# Bijection between a variety $X$ and $\hom(k[X], k)$

I have seen a theorem some time ago, and I don't remember the exact assumptions, so there may be some mistakes. The statement is the following:

If $X$ is an affine variety over an algebraically closed field $k$, then there exists a bijection between $X$ and $\hom(k[X],k)$.

I would like to prove this theorem. What I did so far is to associate for each $x\in X$ the function $\phi_x\in\hom(k[X],k)$, defined by $\phi_x(f) = f(x)$, for each $f\in K[X]$. This association is clearly injective. However, I am not able to prove surjectivity. How can you show that any morphism $\phi: k[X]\to k$ can be written as an evaluation map on some element $x\in X$?

Thanks!

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Choose an embedding of $X$ into affine space, or equivalently, choose generators for the coordinate ring. Then consider the image of the coordinate functions. – Zhen Lin Feb 16 '13 at 20:37

1) Suppose $X=k^n$, so that $k[X]=k[T_1,\cdots,T_n]$.
A $k$-algebra homomorphism $\Phi:k[T_1,\cdots,T_n]\to k$ sends $T_i$ to some $a_i\in k$ and is thus of the form $f(T_1,\cdots,T_n)\mapsto f(a_1,\cdots,a_n)$.
In other words, $\Phi=ev_a$ for $a=(a_1,\cdots,a_n)\in k^n$.

2) If now $X\subset k^n$ is an algebraic set with ideal $I=I(X)\subset k^n$, we have $k[X]=k[T_1,\cdots,T_n]/I$.
A $k$-algebra homomorphism $\phi:k[X]=k[T_1,\cdots,T_n]/I\to k$ is necessarily obtained from a $k$-algebra homomorphism $\Phi:k[T_1,\cdots,T_n]\to k$ vanishing on $I$ by the recipe $$\phi (\overline{f(T_1,\cdots,T_n)})=\Phi(f(T_1,\cdots,T_n))$$
Taking the result in 1) into account we can write $\phi (\overline{f(T_1,\cdots,T_n)})=f(a)$ for some $a\in k^n$, since $\Phi =ev_a$.
But $\Phi$ vanishes on $I$, so that we have $\Phi(f)=f(a)=0$ for all $f\in I$. Hence $a\in X$, since $V(I)=X$.

Conclusion
We have proved that for all affine algebraic sets over $k$, any $k$-algebra homomorphism $\phi:k[X]\to k$ is given by evaluation at some $a\in X$ without using the Nullstellensatz nor the fact that $k$ is algebraically closed !

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I always love to read your answers on this site, very illuminating! – user38268 Feb 16 '13 at 22:35
Thanks, @BenjaLim. – Georges Elencwajg Feb 17 '13 at 0:41