# 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.

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.

• How can I know it is a variety though? Locally at each p in the image of a variety, we have a vanishing set coming from the morphism and the defining polynomials of the domain. But then a union of closed sets is not necessarily closed? May 15, 2015 at 17:05
• @hert3583 I added some more details. Is anything else unclear?
– tsa
May 15, 2015 at 23:22
• I guess what I'm really confused about is why is this isomorphism a homomorphism. Apologies for my badly put together question! I cab understand why if we were looking at affine varieties it would be so, because morphisms is affine varieties are globally polynomial. But morphisms of projective varieties are only locally polynomial, and that's what I'm having trouble dealing with. May 16, 2015 at 10:33
• @hert3583 An isomorphism is defined to be a morphism with an inverse. Any morphism $f$ is continuous, so if it's an isomorphism there is an inverse map $g$ (which must also be continuous, since it's a morphism) so that $f\circ g$ and $g\circ f$ is the identity. In particular, $f$ has a continuous inverse so is definitionally a homeomorphism. Thinks about it as algebro-geometric morphisms are stricter than topological ones--they preserve even more data (i.e. they take polynomial functions to polynomial functions). Thus, an algebro-geometric isom. is a top. isom. May 16, 2015 at 11:17
• @AlexYoucis so is the morphism continuous because any x in the preimage must be s.t. F(x) satisfies a collection of polynomial equations. As F takes these polynomial equations to other polynomial equations, x satisfies a collection of polynomial equations? The variety defined by these will be defined by the polynomials defining F near x. So if we take a union of these over x in X, we get a variety and then intersect it with X to get a subvariety in X which is the preimage of whatever subvariety in Y we are considering? May 16, 2015 at 12:26