the author uses a difference definition (of the dimension of a variety) from yours. He defines the dimension of a variety $\mathbf{V}(I)$ as the degree of the Hilbert polynomial of $\mathbf{I}(\mathbf{V}(I))$. These definitions should be the same although I can't find any reference to prove the equivalence between these definitions. – bfhaha May 2 at 7:50