# Chi exercise, then Divisors, Curves, R-R

### A nice exercise

Claim: Set $X = \mathbb{P}^n$.  For $n\ge 1$$\chi(m) = \chi(\mathcal{O}_{X}(m)) = \binom{n+m}{m}$.  The proof is by induction on n.  For n = 1, its required to show $\chi(m) = m+1$.  For m > -1, $\chi(m) = \dim H^0(X, \mathcal{O}_X(m)) = m+1$ because there are m+1 monomials of degree m in two variables.  For m = -1, all the cohomology vanishes so $\chi(-1) = 0 = m+1$.  For m < -1 there are no global sections, but there is $H^1$.  Serre duality gives $h^1(\mathcal{O}_X(m)) = h^0(\mathcal{O}_X(-2-m)) = -m -1$ so $\chi(m) = - h^1 = m+1$.

Also, for $m=0$, the statement says $\dim H^0(\mathcal{O}_X) = \chi(\mathcal{O}_X) = 1$ which is correct because X only has constant global sections.  So the statement holds for n=1 and all m, and for all n with m = 0.  Now for the induction step, apply the equality $\chi(E) = \chi(E') + \chi(E'')$ to the s.e.s

$0 \to G(m-1) \xrightarrow{\cdot x_n} G(m) \to G'(m) \to 0$

Where $G,G'$ are homogeneous coordinate rings of $\mathbb{P}^n, \mathbb{P}^{n-1}$ respectively.  The binomial identity $\binom{n+m}{m} = \binom{n + (m-1)}{m-1} + \binom{(n-1) + m}{m}$ finishes the proof.

Next Hartshorne went on to prove the equivalent to exercise III.5.2 in Hartshorne (of which the above exercise is a special case).  Unfortunately, I did not take very good notes for the next little bit that Ogus did (this was around the end of march).  But here are some of the things he mentioned.

• Codimension 1 stuff, preparing for Weil divisors
• mentioned GAGA and used it to prove that for $X/\mathbb{C}$ projective, that $h^1(X_an, \mathbb{C}) = h^0(\Omega_{X/\mathbb{C}}) + h^1(\mathcal{O}_X)$
• Blowing up as in II.7 in Hartshorne.  And there was this nice picture

Schematic of blowing up

### Divisors etc.

Divisors on Curves and Riemann-Roch is a popular topic that is treated in many introductory texts on algebraic geometry.  Here’s how Ogus does it.

Cartier divisors were introduced much the same way as in Hartshorne, but Ogus defined Weil divisors for arbitrary schemes.  Namely, for $\eta \in X$, set $ht \eta = \dim \mathcal{O}_{X,\eta}$.  Then

$\mbox{Weil}^+_X$ = free monoid gen. by ht 1 pts.

$\mbox{Weil}_X$ = free group gen. by ht 1 pts.

Then Ogus went on to discuss all the great things that happen with Weil Divisors when X is integral, normal and separated (again, note the lack of Noetherian hypothesis); Ogus used ‘normal’ to replace the condition ‘regular in codimension 1.’  Then he proved the following thm:

Thm: Let $X$ be locally factorial, normal (e.g. regular) and separated.  Then the group of Cartier divisors on X is isomorphic to the group of Weil divisors on X.

Now, some curve stuff.  For Ogus, a curve is a scheme $X$ which satisfies

1. X is quasi compact and finite type over field k
2. $\dim \mathcal{O}_{X,x} = 1$ for every closed pt.
3. $\hom(k(x), \mathcal{O}_{X,x}) = 0$ for every closed pt. (this is something about Cohen-Macaulay)
4. X is separated

Here is some business about the degree of a Weil divisor.  For any $x \in X$, set $\deg x = [k(x): k]$, then

$\deg (\sum_i n_i \cdot x_i) = \sum_i n_i \deg(x_i)$

Thm: For an effective Cartier = Weil divisor D, $\deg D= h^0(D, \mathcal{O}_D) = h^0(X, i_*\mathcal{O}_D)$.

proof: Write $D = \sum_i n_i x_i$ with $n_i \ge 0$.  Set $m_i = m_{x_i} subset \mathcal{O}_{X,x_i}$; it is a principal ideal.  There is a s.e.s

$0 \to I_D \to \mathcal{O}_X \to i_*\mathcal{O}_D \to 0$

Where the ideal sheaf  $I_D = \prod_i m_i^{n_i}$.  By the chinese remainder thm,

$i_*\mathcal{O}_D = \oplus_i \mathcal{O}_X/m_i^{n_i}$

so $h^0(\mathcal{O}_D) = \sum_i h^0(\mathcal{O}_X/m_i^{n_i})$, so I can reduce to the case $D = nx$.  The result holds for $n = 0$.  Also, for $n = 1$, $h^0(\mathcal{O}_D) = h^0(\mathcal{O}_{X,x}/m_x) = [k(x):k]$ which is also correct.  Here’s the induction step.  From the s.e.s of vector spaces

$0 \to m^n/m^{n+1} \to \mathcal{O}_{X,x}/m^{n+1} \to \mathcal{O}_{X,x}/m^n \to 0$

it suffices to prove that $h^0(m^n/m^{n+1}) = \deg x$.  Apparently this follows from smoothness and using the fact that the local rings are DVRs. QED.

