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\begin{document}
\centerline{\sc Bryden Cais. \\ Algebraic Geometry, HW 6.\\   \today}

\medskip\noindent
1.  Let $X=\A_k^1-\{(0,0)\}$, with coordinates $x,y$ and consider $U_1,U_2$ be the affine open cover given by $U_1=(x\neq 0)$ and
$U_2=(y\neq 0)$.  Let $A=k[x,y]$.  Observe that we have $\Gamma(U_1,\O_X)=A_x$, $\Gamma(U_2,\O_X)=A_y$, and $\Gamma(U_1\cap U_2,\O_X)=A_{x,y}$
(where by $A_{x,y}$ we mean the ring $A[x^{-1},y^{-1}]$.  We have
$$C^p(\U,\O_X)=\begin{cases} A_{x}\times A_{y} & \text{if}\ p=0\\ A_{x,y} & \text{if}\ p=1\\ 0 &\text{otherwise}\end{cases}.$$
It follows that $H^1(X,\O_X)=C^1/\im (d:C^0\rightarrow C^1)$, where $d$ is given by $d(f_1,f_2)=f_1-f_2$.
Writing $f_1\in A_x$ and $f_2\in A_y$ as $\hat{f}_1/x^i$ and $\hat{f}_2/y^j$ respectively, where $\hat{f_s}\in A$ for $s=1,2$
and $\hat{f_1}/x\not\in A$, $\hat{f_2}/y\not\in A$, we see that $f_1-f_2=F/x^iy^j$ where every monomial in $F$
either has degree at least $j$ in $y$ or degree at least $i$ in $x$.  On the other hand, every element of $A_{x,y}$
has the form $G/x^ky^l$ with $G$ not divisible by $x$ or $y$.  Writing $G$ as a sum of monomials, we see that we can write
$$\frac{G}{x^ky^l}=\sum c_{i,j}\frac{1}{x^iy^j}+\frac{F}{x^ky^l}$$ with $c_{i,j}\in k$ and $F$
having every monomial with $x$ degree at least $k$ {\em or}
$y$ degree at least $l$.  It readily follows that $H^1(X,\O_X)$ is spanned by $\left\{\frac{1}{x^i}{y^j}:i,j>0 \right\}$
as a $k$-vector space.

\medskip\noindent
2.  Let $X$ be a possibly singular plane curve of degree $d$ so that we have the embedding $i:X\hookrightarrow \P^2$.  Then we have the
exact sequence of sheaves on $\P^2$
$$
\begin{CD}
    0 @>>> \O_{\P^2}(-d) @>>> \O_{\P^2} @>>> i_*\O_X @>>> 0.
\end{CD}
$$
This gives rise to a long exact sequence of cohomology.  In particular, we have
$$
    \ldots \ra  H^1(\P^2,\O_{\P^2}) \ra H^1(\P^2,i_*\O_X )\ra H^2(\P^2,\O_{\P^2}(-d))\ra H^2(\P^2,\O_{\P^2})\ra\ldots.
$$
From our computations of the cohomology of $\P^n$ in class, we know that $H^1(\P^2,\O_{\P^2})=0$.
Now by Serre Duality, we have
$$H^2(\P^2,\O_{\P^2})\simeq H^0(X,\O_{\P^2}(-3))^*=0,$$
since $\O_{\P^2}(-3)$ has no global sections,
so that we have an isomorphism $H^1(\P^2,i_*\O_X )\simeq H^2(\P^2,\O_{\P^2}(-d))^*$.
Again, by Serre Duality we obtain
$$H^2(\P^2,\O_{\P^2}(-d))\simeq H^0(X,\O_{\P^2}(d-3))^*.$$
Since $H^1(\P^2,i_*\O_X )\simeq H^1(X,\O_X)$ we have
$h^1(\O_X)=\dim_k H^0(X,\O_{\P^2}(d-3)).$  But this $k$ vector space is spanned by degree $d-3$ homogenous polynomials in three variables;
it follows that
$$h^1(\O_X)=\binom{d-3+2}{2}=\frac{1}{2}(d-1)(d-2).$$
Since $X$ is a plane curve, we have $\dim X=1$ so that $H^i(X,\O_X)=0$ for $i\geq 2$ whence $h^i(\O_X)=0$ for $i\geq 2$.
Finally, $H^0(X,\O_X)=k$ since $X$ is a projective variety so that $h^0(\O_X)=1$.

