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\begin{document}
\centerline{\sc Bryden Cais. \\ Algebraic Geometry, HW 5.\\   \today}

\medskip\noindent
1.  Let $X$ be a smooth variety over $k$ of dimension $n$ and for a closed point $P$ of $X$,
let $x_1,\ldots ,x_n$ generate $\m=\m_{X,P}$.  Without loss of generality, we may assume that $X$ is affine, say
$X=\spec B$ (by working locally).  Now we have an isomorphism $\m/\m^2\simeq \Omega_{B/k}\otimes_{B} k$ where
$x_i\mapsto dx_i\otimes 1$.  By Nakayama's Lemma, the images of the $x_i$ freely generate $\m/\m^2$
so that under the isomorphism above, we see that $\Omega_X$ is freely generated in a neighborhood of $P$
(since in a neighborhood $\spec B$ of $P$ we have $\Omega_X|U=\tilde{\Omega}_{B/k}$).
It follows that $\omega_X$ is generated by $dx_1\wedge \ldots \wedge dx_n$ in a neighborhood of $P$.

Now suppose that $\omega_X$ is generated by $dx_1\wedge \ldots \wedge dx_n$ over $U$ and by  $dx_1\wedge \ldots \wedge dx_n$
over $V$ where $dx_i=\sum_j \frac{\delta x_i}{\delta y_j}dy_j$.  Then The matrix $L=(\frac{\delta x_i}{\delta y_j})$
gives a linear $L$ transformation from $\omega_X|V$ to $\Omega_X|U$ so that the transition function $g_{UV}$
is given by $\wedge^n L$.  By ordinary linear algebra, this is just multiplication on $\wedge^n \Omega_X$
by $\det L,$ the Jacobian.


\medskip\noindent
3.  Let $X_0,\ldots, X_n$ be homogenous coordinates on $\P^n$ and let $U_i=(X_i\neq 0)$.
Then $U_0$ has coordinates $x_i=X_i/X_0$ for $i\neq 0$ and $U_1$ has coordinates $y_i=X_i/X_1$ for $i\neq 1$.
Observe that $x_i/x_1 = y_i$ for $i\neq 0,1$ and $y_0=1/x_1$.  We then have
$$dy_i = \frac{1}{x_1}dx_i-\frac{x_i}{x_1^2}dx_1$$ for $i\neq 0,1$ and
$$dy_0 = -\frac{1}{x_1^2}dx_1.$$
Now let $s=dx_1\wedge dx_2\wedge\ldots\wedge dx_n\in \Gamma(U_0,\omega_{\P^n})$.
Then we have by problem (1) that
$$s\cap U_1 = -\frac{1}{x_1^2}dx_1\bigwedge_{i>1}\left(\frac{1}{x_1}dx_i-\frac{x_i}{x_1^2}dx_1\right)=-\frac{1}{x_1^{n+1}}dx_1\wedge\ldots\wedge dx_n.$$
It follows that the divisor of zeroes and poles of $s$ is the divisor of zeroes and poles of $1/x_1^{n+1}$,
which is $-(n+1)H$ where $H=(x_1=0)$.  But this is exactly the divisor of zeroes and poles of $\O_{\P^n}(-n-1)$,
from which it follows that $\omega_{\P^n}\simeq \O_{\P^n}(-n-1)$.

\medskip\noindent
4.  Let $X$ be a smooth surface and $\tilde{X}$ the blowup of $X$ at $P$.  Recall that we have the exact
sequence
$$\begin{CD}0@>>> \Z E @>>> \cl\tilde{X} @>\pi_*>> \cl X@>>> 0  \end{CD}.$$
We first claim that $K_{\tilde{X}}=\pi^* K_X +nE$ for some integer $n$.  From the exact
sequence above, it suffices to see that $$\pi_* K_{\tilde{X}} = \pi_*\pi^* K_X = K_X.$$
But this follows from the fact that $\omega_{\tilde{X}-E}$ is the pullback of $\omega_X=\omega_{X-P}$
(since $\tilde{X}-E \simeq X-P$).

Now we show that $n=1$.  Let $U\simeq \A^2$ be an affine subset of $\tilde{X}$
with $P\in \pi(U)$.  We restrict the map $\pi:\tilde{X}\rightarrow X $ to $U$ to obtain a map
$\pi:\A^2_{x,y}\rightarrow \A^2_{u,v}$ where we realize $P$ as $(u,v)=(0,0)$ (in other words, we just
restrict our attention locally about $P$ and coordinatize an affine neighborhood).  Now
since $\pi$ is a blowup, it is given in coordinates on our local patch as $(x,y)\mapsto (x,xy)=(u,v)$.

