$\DeclareMathOperator{\pp}{\mathbb{P}} \DeclareMathOperator{\kk}{\mathbb{K}} \DeclareMathOperator{\cc}{\mathbb{C}} \DeclareMathOperator{\ppp}{\mathfrak{p}}$
Having noticed this question 7 years after it was asked, my answer is that the proof from Mumford's "Algebraic Geometry I" is pretty elementary. The point is that
- Chevalley's theorem (for an arbitrary algebraically closed field $\kk$ in place of $\cc$) is an "elementary" consequence of the following two facts:
- The one dimensional projective space $\pp^1$ is proper. In fact one needs the following fact (which is equivalent to properness): the projection map $\kk^m \times \pp^1 \to \kk^m$ is closed (in Zariski topology) for each $m$.
- The "basic" fact that the dimension of a proper (closed) subvariety of an irreducible variety $X$ is smaller than the dimension of $X$. (In "properness" I meant the property that
- The properness of $\pp^n$ (for arbitrary $n \geq 1$) is an "elementary" consequence of the observation that a linear map $\phi:A \to B$ between vector spaces is surjective if and only if there is an $N \times N$ minor, where $N := \dim(B)$, of the matrix of $\phi$ with nonzero determinant.
- The "basic fact" about dimension quoted above also has an "elementary" proof (if dimension of an irreducible variety is defined as the transcendence degree of its field of rational functions) - see e.g. the proof of Proposition 1.14 in "Algebraic Geometry I".
Below is a sketch of the first two points.
Proof of Chevalley's Theorem (assuming properness of $\pp^1$ plus the "basic fact" about dimension):
Step 1. Reduce the general statement of Chevalley's theorem to the following statement: let $\pi:\kk^{m+1} \to \kk^m$, $m \geq 0$, be the projection onto the first $m$-coordinates.
If $V$ is an irreducible subvariety of $\kk^{m+1}$ and $U$ is a nonempty open subset of $V$, then $\pi(U)$ contains a nonempty open subset of the closure of $\pi(V)$.
Step 2. Prove the above statement as follows: Let $W$ be the closure of $\pi(V)$ in $\kk^m$. Since $V$ is irreducible, $W$ is also irreducible. Moreover, $\kk[V] \cong \kk[W][x_{m+1}]/\ppp$ for some prime ideal $\ppp$ of $\kk[W][x_{m+1}]$. If $\ppp = 0$, then $V \cong W \times \kk$ and $\pi$ is the natural projection, and it is easy to see that the above statement is true. So assume $\ppp \neq 0$. Then since $\pi^*: \kk[W] \to \kk[V]$ is injective, it follows that $\dim(V) = \dim(W)$. Now consider the closure $\bar V$ of $V$ in $\kk^m \times \pp^1$. Then $\bar V$ is irreducible (since $V$ is), and $\pi(\bar V)$ is a closed (since $\pp^1$ is proper) irreducible subset of $\kk^m$ containing $W$. Since $\dim(W) = \dim(V) = \dim(\bar V) \geq \dim(\pi(\bar V))$, it follows that
$$W = \pi(\bar V)$$
Let $V' := \bar V \setminus U$. Then $V'$ is a proper closed subset of $\bar V$, so that $\dim(V') < \dim(\bar V) = \dim(W)$. The properness of $\pp^1$ then implies that $\pi(V')$ is a proper closed subset of $W$. Since $\pi(U) \supset \pi(\bar V) \setminus \pi(V') \supset W \setminus \pi(V')$, the proof is complete.
Proof of properness of $\pp^n$: We prove the following statement (which is equivalent to properness of $\pp^n$):
The projection map $\kk^m \times \pp^n \to \kk^m$ is closed (in Zariski topology) for each $m,n$.
Let $Z$ be a closed subset of $\kk^m \times \pp^n$. Choose affine coordinates $(y_1, \ldots, y_m)$ on $\kk^m$ and homogeneous coordinates $[x_0: \cdots : x_n]$. Then $Z$ is the set of zeroes of polynomials $f_1, \ldots, f_k$ in $(x,y)$ which are homogeneous in $(x_0, \ldots, x_n)$. Given $b = (b_1, \ldots, b_m) \in \kk^m$, $b \not \in \pi(Z)$ if and only if there is $d \geq 1$ such that the vector space $P_d$ of homogeneous polynomials of degree $d$ is contained in the ideal of $\kk[x_0, \ldots, x_n]$ generated by $f_1(x,b), \ldots, f_k(x,b)$. In other words, $\kk^m \setminus \pi(Z) = \bigcup_{d \geq 1} U_d$, where
$$U_d := \{b \in \kk^m: P_d \subseteq \langle f_1(x,b), \ldots, f_k(x,b) \rangle\}$$
(where $\langle \cdot, \cdots, \cdot \rangle$ denotes the ideal generated by the enclosed polynomials). Consequently, it suffices to show that $U_d$ is open for each $d$. Now let $d_j := \deg(f_j)$, $j = 1, \ldots, k$, and consider the map
$$\phi_b:P_{d-d_1} \oplus \cdots \oplus P_{d-d_k} \to P_d$$
given by
$$(g_1, \ldots, g_k) \mapsto \sum_j f_j(x,b)g_j$$
Then $b \in U_d$ if and only if $\phi_b$ is surjective if and only if there is an $m_d \times m_d$-minor of $\phi_b$ with nonzero determinant. Since each of these minors is a polynomial in $b$ (once you choose a fixed basis for each $P_e$), it follows that $U_d$ is Zariski open, as required.
