David K. Zhang — March 2^{nd}, 2021

Let be a field. We denote by the ** affine -dimensional space over ** which is simply the set of -tuples of elements of We write when we wish to emphasize that we are

Any set of polynomials defines a ** zero locus**, denoted by:

Conversely, any subset defines a set of polynomials, as follows:

This defines a pair of maps between subsets of and subsets of

The interplay between these two maps forms the basis of algebraic geometry: relating the geometric properties of a set of points to the algebraic properties of a set of polynomials.

We use the notation because this set is always an *ideal* in thought of as a commutative ring under the usual operations of polynomial addition and multiplication. Similarly, note that where denotes the ideal generated by Therefore, instead of arbitrary subsets we may restrict attention to ideals without loss of generality.

An ** affine algebraic set** is a subset of which is the zero locus of some set of polynomials. The

This shows that finite unions and arbitrary intersections of affine algebraic sets are themselves affine algebraic sets, which confirms that the Zariski topology is a legitimate topology.

It follows from Hilbert's basis theorem that is a Noetherian ring. Thus, any affine algebraic set can be written as the zero locus of a finite set of polynomials. This says something profound about affine algebraic sets: they can all be represented using a finite amount of information (modulo the information content of coefficients in ).

In general, the – round-trip enlarges sets. That is, for all and all we have:

Moreover, both and reverse inclusions.

A quick argument shows that equality holds between and if and only if is algebraic.

Later on, we will see that there is also an if-and-only-if condition for equality between and This condition is stated in a theorem called *Hilbert's Nullstellensatz*, which roughly means “zero-locus-theorem” in German.

*Proof:*
First observe that the ideal is the kernel of the evaluation homomorphism which is a surjective ring homomorphism By the first isomorphism theorem, we can write Since an ideal is maximal if and only if its quotient ring is a field, we may conclude that every ideal of the form is maximal.

To prove the converse, suppose we are given a maximal ideal and let Because is a maximal ideal, the quotient ring is a field. Moreover, is finitely generated (by the residue classes ) as an algebra over It follows* that is an algebraic extension of but since is algebraically closed, Hence, the map

is an isomorphism of with For each let and define Then which implies that However, we already know that is a maximal ideal, so ☐

© David K. Zhang 2016 – 2021