6.4: Properties of the Ring of Germs

Let us prove some basic properties of the ring of germs of holomorphic functions. First some algebra terminology. Given a commutative ring \(R\), an ideal \(I \subset R\) is a subset such that \(f g \in I\) whenever \(f \in R\) and \(g \in I\) and \(g+h \in I\) whenever \(g,h \in I\). An intersection of ideals is again an ideal, and hence it makes sense to talk about the smallest ideal containing a set of elements. An ideal \(I\) is generated by \(f_1,\ldots,f_k\) if \(I\) is the smallest ideal containing \(\{ f_1,\ldots,f_k \}\). We then write \(I = (f_1,\ldots,f_k)\). Every element in \(I\) can be written as \(c_1 f_1 + \cdots + c_k f_k\) where \(c_1,\ldots,c_k \in R\). A principal ideal is an ideal generated by a single element, that is, \((f)\).

For convenience, when talking about germs of functions we often identify a representative with the germ when the context is clear. So by abuse of notation, we often write \(f \in \mathcal{O}_p\) instead of \((f,p) \in \mathcal{O}_p\) and \((f_1,\ldots,f_k)\) instead of \(\bigl((f_1,p),\ldots,(f_k,p)\bigr)\). As in the following exercises.

Exercise \(\PageIndex{1}\)

1. Suppose \(f \in \mathcal{O}_p\) is such that \(f(p) \not= 0\), and \((f)\) is the ideal generated by \(f\). Prove \((f) = \mathcal{O}_p\).
2. Let \(\mathfrak{m}_p = (z_1-p_1,\ldots,z_n-p_n) \subset \mathcal{O}_p\) be the ideal generated by the coordinate functions. Show that if \(f(p) = 0\), then \(f \in \mathfrak{m}_p\).
3. Show that if \(I \subsetneq \mathcal{O}_p\) is a proper ideal (ideal such that \(I \not= \mathcal{O}_0\)), then \(I \subset \mathfrak{m}_p\), that is \(\mathfrak{m}_p\) is a maximal ideal.

Exercise \(\PageIndex{2}\)

Suppose \(n=1\). Show that \({}_1\mathcal{O}_p\) is a pincipal ideal domain (PID), that is every ideal is a principal ideal. More precisely, show that given an ideal \(I \subset {}_1 \mathcal{O}_p\), then there exists a \(k=0,1,2,\ldots\), such that \(I = \bigl( (z-p)^k \bigr)\).

Exercise \(\PageIndex{3}\)

If \(U,V \subset \mathbb{C}^n\) are two neighborhoods of \(p\) and \(h \colon U \to V\) is a biholomorphism. First prove that it makes sense to talk about \(f \circ h\) for any \((f,p) \in \mathcal{O}_p\). Then prove that \(f \mapsto f \circ h\) is a ring isomorphism.

A commutative ring \(R\) is Noetherian if every ideal in \(R\) is finitely generated. That is, for every ideal \(I \subset R\) there exist finitely many generators \(f_1,\ldots,f_k \in I\): Every \(g \in I\) can be written as \(g = c_1 f_1 + \cdots + c_k f_k\), for some \(c_1,\ldots,c_k \in R\). In an exercise, you proved \({}_1\mathcal{O}_p\) is a PID. So \({}_1\mathcal{O}_p\) is Noetherian. In higher dimensions, the ring of germs may not be a PID, but it is Noetherian.

Theorem \(\PageIndex{1}\)

\(\mathcal{O}_p\) is Noetherian.

Proof

Without loss of generality \(p=0\). The proof is by induction on dimension. By an exercise above, \({}_1\mathcal{O}_0\) is Noetherian. By another exercise, we are allowed to do a change of coordinates at zero.

For induction suppose \({}_{n-1}\mathcal{O}_0\) is Noetherian, and let \(I \subset {}_n \mathcal{O}_0\) be an ideal. If \(I = \{ 0 \}\) or \(I = {}_n \mathcal{O}_0\), then the assertion is obvious. Therefore, assume that all elements of \(I\) vanish at the origin (\(I \not= {}_n \mathcal{O}_0\)), and that there exist elements that are not identically zero (\(I \not= \{ 0 \}\)). Let \(g\) be such an element. After perhaps a linear change of coordinates, assume \(g\) is a Weierstrass polynomial in \(z_n\) by the preparation theorem.