Now by problem (1), we have that $\omega_{\tilde{X}}$ is generated over $U$ by $dx\wedge dy$ while
$\omega_X$ is generated over $\pi(U)$ by $du\wedge dv$.  It follows from our description of
$\pi$ that the pullback of $\omega_X$ to $\tilde{X}$ is generated over $U$ by $dx\wedge d(xy)=xdx\wedge dy$.
It follows that $K_{\tilde{X}}=(x)+\pi^* K_{X}=E+\pi^* K_X$ since $E=(x=0)$ in coordinates.  Thus $n=1$
and we are done.

\medskip\noindent
5.  Let $X=\P\times\P$ and let $D\subset X$ be a smooth curve on $X$ in the divisor class $(d_1,d_2)$.
Recall that the canonical divisor $K_X$ on $X$ is in the class $(-2,-2)$ (since it is the pullback of the
canonical divisor on $\P^1$ in each component).  Moreover, by the adjunction formula we have
$$K_D=(K_X+D)|_D.$$  Since $\cl(X)\simeq \Z\oplus\Z$, we see that $K_X+D$ is in the divisor class $(-2,-2)+(d_1,d_2)=(d_1-2,d_2-2)$.
We now use this information to determine the degree of $K_D$.  It will suffice to determine
the degrees of the divisors $H_1|_D,H_2|_D$ with $H_1,H_2$ in the divisor classes $(1,0)$ and $(0,1)$ respectively.
Since $D$ is cut out by a polynomial $F$ homogenous in $X_0,X_1$ of degree $d_1$ and homogenous in $Y_0,Y_1$ of degree
$d_2$, we find that $H_1|_D$ has degree $d_2$ (since $H_1$ is in the class of the hyperplane $(X_0=1)\in\cl( \P^1)$,
and without loss of generality we may suppose that this hyperplane intersects $D$ transversally
so that $F|_{H_1}$ is a polynomial in $k[Y_0,Y_1]$ of homogenous degree $d_2$ and has $d_2$ solutions).
Similarly, $H_2|_D$ has degree $d_1$.  It follows that $K_D=(K_X+D)|_D$ has degree $d_2(d_1-2)+d_1(d_2-2)=2(d_1d_2-d_2-d_1+2)$.
Since $2g-2=\deg K_D$, where $g$ is the genus of $D$, it follows that $D$ has genus
$$g=(d_1-1)(d_2-1).$$

\medskip\noindent
6.  Let $X$ be a compact smooth manifold and let $\O_X$ be the sheaf of smooth functions on $X$.
Let $\F$ be a sheaf of $\O_X$-modules on $X$ and $\U=\{U_i\}_{i\in I}$ a finite open cover of $X$.
Let $\rho_i:X\rightarrow \R$ be a partition of unity subordinate to $\U$.  For $q\geq 1$, let $f\in C^q(\U,\F)$
be a cocycle, i.e. suppose that $df=0$.  We show that $f$ is in fact a coboundary.
Define $K\in C^{q-1}(\U,\F)$
$$K_{i_0\ldots i_{q-1}}=\sum_{i\in I}\rho_i f_{ii_0\ldots i_{q-1}}.$$
Observe that since $\overline{\supp{\rho_i}}\subset U_i$ we have $K_{i_0\ldots i_{q-1}}\in \O_X(U_{i_0\ldots i_{q-1}})$
so indeed $K\in C^{q-1}(\U,\F)$.
We have
\begin{align*}
    (dK)_{i_0\ldots i_{q}}&=\sum_{j=0}^q (-1)^j K_{i_0\ldots \hat{i_j}\ldots i_{q}}\\
    &=\sum_{j=0}^q (-1)^j \sum_{i\in I} \rho_i f_{ii_0\ldots \hat{i_j}\ldots i_{q}}\\
    &= \sum_{i\in I} \rho_i \sum_{j=0}^q (-1)^j f_{ii_0\ldots \hat{i_j}\ldots i_{q}}\\
        &= \sum_{i\in I} \rho_i f_{i_0\ldots  i_{q}}\\
        \intertext{since $df=0$ by assumption,}
        &=f_{i_0\ldots i_{q-1}}\\
        \intertext{since $\rho_i$ is a partition of unity.}
\end{align*}
Thus we have $dK=f$ so that $f$ is a coboundary.  It follows that $H^q(\U,\F)=0$ for $q\geq 1$.


\end{document}
