Dimension is an invariant of isomorphism class of projective varieties I'm trying to understand the solution to the following problem - showing that the dimension of a projective variety is an invariant of it's isomorphism class. I'm struggling a bit though. 
My idea was that if $X$ and $Y$ are two projective varieties, and $F$ an isomorphism between them, then by using the property that F is locally polynomial, I could show an irreducible subvariety $U$ of $X$ has image an irreducible subvariety of $Y$ and vice versa. So then chains of irreducible subvarieties in one would correspond to chains of irreducible subvarieties in the other. 
I'm not sure how to go about finishing it off though - I get to stage where I have a union of quasiprojective varities indexed by points in $F(U)$, but not much else. 
 A: You've shown that an irreducible subvariety of $X$ maps under $F$ to an irreducible subvariety of $Y$.**
Further, a strict inclusion $U\subsetneq V$ of subvarieties maps to a strict inclusion $F(U)\subsetneq F(V)$ of subvarieties of $Y$.
If $\dim X = n$, then there is a strictly increasing chain of irreducible subvarieties $U_0\subsetneq U_1\subsetneq \cdots \subsetneq U_n$ of $X$. By the above, this chain must map under $F$ to a strictly increasing chain $F(U_0)\subsetneq \cdots \subsetneq F(U_n)$ of irreducible subvarieties of $Y$. This shows that $\dim Y \geq n = \dim X$. Use the same reasoning with $F^{-1}$ to get $\dim X \geq \dim Y$.
** Edit: I now understand that the question is also asking how to prove this fact. If $U\subseteq X$ is a subvariety, then $U$ is closed in the Zariski topology. Because $F$ is an isomorphism of varieties, it is also, in particular, a homeomorphism of the underlying topological spaces, so $F(U)$ is also closed in $Y$, i.e., a subvariety.
Assume that $U$ is irreducible and suppose we have $F(U) = V_1 \cup V_2$. Then $U = F^{-1}(V_1)\cup F^{-1}(V_2)$, so irreducibility implies that either $F^{-1}(V_1) = U$ or $F^{-1}(V_2) = U$, so that either $V_1 = F(U)$ or $V_2 = F(U)$, showing that $F(U)$ is also irreducible.