The ring \({}_{n-1}\mathcal{O}_0[z_n]\) is a subring of \({}_n \mathcal{O}_0\). The set \(J= {}_{n-1}\mathcal{O}_0[z_n] \cap I\) is an ideal in the ring \({}_{n-1}\mathcal{O}_0[z_n]\). By the Hilbert basis theorem (see Theorem D.4 in the appendix for a proof), as \({}_{n-1}\mathcal{O}_0\) is Noetherian, the ring \({}_{n-1}\mathcal{O}_0[z_n]\) is also Noetherian. Thus \(J\) has finitely many generators, that is \(J = (h_1,\ldots,h_k)\), in the ring \({}_{n-1}\mathcal{O}_0[z_n]\).

By the division theorem, every \(f \in I\) is of the form \(f = qg+r\), where \(r \in {}_{n-1}\mathcal{O}_0[z_n]\) and \(q \in {}_n\mathcal{O}_0\). As \(f\) and \(g\) are in \(I\), so is \(r\). As \(g\) and \(r\) are in \({}_{n-1}\mathcal{O}_0[z_n]\), they are both in \(J\). Write \(g = c_1 h_1 + \cdots + c_k h_k\) and \(r = d_1 h_1 + \cdots + d_k h_k\). Then \(f = (qc_1 + d_1) h_1 + \cdots + (qc_k + d_k) h_k\). So \(h_1,\ldots,h_k\) also generate \(I\) in \({}_n \mathcal{O}_0\).

Exercise \(\PageIndex{4}\)

Prove that every proper ideal \(I \subset \mathcal{O}_0\) where \(I \not= \{ 0 \}\) is generated by Weierstrass polynomials. As a technicality, note that a Weierstrass polynomial of degree 0 is just 1, so it works for \(I = \mathcal{O}_0\).

Exercise \(\PageIndex{5}\)

We saw above that \({}_1\mathcal{O}_p\) is a PID. Prove that if \(n > 1\), then \({}_n\mathcal{O}_p\) is not a PID.

Theorem \(\PageIndex{2}\)

\(\mathcal{O}_p\) is a unique factorization domain (UFD). That is, up to a multiplication by a unit, every element has a unique factorization into irreducible elements of \(\mathcal{O}_p\).

Proof

Again assume \(p=0\) and induct on the dimension. The one-dimensional statement is an exercise below. If \({}_{n-1}\mathcal{O}_0\) is a UFD, then \({}_{n-1}\mathcal{O}_0[z_n]\) is a UFD by the Gauss lemma (see Theorem D.6).

Take \(f \in {}_n\mathcal{O}_0\). After perhaps a linear change of coordinates \(f = qP\), for \(q\) a unit in \({}_n\mathcal{O}_0\), and \(P\) a Weierstrass polynomial in \(z_n\). As \({}_{n-1}\mathcal{O}_0[z_n]\) is a UFD, \(P\) has a unique factorization in \({}_{n-1}\mathcal{O}_0[z_n]\) into \(P = P_1 P_2 \cdots P_k\). So \(f = q P_1 P_2 \cdots P_k\). That \(P_j\) are irreducible in \({}_n\mathcal{O}_0\) is left as an exercise.

Suppose \(f = \tilde{q} g_1 g_2 \cdots g_m\) is another factorization. The preparation theorem applies to each \(g_j\). Therefore, write \(g_j = u_j \widetilde{P}_j\) for a unit \(u_j\) and a Weierstrass polynomial \(\widetilde{P}_j\). We obtain \(f = u \widetilde{P}_1 \widetilde{P}_2 \cdots \widetilde{P}_m\) for a unit \(u\). By uniqueness part of the preparation theorem we obtain \(P = \widetilde{P}_1 \widetilde{P}_2 \cdots \widetilde{P}_m\). Conclusion is obtained by noting that \({}_{n-1}\mathcal{O}_0[z_n]\) is a UFD.

Exercise \(\PageIndex{6}\)

Finish the proof of the theorem by proving \({}_1\mathcal{O}_p\) is a unique factorization domain.

Exercise \(\PageIndex{7}\)

Show that if an element is irreducible in \({}_{n-1}\mathcal{O}_0[z_n]\), then it is irreducible in \({}_{n}\mathcal{O}_0\).