Varieties II: Quasiprojective Varieties

August 22, 2009

This post contains the definition of (quasi-) projective varieties, regular functions, morphisms, projective coordinate ring, the Nullstellensatz. It also contains the global section of sheaf of regular function over distinguished open sets and the fact that affine neighborhoods form a basis for the Zariski topology

Projective varieties

Let $k$ be an algebraically closed field. We want to parametrize the 1-dimensional subspaces of $k^{n+1}$ . In coordinates, we take away the origin and identify two points $(a_0,\cdots,a_n)$ and $(b_0,\cdots,b_n)$ if there exists $c \in k$ such that $(a_0,\cdots,a_n) = c(b_0,\cdots,b_n)$ . The resulting space is called the projective space, denoted by $k\mathbb{P}^n$ or just $\mathbb{P}^n$ . The coordinates are denoted as $[a_0,\cdots,a_n]$ , called the homogeneous coordinates.

Again we want to consider polynomials as functions. Contrary to the affine case, polynomials aren’t really functions on a projective space, because of the scalar factor. However, if we consider only the homogeneous polynomials (i.e. those with each monomial having the same degree)
we can define the zero set.

Definition 1 A projective variety $X$ is a subset of $\mathbb{P}^n$ , that is the common zeros of a set of homogeneous polynomials.

Remark

The set of polynomials can be taken to be finite. (Hilbert Basis Theorem)

Some terminologies

1. the Zariski topology

Same as the affine case – consider the topology on $\mathbb{P}^n$ with the closed sets being projective varieties. It is easy to verify that this is a topology:

$\emptyset = \{x \in \mathbb{P}^n | \, 1 = 0 \}$ , $\mathbb{A}^n = \{ x \in \mathbb{P}^n | \, 0 = 0 \}$
Union of two closed sets: $\{x \in \mathbb{P}^n | \, f_i(x) = 0 \} \cup \{x \in \mathbb{P}^n | \, g_j(x) = 0 \} = \{ x \in \mathbb{P}^n | \, f_ig_j(x) = 0\}$
Intersection of closed sets: just put all the equations together.

2. the ideal of X $I(X)$

This is the ideal generated by all the homogeneous polynomials in $k[x_1,\cdots,x_n]$ that vanish on $X$ . (Remember that in general, a polynomial as a function in projective space does not make sense)

It is immediate that $I(X)$ is a graded ideal/homogeneous ideal, i.e. for each $f \in I(X)$ , each of its homogeneous components also lies in $I(X)$ .

3. the projective coordinate ring $\displaystyle S(X) = \frac{k[x_0,\cdots,x_n]}{I(X)}$

Consider $x_0,\cdots,x_n$ as the coordinates of the projective space $\mathbb{P}^n$ , and restrict it on $X$ . If we allow free multiplication and addition, this is what we will get.

The Nullstellensatz

Similar to the affine case, we want to develop a correspondence such that:

for a subset $S \subset \mathbb{P}^n$ , we have the homogeneous ideal $I(S)$ generated by all the homogeneous polynomials vanishing on $S$ .
for a homogeneous ideal $\mathfrak{a} \subset k[x_0,\cdots,x_n]$ , take a generating set of homogeneous polynomials, and call their common zeros $Z(\mathfrak{a})$ .

Similarly, we have

If $S \subset T \subset \mathbb{P}^n$ , then $I(T) \subset I(S)$
If $\mathfrak{a} \subset \mathfrak{b}$ are homogeneous ideals of $k[x_o,\cdots,x_n]$ , then $Z(\mathfrak{b}) \subset Z(\mathfrak{a})$
If $\mathfrak{a}$ is a homogeneous ideal of $k[x_o,\cdots,x_n]$ , $\sqrt{\mathfrak{a}} \subset I(Z(\mathfrak{a}))$
If $S \subset \mathbb{P}^n$ , then $Z(I(S)) = \overline{S}$ (closure in Zariski topology)

Again, we want to find a Nullstellensatz such that the third condition is an equality. However, this time it CANNOT be as nice as the affine case, for such an equality would imply a one-one correspondence between projective varieties of $\mathbb{P}^n$ and radical homogeneous ideals of $k[x_0,\cdots,x_n]$ , which is false. (e.g. common zero of $(x_0,\cdots,x_n)$ is empty, but this ideal is not the entire ring)

This counterexample is actually the worse we can get, as we shall see.

Theorem 1 (Hilbert’s Nullstellensatz, projective case) Let $\mathfrak{a} \subset k[x_0,\cdots,x_n]$ be a homogeneous ideal, where $k$ is algebraically closed field. If $Z(\mathfrak{a}) \neq \emptyset$ , then $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$ .