With this result the Riemann-Roch thm can be proved for curves.  The notation is that $\mathcal{O}_X(D)$ is the line bundle associated to the divisor $D$; in particular, $\mathcal{O}_X(D) = I_D^{-1}$.

Thm (R-R): For all $D \in \mbox{Div}_X = \mbox{Weil}_X$, $\deg D = \chi(\mathcal{O}_X(D)) -\chi(\mathcal{O}_X)$

proof: First suppose $D$ is effective.  Consider the s.e.s’

$0 \to I_D \to \mathcal{O}_X \to \mathcal{O}_D \to 0$

$0 \to \mathcal{O}_X \to \mathcal{O}_X(D) \to \mathcal{O}_D(D) \to 0$

Note that $D$ is 0 dimensional so its Euler characteristic is just h^0, and since it has finite support, tensoring with a line bundle does not change the dimension of its global sections; that is $\chi(\mathcal{O}_D) = \chi(\mathcal{O}_D(D))$, in fact $\mathcal{O}_D \cong \mathcal{O}_D(D)$.  So the second s.e.s gives

$\chi(\mathcal{O}_X(D)) = \chi(\mathcal{O}_X) + \chi(\mathcal{O}_D)$

using the fact that $\chi(\mathcal{O}_D) = h^0(\mathcal{O}_D) = \deg D$ gives the result in this case.  For the general case, write the divisor as the difference of two effective divisors: $D = D' - D''$.  Then $\mathcal{O}_D = \mathcal{O}(D')\otimes \mathcal{O}(D'')^{-1}$.  From which it follows that $I_D = \mathcal{O}(D'')\otimes \mathcal{O}(D')^{-1}$, so tensoring the s.e.s for $D''$ with $\mathcal{O}(D')$ I get

$0 \to \mathcal{O}(D) \to mathcal{O}(D') \to \mathcal{O}_{D''}(D') \to 0$

from which I conclude that $\chi(\mathcal{O}(D)) = \chi(\mathcal{O}(D')) - \chi(\mathcal{O}_{D''})$, using the first part I get

$chi(\mathcal{O}(D)) = \chi(\mathcal{O}_X) + \deg D' - \deg D'' = \chi(\mathcal{O}_X) + \deg D$

QED

Combining with Serre duality gives the usual statement.

Next Ogus spent some time on criterion for a morphism into projective space to be a closed immersion (it separates points and tangents); the treatment was not radically different than what is in Hartshorne.  Also spent sometime classifying curves of low genus similar to Hartshorne and Miranda.  Here’s a thm to end this post

Thm: Let $f \colon X \to Y$ be a nonconstant morphism of (smooth) curves.  Then $f$ is

1. surjective
2. affine
3. finite
4. flat

proof: $X,Y$ are both projective so $f$ is proper meaning $f(X)$ is connected closed subset that is more than a point, hence all of $Y$.  That the morphism is affine is the statement that a curve minus a finite number of points is affine.  Here’s one way of doing this.  Use global sections $1 = s_0, s_1, ..., s_n \in \Gamma(X, \mathcal{O}(kp))$ for k huge to get a closed immersion $\phi$ into projective space.  The sections $s_i, i > 0$ are nonconstant so they have poles at p, so the way this morphism works (think!) means that $\phi(p) = [0:z_p^{m_1} s_1(p): ...: z_p^{m_n}s_n(p)]$ where $z_p$ is a function defined locally that vanishes at $p$.  The points is the affine $U_o \subset \mathbb{P}^n$ intersected with $\phi(X)$ gives another affine that as a point set is $X - p$.  For the general case, its another exercise to show that a one dimensional affine minus a finite number of points is still affine.

Another general result is that a proper affine map is finite (this is essentially exercise II.4.6 in Hartshorne).  Finally $f_* \mathcal{O}_X$ is a loc. free $\mathcal{O}_Y$ module, which gives flatness, or also one could apply result III.9.7 in Hartshorne.  It says is $Y$ is integral and regular of dimension 1, then f is flat iff all the associated points of $X$ map to the generic point (these are points where $m_x$ consists of zero divisors).  As $X$ is integral, only the generic point of X satisfies this criterion, so the morphism is flat. QED.