(One version of) Sard's theorem states that:
Theorem (Sard): Given $M$ and $N$ smooth manifolds of dimensions $m$ and $n$ respectively, and a smooth map $f:M\to N$, then the set of singular values of $f$ has measure zero.
A corollary of this is:
Corollary: If $m<n$, then there exists no smooth surjective map $f:M\to N$.
Now, I am quite certain that my proof of the following fact is correct:
Claim: Let $X$ and $Y$ be two quasi-projective, irreducible algebraic varieties (over an algebraically closed field $k$) of dimensions $m$ and $n$ respectively. If $m<n$, then there exists no smooth surjective map $f:X\to Y$.
Proof: Assume there was such a map $f:X\to Y$. Then it would induce an injective map $f^*:K(Y)\to K(X)$ between the fields of rational functions of the two varieties. We know that: $$\begin{array}{l}K(X)=k(x_1,\ldots,x_m)[u]\\K(Y)=k(y_1,\ldots,y_n)[v]\end{array}$$ where $k(x_1,\ldots,x_m)$ is a purely transcendental field extension of $k$, and $u$ is algebraic over $k(x_1,\ldots,x_m)$, and similarly for $K(Y)$. Consider the elements $f^*(y_i)\in K(X)$. They are transcendental over $k$, indeed if there was a polynomial $p\in k[z]$ with: $$0=p(f^*(y_i))=f^*(p(y_i))$$ then by injectivity of $f^*$ we would have $p(y_i)=0$, which is a contradiction. Similary, these elements are algebraically independent, since: $$0=p(f^*(y_1),\ldots,f^*(y_n))=f^*(p(y_1,\ldots,y_n))$$ would imply algebraical dependence of the $y_i$'s. So, the elements $f^*(y_i)$ form an algebraically independent set of transcendental element of cardinality $n$ in $K(X)$, which contradicts the assumption that $X$ had dimension $m<n$.
This last statement looks almost exactly the same as the previously stated corollary to Sard's theorem, so I was wondering: is there a version of Sard's theorem for algebraic varieties, or a similar result that would imply the claim above?