Proof

It suffices to show that $I(Z(\mathfrak{a})) \subset \sqrt{\mathfrak{a}}$ .

Let $f$ be a homogeneous polynomial such that for some $r$ , $f^r$ vanish on $Z(\mathfrak{a})$ . If $f = 0$ we are done. Suppose not, then we see that $f$ cannot be a constant since $Z(\mathfrak{a}) \neq \emptyset$ . So $f$ is homogeneous of degree at least 1, which means that it would vanish at $(0,\cdots,0)$ when regarded as a function on $k^{n+1}$ . Regard $\mathfrak{a}$ as an usual ideal in $k[x_0,\cdots,x_n]$ and apply the usual Nullstellensatz, we are done. $\Box$

What if $Z(\mathfrak{a}) = \emptyset$ ? Regarding $\mathfrak{a}$ as an ideal of $k[x_0,\cdots,x_n]$ and $Z(\mathfrak{a})$ as a subset of $\mathbb{A}^{n+1}$ , we see that $Z(\mathfrak{a}) = \emptyset$ or $\{0\}$ .

$Z(\mathfrak{a}) = \emptyset \Leftrightarrow \sqrt{\mathfrak{a}} = k[x_0,\cdots,x_n] \Leftrightarrow \mathfrak{a} = k[x_0,\cdots,x_n]$
$Z(\mathfrak{a}) = \{0\}\Leftrightarrow \sqrt{\mathfrak{a}} = (x_0,\cdots,x_n)$

So any homogeneous ideal that contains a power of $(x_0,\cdots,x_n)$ would have empty common zero set in projective space. As in the affine case, it seems more natural to correspond $\emptyset$ and $k[x_0,\cdots,x_n]$ . So $(x_0,\cdots,x_n)$ is called the irrelevant maximal ideal, and if we ignore all the radical ideals that contain $(x_0,\cdots,x_n)$ , we have

Corollary 1 Let $k$ be algebraically closed. The association $X \rightarrow I(X)$ is a 1-1 correspondence between projective subvarieties of $\mathbb{P}^n$ and radical homogeneous ideals of $k[x_0,\cdots,x_n]$ that do not contain $(x_0,\cdots,x_n)$ . Moreover, $X$ is irreducible iff $I(X)$ is prime.

Affine cover of projective varieties

The projective space has a cover by affine space: for the piece $P_i = \{x_i \neq 0 \} \subset \mathbb{P}^n$ , every point $(x_0,\cdots,x_i,\cdots,x_n)$ corresponds to $\displaystyle \left(\frac{x_0}{x_i},\cdots,\frac{x_n}{x_i} \right)$ in $\mathbb{A}^n$ . In the other direction, a point $(x_1,\cdots,x_n) \in \mathbb{A}^n$ corresponds to $(x_1,\cdots,1,\cdots,x_n)$ where 1 is inserted to the $i$ -th position.

We can also carry this cover to a projective subvariety. Let $X \subset \mathbb{P}^n$ be an irreducible projective subvariety. Then $X_i = P_i \cap X$ is closed in $P_i$ , which is like an affine space. If $X_i$ is an affine variety, how would its defining equations be related to that of $X$ ?

Let $I(X) = (f_1,\cdots,f_k)$ , where each $f_j$ are homogeneous polynomials. $X_i$ is then defined as

$\displaystyle X_i = \{\left(x_1,\cdots,x_n\right): \, f_j(x_1,\cdots,1,\cdots,x_n) = 0 \,\, for \, j = 1,\cdots,k\}$

where 1 is inserted to the $i$ -th position. This is exactly the process of dehomogenenization. It is then clear that $I(X_i)$ is exactly the dehomogenization of $I(X)$ .

Notice that if $X$ is irreducible, then the closure of $X_i$ in $\mathbb{P}^n$ is exactly $X$ . (More generally, an open subset $U$ of an irreducible set $X$ is dense. If not, $X = (X - U) \cup \bar{U}$ is a nontrivial decomposition) So we can also work on the “inverse problem”: Given an irreducible affine variety $Y \subset \mathbb{A}^n$ , embed it into $\mathbb{P}^n$ (say, by identifying the affine space with $P_0$ ), then how would the defining equations of $\bar{Y}$ be related to that of $Y$ ?

We can guess the answer – it should be the inverse process of dehomogenization. Given a polynomial in $x_1,\cdots,x_n$ , we can of course insert the powers of $x_0$ in each term such that the polynomial becomes homogeneous. This is called homogenization.

