Holomorphic functions of several variables - 3

In this post, we explore some properties of the ring and analytic germs in .

First we can show that the local ring is a noetherian UFD. Note that we have the results in algebra that

1. if is a UFD, then is a UFD;
2. if is a UFD and are relatively prime, then there exist and nonzero such that .

Theorem. The local ring is a UFD.

Proof. Prove by induction on . When , is a field and clearly a UFD. Suppose is a UFD. Then is a UFD. By Weierstrass preparation theorem, it remains to show that any irreducible factor of a Weierstrass polynomial in is also irreducible in . However, this follows from the facts that any Weierstrass polynomial can be written as a product of irreducible Weierstrass polynomials, and that any irreducible Weierstrass polynomial is irreducible in . These facts can be proved by direct applications of Weierstrass preparation theorem.

Recall that a commutative ring is called noetherian if every ideal in is finitely generated. Hilbert’s basis theorem tells us that if is noetherian then is also noetherian.

Theorem. The local UFD is noetherian.

Proof. Prove by induction on . The result is trivial when . Now suppose that is noetherian. Then is noetherian. Consider any non-trivial ideal and take a nonzero . Up to a change of coordinates, we may assume with a Weierstrass polynomial and a unit. Then . Since is noetherian, the ideal is generated by finitely many elements . For any , Weierstrass division theorem gives with , and then . It follows that . We conclude that is finite generated.

We can consider the zero set of a germ . In fact, what we consider is a germ of sets. Define an equivalence relation on the collection of subsets of by that if and only if there exists an open neighborhood of such that . An equivalence class of with respect to this equivalence relation is called a germ of sets in . Now we can define to be the germ given by the zero set of . Moreover, we can define for a subset its zero set to be the germ given by the common zero set of functions in . Note that the inclusion, union and intersection of germs are all well-defined. We can see that

Proposition. Suppose is irreducible. If vanishes on , i.e., , then divides .

Proof: We may assume is a Weierstrass polynomial with degree . Since is irreducible, and are relatively prime in , and hence there exists a nonzero such that It follows that if has a multiple root in , then is a zero of . Meanwhile, from the identity theorem we see that has an empty interior.

Weierstrass division theorem yields with a Weierstrass polynomial of degree less than . We must have as . Therefore for any outside , has at least distinct roots, implying that for the degree reason. It follows that is identically zero and divides .

Proposition. If is a holomorphic function on a neighborhood of such that is irreducible in , then is irreducible in for in a sufficiently small neighborhood of .

Proof: We may assume is a Weierstrass polynomial. If is reducible in , then with , and hence This shows that the germ of the set of points such that is reducible in is contained in . It suffices to show that is a proper subset of . This is direct from the preceding proposition.

Proposition. If and are holomorphic functions in a neighborhood of such that they are relatively prime in , then and are relatively prime in for in a sufficiently small neighborhood of .

Proof: Still we may assume are both Weierstrass polynomials. Then they are relatively prime as elements in and hence there exist and nonzero such that . This holds on a sufficiently small neighborhood around .

If have a non-unit common factor in , then divides and hence belongs to . It follows that is zero, and hence and are both zero, a contradiction.

Now we consider the relations between ideals of and germs in . A germ in is called analytic if there exist such that as germs. For a germ in , we let to be the set of elements satisfying .

Using the definitions and the fact that is noetherian, we can see that is an ideal of for any germ , and is an analytic germ in for any , where is the ideal generated by . Moreover, the assignments and are both inclusion-reserving. We also have for any analytic germ in and for any ideal .

An analytic germ is called irreducible if any decompostion such that and are analytic germs implies either or .

Proposition. An analytic germ in is irreducible if and only if is a prime ideal in .

Proof: First suppose that is irreducible and . Then we have and hence . Since is irreducible, we must have either or , which implies either or .

Now suppose that is a prime ideal and with analytic germs. If , then , which suggests that is a proper subset of . If we further have , then we can take and . Since , , contradicting to that is a prime ideal.

The final conclusion of the relations between ideals and germs is given by the analytic version of Nullstellensatz.

Theorem (Nullstellensatz). If is an ideal, then where is the radical of , i.e., the ideal of all elements such that for some .

The complete proof is too involved to be presented, so we will just prove the pricipal ideal case of Nullstellensatz.

Proof (of principal case): Suppose is nonzero and . We may assume with a unit and irreducible. Then it suffices to show that each divides . However, this is clear since vanishes on and hence .