# Dominant morphisms with not both varieties irreducible

Why is it that if there is a morphism of algebraic varieties $f: X \longrightarrow Y$ which is dominant and finite, with just $Y$ irreducible, we have $\dim(X) = \dim(Y)$? I know that $\dim(X) \geq \dim(Y)$, but I can't seem to get the other direction. I am trying to prove that the fibres of $f$ are always finite and that implies that $\dim(X) \leq \dim(Y)$. Is this correct?

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Well, you have that the dimension of a general fiber is equal to $\dim X-\dim Y$, and so if you prove that the fibers are finite, you're all set (see Shafarevich for example). Assume that $X$ and $Y$ are affine, and so finiteness means that $k[X]$ is integral over $f^*k[Y]$. Take, then, $t_i$ to be the coordinate functions of $k[X]$. They then satisfy an equation $$t_i^{n_i}+f^*p_{n_i-1,i}t_i^{n_i-1}+\cdots+f^*p_{1,i}t_i+f^*p_{0,i},$$ where the $p_{k,i}$ are in $k[Y]$. If $y\in Y$, then a point $x=(x_1,\ldots,x_n)\in f^{-1}(y)$ (assuming $X$ is in $\mathbb{A}^n$) is going to satisfy the equation $$x_i^{n_i}+p_{n_i-1,i}(y)x_i^{n_i-1}+\cdots+p_{1,i}(y)x_i+p_{0,i}(y).$$ Since $y$ is fixed, the coordinates of $x$ must satisfy these polynomial equations, and so there are only finitely many possibilities. This shows that the fibers are finite.
Thank you! So does the fact that the fibres are finite imply that the dimension of the fibre is zero? Also, was the assumption $Y$ irreducible and $f$ dominant necessary? –  Math2012pc Feb 6 '13 at 0:11
Yes, for varieties dim 0 is equivalent to being a finite set. The assumption that $Y$ is irreducible and $f$ dominant is necessary for saying that the general fiber has dimension $\dim X-\dim Y$. –  rfauffar Feb 6 '13 at 1:14
Robert, after looking through this again, I still have a problem. The fiber dimension theorem (at the least the one I have) is an extension of Chevally's theorem and is valid only for X and Y irreducible. In general, I know that the dimension of the fiber of X when X is reducible is $\max{\{X_i\}}$ where $X_i$ are the irreducible components. I realize that this makes things more complicated in the sense I have to look at the coordinate ring $k[X_i]$, $X_i$ irreducible in order to use the fiber dimension theorem. I am not even sure $X_i$ is algebraic over $Y$. Am I complicating things? –  Math2012pc Feb 6 '13 at 22:58
You're right that I'm considering $X$ to be irreducible. If $X$ is reducible, take $X_1,\ldots,X_r$ to be its irreducible components. Because of what I said earlier, the fibers must be finite (using that the coordinate ring of $Y$ is integrally closed in $k[X]$, etc.). Take $X_i$ to be an irreducible component with $\dim X_i=\dim X$ and such that $f$ restricted to $X_i$ is surjective. This $X_i$ has to exist, because if it doesn't, then $Y$ would be the union of subvarieties of smaller dimension which is a contradiction. Using what I said above, we have that $\dim X=\dim X_i=\dim Y$. –  rfauffar Feb 7 '13 at 19:14