Proposition 1 Given $Y \subset \mathbb{A}^n$ an irreducible affine variety, and embed it into $\mathbb{P}^n$ via $P_0$ . Then $I(\bar{Y})$ is the ideal generated by the homogenization of each term of $I(Y)$ .

Proof

Denote the ideal generated by the homogenization of each term of $I(Y)$ as $\beta(I(Y))$ . We want to show that it is exactly $I(\bar{Y})$ . Note that $\beta(I(Y))$ is generated by homogeneous polynomials, so it is a homogeneous ideal.

Clearly, $Y \subset Z(\beta(I(Y)))$ , so $\bar{Y} \subset Z(\beta(I(Y)))$ , which implies $I(\bar{Y})$ contains $\beta(I(Y))$ .

For the other direction, consider $f \in I(\bar{Y})$ . If we dehomogenize this polynomial by substituting $x_0$ by 1, the new polynomial $\bar{f} \in I(Y)$ . Then when we homogenize back, $f \in \beta(I(Y))$ , implying that $I(\bar{Y}) \subset \beta(I(Y))$ , showing the equality. $\Box$

So far we talked about projective varieties and affine varieties. Just now we have seen that affine varieties can be embedded into the projective space and is a locally closed subset of $\mathbb{P}^n$ . (meaning that it is an intersection of closed subset and open subset) This is the class of varieties we would focus on.

Definition 2 A quasi-projective variety in $\mathbb{P}^n$ is an intersection of a closed subset and an open subset in Zariski topology.

From now on a variety would mean a quasi-projective variety.

Regular functions of quasi-projective varieties

In the affine variety case, we first looked at the quotient of polynomials (with the denominator not vanishing anywhere on the variety), and proved that it’s the same as polynomial function.

In the projective case, our analogue of polynomial is homogeneous polynomials. However, when we want to do quotient, we would want the numerator and the denominator to have the same degree – otherwise the function is still not well defined. So our first attempt is, for a quasi-projective variety $X \subset \mathbb{P}^n$ , a regular function is a quotient of homogeneous polynomials of the same degree with denominator nonvanishing on $X$ . This, however, has some deficiencies.

Motivating case

Consider $\mathcal{C}: X^2+Y^2 - Z^2 \subset \mathbb{P}^2$ minus the point $[0,1,1]$ . The function $\frac{Y+Z}{X}$ defines a regular map from $\mathcal{C}$ minus $[0,1,1]$ , for the denominator vanishes when $X = 0$ , i.e. $Y=Z$ ([0,1,1]) or $Y = -Z$ ([0,1,-1]).

$\frac{Y+Z}{X}$ has another expression on $X^2+Y^2 = Z^2$ , that is $\frac{X}{Z-Y}$ . It is unnatural to distinguish these two as functions, while their denominators do vanish at different places. In particular, the expression $\frac{X}{Z-Y}$ “extends” the definition $\frac{Y+Z}{X}$ to the point $[0,1,-1]$ , so $\frac{Y+Z}{X}$ should make sense as a function on $\mathcal{C}$ , even though the value at $[0,1,-1]$ is not initially defined.

This suggests the following definition,

Definition 3 Let $X \subset \mathbb{P}^n$ be a variety. A regular function $f$ from $X$ to $k$ is a map such that for each $x \in X$ , there exists a neighborhood $U$ around $x$ and homogeneous polynomials $g,h$ of the same degree such that $h$ is nonvanishing on $U$ and $f = \frac{g}{h}$ on $U$ .

We can show that this is consistent with our definition for the affine case. In fact,

Proposition 2

For $D(f) = \{f(x_1,\cdots,x_n) \neq 0 \}\cap X \subset \mathbb{A}^n$ ( $f$ is a polynomial, and $X$ an affine variety), a regular map is of the form $\displaystyle \frac{polynomial}{f^k}$

For $D^{+}(g) = \{g(x_0,\cdots,x_n) \neq 0 \} \cap Y \subset \mathbb{P}^n$ ( $g$ is a homogeneous polynomial, and $Y$ is a projective variety), a regular map is of the form $\displaystyle \frac{homo \, polynomial}{g^k}$ , where the numerator has the same degree as the denominator.

Proof

We will prove the first statement only as these statements are analogous.

First notice that the $D(h)$ sets (as $h$ varies) actually form a basis of the Zariski topology on $X$. (For this reason they are called the distinguished open sets)

A better local representation of regular function

Consider an arbitrary regular function $\alpha: D(f) \rightarrow k$ . For each $x \in D(f)$ , there exists a neighborhood $U_x$ such that $\displaystyle \alpha = \frac{p_x}{q_x}$ on $U_x$ . Since $D(h)$ form a basis, we can assume that $U_x = D(h_x)$ by shrinking if needed.

