One-One correspondence between elements of G (algebraic group) and maximal ideals of K[G] We are given that $G$ is an algebraic group over $K$ and $K[G]= K[x_1,...,x_n]/I(G)$ where $I(G)$ is the ideal consisting of all polynomials of which elements of $G$ are the common zeroes. Now,if we consider any closed subset $H$ of $G$, we can find an ideal $J$ of $K[x_1,...,x_n]$ containing  $I(G)$ or equivalently an ideal $<J/I(G)>$ in $K[G]$ of which elements of $H$ are precisely the common zeroes. We can check that $<J/I(G)>$ is a radical ideal. Similarly, if we are given an irreducible closed set in $G$, the ideal $<J/I(G)>$ will turn out to be a prime ideal. (please correct if I am wrong in understanding what I have stated above).
Now, I read that elements of $G$ correspond to maximal ideals in $K[G]$. I think it goes this way :
Given an element $g \in G$ consider the ideal $K= <p+ I(G)>$ where $p$ is the one degree polynomial satisfied by $g$. Is this the required maximal ideal and how do we ensure that it is proper ? Also, conversely if we are given a maximal ideal in $K[G]$, how do we see that the common zeroes of the ideal is a single point ? I guess it is like if the common zeroes are 2 or more, then we can find an ideal of the above type containing the given maximal ideal. Please help me by providing a clear explanation.
 A: First, prove things when $G$ is all of $K^n$.  Prove, using the fact that $K$ is algebraically closed, that these three things are in bijection:
-Points in $K^n$.
-Maximal ideals in $K[X_1, ... , X_n]$
-$K$-algebra homomorphisms $K[X_1, ... , X_n] \rightarrow K$
For example, if $x = (x_1, ... , x_n)$ is a point in $K^n$, then evaluation at $x$ gives a surjective homomorphism $K[X_1, ... , X_n] \rightarrow K$, whose kernel must be a maximal ideal.
On the other hand, if $\mathfrak m$ is any maximal ideal, then $F := K[X_1, ... , X_n]/\mathfrak m$ is a field which is finitely generated as an algebra over $K$.  Zariski's lemma then tells us that $F$ is algebraic over $K$, hence $K = F$.  In other words, the natural injection $K \rightarrow K[X_1, ... , X_n]/\mathfrak m, x \mapsto x + \mathfrak m$ is actually an isomorphism, and in particular, there are points $x_i \in K$ such that $x_i + \mathfrak m = X_i + \mathfrak m$.  Hence $X_i - x_i \in \mathfrak m$ implies $(X_1 - x_1, ... , X_n - x_n) \subseteq \mathfrak m$, which implies $(X_1 - x_1, ... , X_n - x_1) = \mathfrak m$.
Once you're comfortable with these bijections, suppose $G$ is a closed set in $K^n$ with corresponding ideal $\mathfrak a = I(G)$, and prove that the bijections above induce new bijections between:
-Points in $G$
-Maximal ideals of $K[X_1, ... , X_n]$ containing $\mathfrak a$ (that is, maximal ideals of $K[G]$)
-$K$-algebra homomorphisms $K[G] \rightarrow K$.
For example if $g = (x_1, ... , x_n)$ is a point of $G$, then we know that $g$ corresponds to the maximal ideal $\mathfrak m = (X_1 - x_1, ... , X_n-x_n)$.  But $\{g\} \subseteq G$ implies $I(G) \subseteq I(\{g\})$, or $\mathfrak a \subseteq \mathfrak m$.
