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!
 A: 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 !
