Several Complex Variables (VIII): Varieties in $\bb C^n$

18 Sep 2018

• several complex variables

In this post, we introduce the notion of a variety, which is (locally) the common zero set of some holomorphic functions. This geometrical object will be the focus of much of our subsequent study.

Let $U\subseteq\bb C^n$ be a open set. If $S$ is a set of holomorphic functions on $U$, define $V(S)$ to be their common zero set, ie.

$$V(S)=\{z\in U\,:\,f(z)=0\text{ for all }f\in S\}.$$

A holomorphic subvariety of $U$ is a closed subset $V\subseteq U$ which is locally equal to some $V(S)$. In other words, for each $\lambda\in U$ there is a neighbourhood $W_\lambda$ of $\lambda$ and a set $S$ of holomorphic functions on $W_\lambda$ such that

$$V\cap W_\lambda=V(S).$$

We can say that $V$ is “cut out” by the functions in $S$ near $\lambda$.

Note that we may replace $\lambda\in U$ in the definition by $\lambda\in V$, since if $\lambda\not\in V$ then some neighbourhood of $\lambda$ does not intersect $V$, so $V$ is cut out near $\lambda$ by the set of functions $\{1\}$.

Proposition: A finite union (resp. finite intersection) of holomorphic subvarieties of $U$ is also a holomorphic subvariety of $U$.

Proof: It suffices to prove that for two subvarieties $V,\tilde V$, we have $V\cup\tilde V,V\cap\tilde V$ are also subvarieties. The general case then follows by induction.

For $\lambda\in U$, take $W_\lambda,\tilde W_\lambda$ and $S,\tilde S$ as in the definition of a subvariety. Then on $W=W_\lambda\cap\tilde W_\lambda$, we have

$$\begin{aligned} (V\cap\tilde V)\cap W&=V(S)\cap V(\tilde S)\\ &=V(S\cap\tilde S), \end{aligned}$$

so $V\cap\tilde V$ is a subvariety. Similarly,

$$\begin{aligned} (V\cup\tilde V)\cap W&=(V(S)\cup V(\tilde S))\cap W\\ &=V(\{f|_W\tilde f|_W\,:\,f\in S,\,\tilde f\in\tilde S\}), \end{aligned}$$

so $V\cup\tilde V$ is a subvariety. $\qed$

Why do we want to define subvarieties locally, instead of as the common zero set of some globally defined holomorphic functions? We are anticipating the extension of this definition to complex structures other than $\bb C^n$. For example, every subvariety on $\bb C$ should also be a subvariety on the Riemann sphere $\widehat\bb C$. However, all global holomorphic functions on $\widehat\bb C$ are constant, so the only “global subvarieties” are $\emptyset$ and $\widehat\bb C$ itself!

Germs of varieties

Another reason to define subvarieties locally is to the notion of the germ of a subvariety, like we did with the germs of holomorphic functions, which we can then study algebraically. Two subvarieties $V,\tilde V$ are equivalent at $\lambda\in V\cap\tilde V$ if for some neighbourhood $U_\lambda$ of $\lambda$, we have $V\cap U_\lambda=W\cap U_\lambda$. The germ of a subvariety $V$ at $\lambda\in V$ is the equivalence class containing $V$. Now we say that $\bf V$ is the germ of a holomorphic variety at $\lambda\in\bb C^n$ if $\bf V$ is the germ of a subvariety $V$ of some neighbourhood of $\lambda$, where $\lambda\in V$.

Operations and relations on subvarieties carry over to germs as usual; we only elaborate on one example. Given germs of varieties $\bf V,\bf{\tilde V}$ at $\lambda$, choose a common (“actual”) neighbourhood $U$ of $\lambda$ where they have representatives $V,\tilde V$; we then define the intersection $\bf V\cap\bf{\tilde V}$ as the germ of the intersection $V\cap\tilde V$. Similarly, we may define finite intersections, finite unions, and the inclusion relation between subvarieties.

There is also a natural relation between holomorphic functions $f$ and holomorphic subvarieties $V$, namely that $f$ can vanish on $V$ on a neighbourhood of $\lambda$. This notion passes to germs: we say that a germ $\bf f$ of a holomorphic function at $\lambda$ vanishes on the germ $\bf V$ of a variety at $\lambda$ if there is an (“actual”) neighbourhood $U$ of $\lambda$, and representatives $f,V$ in $U$, such that $f$ vanishes on $V$.

As usual, it is routine to check that the above definitions do not depend on the choice of representatives or neighbourhood.

Ideal and locus

The interplay between germs of holomorphic functions and germs of varieties suggests natural algebraic constructions, as follows.

If $\bf V$ is the germ of a holomorphic variety at 0, define the ideal of $\bf V$, denoted $\operatorname{id}\bf V$, as the set of germs in ${}_n\mc H_0$ which vanish on $\bf V$. It is easily checked that this set is in fact an ideal of ${}_n\mc H_0$. Moreover, by definition, any other ideal of ${}_n\mc H_0$ whose elements all vanish on $\bf V$ must be a subset of $\operatorname{id}\bf V$.