$q_x$ does not vanish at all on $D(h_x)$ , meaning that $Z((q_x) + I(X)) \subset Z((h_x) + I(X))$ . Taking $I$ on both sides, $h_x \in \sqrt{(q_x) + I(X)}$ , i.e. For some $r_x$ , $h_x^{r_x} = c_xq_x$ on $X$ . Therefore the regular function has another local expression

$\displaystyle \frac{p_x}{q_x} = \frac{c_xp_x}{c_xq_x} = \frac{c_xp_x}{h_x^{r_x}}$

Notice furthermore that $D(h_x) = D(h_x^{r_x})$ . So let us replace $h_x^{r_x}$ by $h_x$ . That means that for any $x$ , there exists a neighorhood $D(h_x)$ such that $\displaystyle \alpha = \frac{p_x}{h_x}$ on $D(h_x)$ .

Putting the presentations together

Notice that $D(f)$ is compact, for if $\displaystyle D(f) = \cup_{a \in A} D(l_a)$ where $l_a$ are polynomials and $A$ is an index set, then $\displaystyle Z(f + I(X)) = \cap_{a \in A} Z(l_a + I(X)) = Z\left(J + I(X)\right)$ , where $J$ is the ideal generated by all $l_a$ . Applying $I$ on both sides, $f \in \sqrt{J + I(X)}$ , i.e. there exists $r$ , $f^r \in J + I(X)$ . Say $f^r = m_1l_1 + \cdots + m_kl_k$ , where $m_i \in k[x_1,\cdots,x_n]$ , as functions on $X$ . Then $\displaystyle D(f) = \cup_{j=1}^k D(l_j)$ , showing compactness.

So suppose that $D(f)$ is covered by $D(h_1) \cup \cdots \cup D(h_k)$ , where $h_i$ are chosen as before. As before, $\sqrt{(h_1,\cdots,h_k) + I(X)} = \sqrt{(f) + I(X)}$ , so there exists polynomials $a_i$ such that

$f^r = a_1h_1 + \cdots + a_kh_k$

This tells us how to patch the local functions $\displaystyle \frac{p_x}{h_x}$ together. They are supposed to be same fraction except that they are defined on different domains, so these functions should all be the same as

$\displaystyle \frac{a_1p_1 + \cdots + a_kp_k}{a_1h_1 + \cdots + a_kh_k} = \frac{a_1p_1 + \cdots + a_kp_k}{f^r}$

which finishes the proof. $\Box$

Remarks

For the projective case, $D^+(g)$ is also a basis.
The regular functions for a variety $X$ clearly form a ring, denoted as $\mathcal{O}(X)$ .

Corollary 2

For an affine variety, a regular function is a polynomial map.

For an irreducible projective variety, a regular function is constant.

Proof

Take $f,g = 1$ . $\Box$

Proposition 3 Let $X$ be an irreducible variety and $f,g: X \rightarrow k$ be two regular maps. If they agree on an open set, then they are the same.

Proof

The set $(f-g)^{-1}(0)$ is closed because $f-g$ is regular, and it contains an open set, which must be dense in $X$ . So $X = (f-g)^{-1}(0)$ . $\Box$

Maps between quasi-projective varieties

The definition of regular functions suggests a local definition for regular maps as well, so let us define

Definition 4 Let $X,Y$ be varieties. For a map $f: X \rightarrow Y$ ,

If $Y$ is quasi-affine (intersection open and closed set in $\mathbb{A}^n$ ), $f$ is regular if each component function is regular.

If $Y$ is quasi-projective, consider the affine cover of $\displaystyle Y = \cup Y_i$ . $f$ is regular if the restrictions of $f$ to $f^{-1}(Y_i) \rightarrow Y_i$ is regular.

There is another definition of regular maps, that makes use of the regular functions and is analogous to differentiable functions.

Proposition 4 Let $X,Y$ be varieties. A map $f: X \rightarrow Y$ is regular if and only if $f$ is continuous and for any regular function $g:V \rightarrow k$ ( $V$ open in $Y$ ), the composition $g \circ f : f^{-1}(V) \rightarrow k$ is also regular.

Proof

$\Rightarrow)$ We first show that if $Y$ is quasi-affine, a regular map is continuous.

Let $C$ be a closed subset of $Y$ , and we want to show that $f^{-1}(C)$ is closed. Consider any $x \in f^{-1}(C)$ . We will show that there exists a neighborhood $U_x$ such that $U_x \cap f^{-1}(C)$ is closed in $U_x$ , then we are done by this lemma.

