# 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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## 1 Answer

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.

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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
Does that make sense? – rfauffar Feb 7 '13 at 19:16