# 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: $V = V_d \supsetneq V_{d-1} \supsetneq \dots \supsetneq V_0$.

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.
• 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