\medskip\noindent
3.  Let $X$ be a curve and $D$ a divisor on $X$.

\smallskip\noindent
    (a) Let $f\in H^0(D)=\Gamma(X,\O_X(D))$.  Then $((f)+D)\cap X \geq 0$, so setting $D'=((f)+D)\cap X$
    gives $D'\sim D$ with $D\geq 0$.  This gives a map $H^0(D)\rightarrow |D|$ given by $f\mapsto ((f)+D)\cap X$.  Observe that the image
    of $\tilde{f}$ coincides with the image of $f$ under this map if and only if $\tilde{f}=cf$ for some $c\in k$.
    This gives an isomorphism $\P H^0(D)\simeq |D|$.

\smallskip\noindent
    (b) Recall that for any $f\in k(X)$ we have $\deg (f)=0$ (since $(1:f)$ gives a map $\pi:X\rightarrow \P^1$, we have
    $\deg (f)=\deg \pi \deg ((1:0)-(0:1))=0$).  It follows that if $\deg D <0$ then $\deg (((f)+D)\cap X)=\deg (f)+\deg D=\deg D$
    so that $D$ cannot be equivalent to an effective divisor; hence, by part (a), since $|D|=0$ we have $h^0(D)=0$.

\smallskip\noindent
    (c) Suppose that $\deg D=0$.  Since for any effective divisor $D'$ we have $\deg D'>0$ unless $D'\sim 0$, we see that
    $H^0(D)=0$ unless $D'=D+(f)\sim 0$ for some $f$, i.e. $D\sim 0$.  In this latter case, there is a unique (up to scalar multiplication)
    $f\in k(X)^{\times}$ satisfying $D+(f)=0,$ namely the function $f$ with divisor $-D$ (since $D$ has degree 0, we know such
    a function exists).  Thus, $H^0(D)=kf$ and $h^0(D)=1$, as required.

\smallskip\noindent
    (d) We proceed by induction on the degree of $D$.  Suppose that $h^0(D)\leq \deg D+1$ for all divisors $D$ of degree $k$.
    Now any divisor of degree $k+1$ has the form $D+P$ for some divisor $D$ of degree $k$ and some point $P$.
    Let $h,g\in H^0(D+P)/H^0(P)$ (the quotient of $k$ vector spaces makes sense as $H^0(D)\subset H^0(D+P)$).  Then we may assume
    that $v_P(h)=v_P(g),$ for if not, then $v_P(h)<v_P(g),$ say, so that in fact $h\in H^0(D)$ and $h$ is hence trivial in the quotient.
    But now since $v_P(h)=v_P(g)$, there exist $a,b\in k$ such that $v_P(ah-bg)<v_P(h)$ so that $ah-bg\in H^0(D)$.  It follows that
    $g,h$ are linearly dependent in $H^0(D+P)/H^0(D)$ so that $\dim_k(H^0(D+P)/H^0(D))\leq 1$.  This gives
    $h^0(D+P)\leq h^0(D)+1$, which is the required induction step.  Since the base case holds by part (c) we have shown that
    $$h^0(D)\leq \deg D+1.$$  Now we have already seen that equality holds if $X\simeq \P^1.$  For the converse, suppose that
    equality holds.  Let $P$ be any prime divisor of $D$.  We claim that there exists a function $f\in H^0(D)$ with a single simple pole
    at $P$.  Indeed, if $h,g\in H^0(D)$ are linearly independent with $v_P(h)=v_p(g)$ then there exist $a,b\in k$
    such that $v_P(ah-bg)<v_P(H)$ and $f=ah-bg$ is non-constant since $h,g$ are linearly independent.  The existence of $f$ now follows by descent.
    But then the map $\phi:X\rightarrow \P^1$ given by $(1:f)$ is surjective (since $f$ is nonconstant) and degree 1 (since $f$ has a single simple
    pole) and is hence an isomorphism.