Lemma 1 Let $Y$ be a topological space with an open cover $\cup U_i$ . A subset $C$ is closed iff $C \cap U_i$ is closed in $U_i$ for all $i$ .

Let $f$ has coordinate functions $f_1,\cdots,f_m$ . We can pick a neighborhood $U_x$ such that all these $f_i$ are of the form $\displaystyle \frac{P_i}{Q_i}$ , where $P_i,Q_i$ are homogeneous of the same degree. If $C$ is the common zeros of polynomials $g_1,\cdots,g_k$ , then $U_x \cap f^{-1}(C)$ is the common zeros of these $g_j$ compose with $latex\displaystyle \frac{P_i}{Q_i}$. This is closed once we clear the denominators.

For the regular function part, take any regular function $g:V \rightarrow k$ . We have an open cover of $V = \cup V_i$ such that each $V_i$ is quasi-affine, and by definition it is clear that $g$ is regular if and only if each restriction to $V_i$ is regular. Thus we may assume that $V$ is quasi-affine by shrinking $V$ if needed. The regularity of $g \circ f$ can be shown using a similar argument of closedness above.

$\Leftarrow)$ If $Y$ is quasi-affine, note that the projection maps $\pi_1,\cdots,\pi_m$ are regular functions, so $f_i = \pi_i \circ f$ is also regular by the hypothesis.

For quasi-projective case, we are done by the following lemma,

Lemma 2 Let $X,Y$ be varieties, and $f:X \rightarrow Y$ be a continuous map such that for any regular function $g: V \subset Y \rightarrow k$ ( $V$ is open in $Y$ ), $g \circ f$ is also regular. If $W \subset Y$ is open and $U \subset X$ is open such that $f(U) \subset W$ , then the restriction of $f$ on $U \rightarrow W$ also satisfies the fore mentioned property.

Proof of Lemma 2

The key is that the restriction of a regular function on an open subset is still regular, which is clear. $\Box$

Corollary 3 The composition of regular maps is regular.

Knowing how to define a regular map, we now have the notion of isomorphism of varieties. From now on a variety that is isomorphic to an affine variety will be called affine. Similarly a variety that is isomorphic to a projective variety will be called projective.

Regular maps and ring of regular functions

A regular map $f: X \rightarrow Y$ induces a $k$ -algebra homomorphism $f^*: \mathcal{O}(Y) \rightarrow \mathcal{O}(X)$ .

Can we generalize Proposition 2 in the last post? Examining the proof, we see that we used the coordinates of $A(Y)$ . For $A(X)$ there, we only regard it as regular functions on $X$ . Thus the proof generalizes to give

Proposition 5 For two varieties $X, Y$ , where $Y$ is affine, we have a natural isomorphism

$\mathrm{Hom}(X,Y) \rightarrow \mathrm{Hom}_{k-algebra}(A(Y), \mathcal{O}(X))$

Projective coordinate ring and ring of regular functions

For affine varieties we have seen that $A(X) \simeq \mathcal{O}(X)$ . However for projective coordinate ring, this is not the case. For example in proposition 2, we see that $\mathcal{O}(\mathbb{P}^n) = k$ , while $S(\mathbb{P}^n) = k[x_0,\cdots,x_n]$ . This shows that affine varieties are quite special.

In general, the ring of regular functions could be wild. It is mentioned in Shafarevich’s book that Rees and Nagata constructed examples of quasiprojective varieties such that $\mathcal{O}(X)$ is not finitely generated, though I can’t find these examples anywhere.

Distinguished open sets

We used this basis in the proof of proposition 2. Since the ring of regular functions on the basis is nice (as shown in the proposition), these sets are important.

Proposition 6 Every point of a variety $X$ has an affine neighborhood.

Proof

If $X$ is quasi-projective, it has an affine open cover, so WLOG assume that $X$ is quasi-affine. As we have seen that $D(f)$ is a basis, we would be done if we can show that $D(f)$ is affine.

Let $D(f) = \{f(x_1,\cdots,x_n) \neq 0\} \cap X$ , where $X$ is affine. Suppose that $X \subset \mathbb{A}^n$ is the common zeros of $f_1,\cdots,f_m$ . Then notice that $D(f)$ is isomorphic to the set $Y \subset A^{n+1}$ defined by $f_1(x_1,\cdots,x_n),\cdots,f_m(x_1,\cdots,x_n) = 0$ , $f(x_1,\cdots,x_n)x_{n+1} = 1$ , which is affine. $\Box$

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recorder smith permalink

November 6, 2009 9:04 pm

dude, where do u go to school ? Cuhk ? or HKU ?

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- soarerz permalink*
  
  November 6, 2009 9:22 pm
  
  I study at UST.
  
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