Naively, we expect that the analogous definition works when we switch the roles of functions and varieties: if $\mc I$ is an ideal of ${}_ n\mc H_0$, we want to define the locus of $\mc I$ as “the germ of a variety on which all functions of $\mc I$ vanish,” or just “the intersection of the germs of all $V(\{f\})$ for $f\in\mc I$.” The problem is that there are functions in $\mc I$ defined on arbitrarily small neighbourhoods of 0, so there is no neighbourhood on which every function in $\mc I$ has a representative!

The solution is to use the nontrivial result that ${}_n\mc H_0$ is a Noetherian ring, so any ideal $\mc I$ is generated by a finite set, say $\{\bf g_1,\ldots,\bf g_r\}$. Then we can pick a common neighbourhood $U$ on which these have representatives $S=\{g_1,\ldots,g_r\}$, and we define the locus of $\mc I$, denoted $\operatorname{loc}\mc I$, as the germ of $V(\{g_1,\ldots,g_r\})$.

Proposition: $\operatorname{loc}\mc I$ is indeed a germ of a variety on which every element of $\mc I$ vanishes, and it contains every germ of a variety with this property.

This last statement implies that $\operatorname{loc}\mc I$ is well-defined, independent of the choice of neighbourhood and generating set.

Proof: If $\bf g\in\mc I$, then it can be written as

$$\bf g=\bf f_1\bf g_1+\cdots+\bf f_r\bf g_r.$$

On some neighbourhood of 0, all the above functions have holomorphic representatives

$$g=f_1g_1+\cdots+f_rg_r.$$

By shrinking the neighbourhood, we may take a representative subvariety $V$ of $\operatorname{loc}\mc I$ such that every $g_k$ vanishes on $V$; then $g$ vanishes on $V$ also.

If $\bf W$ is a germ of a variety on which every element of $\mc I$ vanishes, it has a representative subvariety $W$ on some neighbourhood of 0. By shrinking the neighbourhood, we may take representatives $g_1,\ldots,g_r$ of the generating set of $\mc I$, which vanish on $W$. By shrinking further, we may take a representative subvariety $V$ of $\operatorname{loc}\mc I$ which is precisely the intersection of the zero sets of $g_1,\ldots,g_r$ in the neighbourhood; hence $W\subseteq V$ in this neighbourhood, so $\bf W\subseteq\operatorname{loc}\mc I$. $\qed$

Now we list some elementary properties of $\operatorname{id}$ and $\operatorname{loc}$.

Proposition:

1. $\bf V_1\subseteq\bf V_2\implies\operatorname{id}\bf V_1\supseteq\operatorname{id}\bf V_2$;
2. $\mc I_1\subseteq\mc I_2\implies\operatorname{loc}\mc I_1\supseteq\operatorname{loc}\mc I_2$;
3. $\bf V=\operatorname{loc}\operatorname{id}\bf V$;
4. $\mc I\subseteq\operatorname{id}\operatorname{loc}\mc I$;
5. $\operatorname{id}(\bf V_1\cup\bf V_2)=(\operatorname{id}\bf V_1)\cap(\operatorname{id}\bf V_2)\supseteq(\operatorname{id}\bf V_1)\cdot(\operatorname{id}\bf V_2)$;
6. $\operatorname{id}(\bf V_1\cap\bf V_2)\supseteq(\operatorname{id}\bf V_1)+(\operatorname{id}\bf V_2)$;
7. $\operatorname{loc}(\mc I_1\cdot\mc I_2)=\operatorname{loc}(\mc I_1\cap\mc I_2)=\operatorname{loc}(\mc I_1)\cup\operatorname{loc}(\mc I_2)$;
8. $\operatorname{loc}(\mc I_1+\mc I_2)=\operatorname{loc}(\mc I_1)\cap\operatorname{loc}(\mc I_2)$.

Proof: Once we choose neighbourhoods and representatives, many of these statements follow from definition and some set theory. We sketch the proofs of the more substantial statements.

(3): ($\supseteq$) Pick a representative subvariety $V$ in a neighbourhood $U$ of 0, so $V$ is the common zero set of some set of holomorphic functions on $U$. But the germs of these functions lie in $\operatorname{id}\bf V$, so $\bf V\supseteq\operatorname{loc}\operatorname{id}\bf V$ by (1).

(7): $\operatorname{loc}(\mc I_1)\cup\operatorname{loc}(\mc I_2)\subseteq\operatorname{loc}(\mc I_1\cap\mc I_2)$ is clear, and $\operatorname{loc}(\mc I_1\cap\mc I_2)\subseteq\operatorname{loc}(\mc I_1\cdot\mc I_2)$ follows from (2).

Pick finite sets of holomorphic functions $S_1,S_2$ on some neighbourhood of 0, whose germs generate $\mc I_1,\mc I_2$ respectively. Then

$$S_1\cdot S_2=\{fg\,:\,f\in S_1,\,g\in S_2\}$$

is a set of holomorphic functions whose germs generate $\mc I_1\cdot\mc I_2$. Now $V(S_1\cdot S_2)=V(S_1)\cup V(S_2)$, so $\operatorname{loc}(\mc I_1\cdot I_2)=\operatorname{loc}(\mc I_1)\cup\operatorname{loc}(\mc I_2)$.

