If is a Noetherian ring, then is a Noetherian ring.

Let an ideal. It suffices to prove that is finitely-generated. For each , set

where is the “leading coefficient”, the coefficient of the top degree monomial in . For example, if we have

where for all and . Certainly, is an ideal. It has a finite set of generators (as is Noetherian), which we denote by

Note that for all . Also, we have the containments

Since is Noetherian, this stabilises at some .

Claim: generates .
Proof. Clearly we have . Conversely, let be a polynomial of minimal degree in . Let . Then there exist such that

If , then and therefore . Otherwise, and , and therefore . Set

Then and (we have peeled off the leading term of ), but . This contradicts the minimality of . □

Let be a Noetherian ring (e.g. a field or PID). If is a finitely-generated -algebra then is Noetherian and finitely-presented.

Note that the converse is not true: not every Noetherian ring is finitely-generated as an algebra. For example, the complex numbers is not finitely-generated as a -algebra.

Let be a Noetherian ring. Let be a multiplicative subset. Then is Noetherian.

If is an ideal, then .

Recall that for an ideal , we defined the vanishing locus

The problem is that we don’t know . Note however that we know for the other definition is non-empty by localisation(?).

Let be a field and and integer. Let . Let and define

Then is maximal.

Conversely, if is algebraically closed and is a maximal ideal, then there exists such that

From this, we know that . For if is a proper ideal, it is contained in a maximal ideal , and therefore

But means precisely that for all . So .

First, let . Let

So . Also, is a surjection: we have a splitting

Therefore is maximal ( is a field). Now for the other inclusion, observe that every polynomial for some . So if , then . The general case can be handled by a change of coordinates.

Conversely, let . Then is a finitely-generated -algebra that is also a field. By Theorem (Zariski’s Lemma)Theorem (Zariski’s Lemma), is a finite extension. But because is algebraically closed, . Consider now the composition

Let . Then

Therefore . So . This inclusion is actually equality because is maximal.

Let be an algebraically closed field. Let be an ideal. Then there is a bijection

Use the ideas in the remark aboveremark above.

Let be a field. Let be a finitely-generated -algebra. If is a field then is a finite extension of .

Being a field means that we have lots of units: we can put many things into denominators. We will show that only way we can have both finitely-generated and many denominators is if we have a finite extension.

If is uncountable, let be transcendental (if it exists). So and is a unit in for all .

Claim: The set is linearly independent.
Proof. Why? Because if we have a linear combination

We have the evaluation homomorphisms

and so for each , we have

This gives us an uncountable set of linearly independent elements in □

However, is a finitely-generated -algebra, and so there exists a presentation

If is countable, then we have a tower

where are algebraically independent, and is finite.

Let be a Noetherian ring. Let be an -algebra homomorphism where is a finitely-generated -algebra and is a finitely-generated -module. Then is a finitely-generated -algebra.

A finite group acting on a finitely-generated algebra: the ring of invariant functions is finitely-generated.