Indeed, let $H'$ be the set of algebra homomorphisms from $k[X_1, . . . ,X_n]$ to $k$. The map $\chi \rightarrow (\chi (X_1), . . . , \chi (X_n))$ is clearly a bijection from $H'$ to $k^n$.

A subset F of V is said to be closed if there exists a family $(f_i)_{i\in I}$ of elements of $A_V$ such that $x\in F\Leftarrow \Rightarrow x\in V$ and $f_i(x) = 0$ for all $i\in I$.

It is clear that $\emptyset$ and V are closed, and that any intersection of closed sets is closed. If F is defined by the vanishing of the $f_i$ and $F'$ by that of the $f'_j$, $F\cup F'$ is defined by the vanishing of the $f_if'_j$, and hence is closed. Thus, there exists a topology on V such that the closed sets for this topology are exactly the closed sets in the above sense. This topology is called the Zariski topology on V. For any $f\in A_V$, we denote by $V_f$ the set of $x\in V$ such that $f(x)\not= 0$; this is an open subset of V. It is clear that the $V_f$ form a base of the Zariski topology. (If $k$ is a topological field, the canonical topology of V is finer than the Zariski topology.)

The map $x \rightarrow \mathfrak{m}_x$ of the Cor. of Prop. 1 can be considered as a map $\varepsilon$ from V to the prime spectrum Spec(A$_V$) of $A_V($Commutative Algebra, Chap. II, §4, no. 3, Def. 4). It is immediate that the Zariski topology is the inverse image under $\varepsilon$ of the topology of Spec(A$_V$).

Since $A_V$ is noetherian, Spec(A$_V$) is noetherian (Commutative Algebra, Chap. II, §4, no. 3, Cor. 7 of Prop. 11), and every subspace of a noetherian space is noetherian (loc. cit., no. 2, Prop. 8). With the notation of the Cor. of Prop. 1, the intersection of the $\mathfrak{m}_x$ is $\{0\}$, and $\{0\}$ is a prime ideal of $A_V$; thus V is irreducible (loc. cit., no. 3, Prop. 14).

It suffices to prove that, for every non-zero element $\varphi$ of $A_V,f(V_{\varphi})$ contains a dense open subset of W. Identify $A_W$ with a subalgebra of $A_V$ by means of the homomorphism associated to $f$. There exists a non-zero element $\psi$ of $A_W$ such that every homomorphism $w: A_W\rightarrow k$ which does not annihilate $\psi$ extends to a homomorphism $v: A_V\rightarrow k$ which does not annihilate $\varphi ($Commutative Algebra, Chap. V, §3, no. 1, Cor. 3 of Th. 1). Now such a $w$ (resp. $v$) can be identified with an element of $W_{\psi}$ (resp. of $V_{\varphi}$) and to say that $v$ extends $w$ means that $f(v) =w$. Hence, $W_{\psi}\subset f(V_{\varphi})$. Q.E.D.

Let $f: V\rightarrow W$ be a polynomial map, and $x_0\in V$. The map $h \rightarrow f(x_0+h)$ from V to W is polynomial. Decompose it into a finite sum of homogeneous polynomial maps:

where $D_i: V\rightarrow W$ is homogeneous of degree $i($Algebra, Chap. IV, §5, no. 10, Prop. 19). The linear map $D_1$ is called the tangent linear map of $f$ at $x_0$. We denote it by $Df(x_0)$.

Applying a translation in V and one in W, we can assume that $x_0= 0$ and $f(x_0) = 0$. The decomposition of $f$ as a sum of homogeneous elements can then be written

$f=f_1+f_2+\cdots$ with deg $f_i=i$,

and the linear map $f_1$ is surjective by hypothesis. Suppose that $f$ is not dominant. Then there exists a non-zero element $\psi$ of $A_W$ such that $\psi \circ f= 0$. Let $\psi =\psi_m+\psi_{m+1}+\cdots$ be the decomposition of $\psi$ into homogeneous elements, with deg $\psi_i=i$ and $\psi_m\not= 0$. Then

where $\rho$ is a sum of homogeneous polynomial maps of degrees $> m$. It follows that $\psi_m\circ f_1= 0$. Since $f_1$ is surjective, $\psi_m= 0$, a contradiction.

This follows from Props. 3 and 4.