(8): ($\supseteq$) Similar to the above, but with $V(S_1+S_2)=V(S_1)\cap V(S_2)$. $\qed$

Part (4) of the above proposition can be refined into the following (compare with Hilbert’s Nullstellensatz from algebraic geometry):

Rückert’s Nullstellensatz: If $\mc I$ is an ideal in ${}_n\mc H_0$, then

$$\operatorname{id}\operatorname{loc}\mc I=\sqrt{\mc I}\coloneqq\{\bf f\in{}_n\mc H_0\,:\,\bf f^k\in\mc I\text{ for some }k\}.\qed$$

Despite its short statement, we will not be proving the above theorem in these notes, due to its technical nature.

Irreducible varieties

We conclude with a short example of how the algebraic constructions of $\operatorname{id}$ and $\operatorname{loc}$ can give us geometric insight into varieties.

A subvariety $V$ of an open set $U$ is said to be reducible if $V=V_1\cup V_2$, where $V_1,V_2\neq V$ are proper subvarieties of $V$. Similarly, a germ $\bf V$ of a variety at 0 is reducible if $\bf V=\bf V_1\cup\bf V_2$, where $\bf V_1,\bf V_2\neq\bf V$ are germs of varieties at 0 which are properly contained in $\bf V$.

Proposition: A germ of a variety $\bf V$ is irreducible if and only if $\operatorname{id}\bf V$ is a prime ideal.

Proof: ($\Leftarrow$): If $\bf V$ is reducible, say $\bf V=\bf V_1\cup\bf V_2$ with $\bf V_1,\bf V_2\neq\bf V$, then (5) from the previous proposition gives $(\operatorname{id}\bf V_1)\cdot(\operatorname{id}\bf V_2)\subseteq\operatorname{id}\bf V$, while (1) and (3) imply that $\operatorname{id}\bf V_1,\operatorname{id}\bf V_2\neq\operatorname{id}\bf V$. Hence $\operatorname{id}\bf V$ is not a prime ideal.

($\Rightarrow$): If $\operatorname{id}\bf V$ is not a prime ideal, then it contains $\mc I_1\cdot\mc I_2$ for some ideals $\mc I_1,\mc I_2$ not contained in $\operatorname{id}\bf V$. By (2), (3) and (7), $\bf V$ is contained in $\operatorname{loc}(\mc I_1\cdot\mc I_2)=\operatorname{loc}\mc I_1\cup\operatorname{loc}\mc I_2$, but is contained in neither of $\operatorname{loc}\mc I_1$ or $\operatorname{loc}\mc I_2$. Now taking $\bf V_i=\bf V\cap\operatorname{loc}\mc I_i$, we have $\bf V_i\neq\bf V$ and $\bf V=\bf V_1\cup\bf V_2$; hence $\bf V$ is reducible. $\qed$

Proposition: Every germ of a variety can be uniquely (up to order) decomposed into the union of finitely many irreducible germs of varieties

$$\bf V=\bf V_1\cup\cdots\cup\bf V_k,$$

where $\bf V_i\not\subseteq\bf V_j$ if $i\neq j$.

Proof: Suppose that $\bf V$ does not have such a decomposition. In particular, $\bf V$ is not irreducible, so we can write it as $\bf V=\bf V_1\cup\bf V_1’$. At least one of these factors, say $\bf V_1$, does not have a decomposition, and is hence reducible, say $\bf V=\bf V_2\cup\bf V_2’$. Continuing in this way, we get an infinite descending chain of germs of varieties

$$\bf V\supsetneq\bf V_1\supsetneq\bf V_2\supsetneq\cdots$$

By (2) and (3), this gives an infinite increasing sequence of ideals

$$\operatorname{id}\bf V\subsetneq\operatorname{id}\bf V_1\subsetneq\operatorname{id}\bf V_2\subsetneq\cdots,$$

contradicting the fact that ${}_n\mc H_0$ is a Noetherian ring. This proves that $\bf V$ does have a decomposition as a finite union of irreducibles. Deleting redundant terms yields the other required condition.

If $\bf V$ has two such decompositions

$$\bf V=\bf V_1\cup\cdots\cup\bf V_k=\bf V_1'\cup\cdots\cup\bf V_m',$$

then $\bf V_i=(\bf V_i\cap\bf V_1’)\cup\cdots\cup(\bf V_i\cap\bf V_m’)$; since $\bf V_i$ is irreducible, this means that $\bf V_i$ is contained in some $\bf V_j’$. Similarly, this $\bf V_j’$ is contained in some $\bf V_k$. Now by the non-redundancy condition we have $i=k$, so $\bf V_i=\bf V_j’$. Repeating this for all $i$ shows that the $\bf V_j’$ are equal to the $\bf V_i$ up to reordering. $\qed$