# Is evaluation homomorphism surjective?

Let $A^n$ be an affine space over $\mathbb{C}$ and let $\mathbb{C}[X_1,\cdots,X_n]$ be the polynomial ring of $n$ variables. Then $A^n\to (\mathbb{C}[X_1,\cdots,X_n])^*$ by evaluation homomorphism, where $(\mathbb{C}[X_1,\cdots,X_n])^*$ is the set of ring homomorphisms from $\mathbb{C}[X_1,\cdots,X_n]$ to $\mathbb{C}$. Is this homomorphism surjective?

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By ring homomorphism, do you mean $\mathbb{C}$-algebra homomorphism? If so, a $\mathbb{C}$-algebra homomorphism $\mathbb{C}[X_1,\ldots, X_n]\to \mathbb{C}$ is determined by where you send the elements $X_i$, which is just evaluation, like you want. If you actually mean ring homomorphism, then any ring homomorphism $\mathbb{C}[X_1,\ldots, X_n]\to \mathbb{C}$ that is not a $\mathbb{C}$-algebra homomorphism cannot come from evaluation, since evaluation is a $\mathbb{C}$-algebra homomorphism. – froggie Feb 28 '14 at 9:52

An example ring homomorphism $\mathbb{C}[X_1,\ldots, X_n]\to \mathbb{C}$ that is not a $\mathbb{C}$-algebra homomorphism is the map $\phi(P) = \overline{P(0)}$. This does cannot arise from evaluation, since evaluation maps are $\mathbb{C}$-algebra homomorphisms.
On the other hand, as explained in the comments, if $\psi\colon \mathbb{C}[X_1,\ldots, X_n]\to \mathbb{C}$ is a $\mathbb{C}$-algebra homomorphism, then $\psi$ is determined by the values $\psi(X_i) = z_i$, and $\psi$ is the evaluation map at the point $(z_1,\ldots, z_n)\in \mathbb{A}^n$.
It is not surjective. There are bijections $$\mathbb A^n\cong \textrm{Hom}_{\mathbb C\textrm{-Var}}(\textrm{Spec }\mathbb C,\mathbb A^n)\cong \textrm{Hom}_{\mathbb C\textrm{-Alg}}(\mathbb C[X_1,\dots,X_n],\mathbb C),$$ and the latter is strictly contained in $\textrm{Hom}_{\textrm{Ring}}(\mathbb C[X_1,\dots,X_n],\mathbb C)$. Now, the way the isomorphisms are constructed is precisely by means of evaluations. So evaluation cannot surject onto the set of ring homomorphisms.
I'm a little unsure about explanations (1) and (2) here.... For (1): How is the set of ring homomorphisms $\mathbb{C}[X_1,\ldots, X_n]\to \mathbb{C}$ a complex vector space? For (2): I don't think what you've described is actually a ring homomorphism. For instance, if $\phi$ is the map you've defined, then $$0 = \phi(X_1 - 2) = \phi(X_1 - 1 + 1) = \phi(X_1-1)+\phi(1) = 0 + 1 = 1.$$ – froggie Feb 28 '14 at 10:20