This post is an elementary discussion of understanding points in as points in , where is a (not necessarily algebraically closed) field. The reference is Ravi Vakil’s lecture notes, though the post is not identical to the notes.
Theorem. Given a point , the kernel of the evaluation homomorphism (at the point) given by is . In particular, the ideal is maximal, so .
Remark. It is rather hard to compute the kernel in a naive way. Let’s try it. Suppose that . To show that directly, we need to find such that
But there is a way to get around this difficulty: to get help from morphisms!
Proof. Define by . This is well-defined because if , then so that .
Now, define by . Then
and , so is an isomorphism. Let be the kernel of evaluation. Then we have constructed the following isomorphism
given by . Since zeros of each side correspond, we have . Q.E.D.
Remark. The key of the above proof was to recognize that
in the quotient ring where is a given ideal of the polynomial ring. Let . The above identity can be also understood as
so it is acceptable to write
which is given by the restriction of scalars given by . Notice that our arguments do not need to be isomorphic to ! That’s why I believe that this can be generalized as much as we want (see below).
There is another reason that it is legitimate to write : if we look at the quotient as a -algebra, this is how one defines -action on the quotient module: if is an -submodule of , we define . Both views are valid here!
Question. Can we generalize this to , where is an arbitrary commutative ring with unity?
Answer (not professional). I think we can, but it is more reasonable to assume that is an integral domain so that we can conclude that are prime ideals given by the isomorphism similar to above. However, I think that the isomorphism is given over any commutative rings.
Consider an example: given by . Then we can construct
by . Then we have
. This clearly generalizes to the case where is any commutative ring and we have arbitrarily many finite number of indeterminates. Again, the key is to see . As we saw before, this is merely because polynomials are “constructed by addition and multiplication”. Just like in Theorem, this argument is not about merely giving an isomorphism, but it also shows that
if and only if .
To put it differently, we have
if and only if .
A more ambitious question. If I have not made any incorrect remark, it is natural to ask what is the notion of ideals that is the kernel of . If the ground ring is not an integral domain, we necessarily have , so we might want to make bigger! Let us call such an ideal as a type C ideal. Now we may ask what are the type C ideals in an arbitrary ring and if it enriches the theory! However, this might be just a very shallow question.
Remark. Notice that we have not used Nullstellensatz (weak nor strong) for this discussion. Weak Nullstellensatz is only used to say that over an algebraically closed field, classical points are all the maximal ideals of a polynomial ring over the field.