\medskip\noindent
5.  Let $X$ be a curve of genus $g \geq 2$ and let $G=\aut X$ have order $n$.  Then $G$ acts on $K(X)$ with fixed field
$K(X)^G=L$; let $Y$ be the curve corresponding to $L$.  Then the Galois field extension $K(X)/L$ gives a morphism
$f:X\rightarrow Y$ of degree $n$.  Now let $P\in X$ be a ramification point with $e_P=r$. Then for any $Q\in f^{-1}f(P)$
we have $Q=gP$ for some $g\in G$ (since $K(X)/L$ is Galois with Galois group $G$) so that the isotropy subgroups of $G$ at
$P,Q$ are conjugate and therefore have the same index in $G$; since this index is exactly $e_P=e_Q,$ we see that every point
in $f^{-1}f(P)$ has the same ramification index.  Moreover, from the formula $n=\sum_{Q\in f^{-1}f(P)} e_p$, we see that
$f^{-1}f(P)$ consists of $n/r$ points, each having ramification index $r$.  Letting $P_1,\ldots,P_s$ be a maximal set of
ramification points of $X$ over distinct points of $Y$ we obtain, by the Riemann Hurwitz formula
\begin{align*}
    2g-2 &= n(2g(Y)-2)+\sum_{Q} (e_{Q}-1)\\
        &= n(2g(Y)-2)+\sum_{i=1}^s \sum_{Q\in f^{-1}f(Q)}(e_{P_i}-1)\\
        &= n(2g(Y)-2)+\sum_{i=1}^s \frac{n}{r_i}(r_i-1).
\end{align*}
It follows that
$$(2g-2)/n = 2g(Y)-2+\sum_{i=1}^s \left(1-\frac{1}{r_i}\right).$$

We now establish a lower bound for the right hand side.  Indeed, the left hand side is positive so that the right hand side must be also.
Moreover, we have $r_i\geq 2$ and $g(Y)\geq 0$.  Certainly, the minimal value will occur when $g(Y)=0$ since this is always an integer.
Since $r_i\geq 2$, we find that $1/2\leq (1-1/r_i) <1$.  From this, it is clear that in order to have the right hand side above greater than 0,
we must have $2<s$.  Moreover, with $s=3$ and $r_1=r_2=r_3=6$ we get a positive value for the RHS; it follows that $s\leq 3$
so that $s=3$.  We then seek to minimize $1-1/r_1-1/r_2-1/r_3$.  We use the greedy algorithm so that $r_1=2$ and $r_2=3$ whence
$1-1/r_2-1/r_2=1/6$ so we must take $r_3=7$.  It follows that the minimum value is $1-1/2-1/3-1/7=1/42$.
This gives $(2g-2)/n\geq \frac{1}{42},$ or equivalently,
$$n\leq 84(g-1).$$


\medskip\noindent
6.  Let $X$ be a compact differentiable manifold and let $\A$ denote the sheaf of smooth $\C$-valued functions on $X$.  Then the
exponential sequence
$$
\begin{CD}
    0 @>>> \Z @>>> \A @>>> \A^{\times} @>>> 0.
\end{CD}
$$
gives rise to a long exact sequence of cohomology:
$$
    \ldots \ra  H^1(X,\A) \ra H^1(X,\A^{\times} )\ra H^2(X,\Z)\ra H^2(X,\A)\ra\ldots.
$$
Now on Example Sheet 5 we showed using the existence of partitions of unity that $H^i(X,\A)=0$ for all $i>0$.
Thus we obtain the isomorphism $H^1(X,\A^{\times} )\simeq H^2(X,\Z)$ given by the boundary map $c_1$ (the first Chern class).
Recall that we have shown that $\pic X\simeq H^1(X,\A^{\times} )$.  It follows that the first Chern class yields an isomorphism
$$\pic X\simeq H^2(X,\Z).$$

\end{document}
