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\begin{document}
\centerline{\sc Bryden Cais. \\ Algebraic Geometry, HW 4.\\   \today}

\medskip\noindent
1.  Let $X=(xy=z^2\subset \A^3)$ and let $D=(x=z=0)\subset X$.  We show first that $D$ is not Cartier by showing that it is
not locally principal at $0$.  Indeed, let $\m=(x,y,z)$ be the ideal corresponding to the origin.  Now let $A=k[x,y,z]/(xy-z^2)$
be the coordinate ring of $X$ and set $\p=(x,z)$.  Then $pA_{\m}$ is not principal: indeed, $\dim_k(\m/\m^2)=3$, generated by
the images of $x,y,z$ (since $X$ is not smooth at the origin) and the image of $\p$ in $A_{\m}$ is generated by the images
of $x$ and $z$ and is therefore not principal.  It follows that $D$ is not Cartier.

Now we show that $\cl(X)$ is generated by $D$.  Observe that we have an exact sequence
$$\begin{CD}\Z @>>> \cl(X) @>>> \cl(X-D)\end{CD}$$ where $\Z\rightarrow \cl(X)$ is given by $1\mapsto D$.
Exactness is clear since if $Y$ is a prime divisor of $X$ then $Y\cap(X-D)=0$ if and only if $Y\sim D$.
Now we have $X=\spec A$ as above, and $D$ is given as a subset of $X$ by the single
equation $x=0$ so that $X-D=\spec A_x=\spec k[x,y,z,x^{-1}]/(xy-z^2)=\spec k[x,z,x^{-1}]$.  Now $k[x,z,x^{-1}]$
is a UFD so that $\cl(X-D)=0$.  It follows that we have an exact sequence
$$\begin{CD}\Z @>>> \cl(X) @>>> 0\end{CD}$$
which shows that $\cl(X)$ is generated by $D$.  Now we have shown that $D$ is not locally principal (and hence not principal)
which in particular shows that $\pic(X)=0$.

We finally show that $2D=0$ in $\cl(X)$.  Indeed, the ideal of $D$ is $\p=(x,z)$ so that $y$ is a unit in $A_{\p}$.
Since $xy=z^2$ and $y$ is a unit, it follows that $x\in zA_{\p}$ and hence that $z$ generates the ideal of $D$ in
$A_{\p}$.  Since $D=(x=0)$ and $x\in (z)^2$, we have $v_z(D)=2$ so that $2D=(x)$ is principal, as claimed.
Thus, we have $\cl(X)\simeq \Z/2\Z$.

\medskip\noindent
2.  Let $X$ be a smooth surface and $P\in X$ a closed point.  Let $\pi:\tilde{X}\rightarrow X$ be the blowup of the point $P$
and let $E=\pi^{-1}(E)$.  We then have a map $\pi_*:\cl{\tilde{X}}\rightarrow \cl{X}$
given on prime divisors by $\pi_*(Y)=\pi(Y)$ if $\pi(Y)$ is a divisor (that is, if $Y\neq E)$ and $\pi_*(Y)=0$ if $\pi(Y)$ is not a divisor.
It is clear that $\pi_*$ is surjective as for any prime divisor $Z\subset X$ we have $\pi^{-1}(Z-P)^-$ a prime divisor of $\tilde{X}$
that maps to $Z$ under $P$ (here we are using the fact that $\pi^{-1}:x-P\rightarrow \tilde{X}-E$ is an isomorphism).
Now if $Y\neq E$ is a prime divisor of $\tilde{X}$ that maps via $\pi_*$ to 0 then $\pi(Y)\sim (f)$ for some $f\in k(X)$.
But we have $k(X)=k(X-P)$ so that, since $\pi^{-1}:X-P\rightarrow \tilde{X}-E$ is an isomorphism (in particular a morphism)
we obtain a function $g\in k(\tilde{X}-E)=k(\tilde{X})$ (since $\tilde{X}-E$ is dense in $\tilde{X}$) with $Y=(g)$ so that $Y\sim 0$
as an element of $\cl(\tilde{X})$.  Now if $Y=E$ then by definition of $\pi_*$ we have $\pi_*(Y)=0$.  We have thus shown
that $\ker(\pi_*)$ is generated by $E$.  It remains to show that the map $\Z\rightarrow \cl(\tilde{X})$ given by $1\mapsto E$
is injective.  Again, suppose that $n\mapsto 0$ for some integer $n$ and for the sake of simplicity, suppose that $n>0$.
Then there is some $f\in k(\tilde{X})=k(\tilde{X}-E)$ with $v_E(f)=n$ and $v_Y(f)=0$ for all other prime divisors $Y\subset \tilde{X}$.  Since $\pi:\tilde{X}-E\rightarrow X-P$
is an isomorphism, this gives a function $g=f\circ \pi^{-1}\in k(X-P)=k(X)$ vanishing to order $n$ at the point $P$.
But $(g=0)\subset X$ has codimension 1 so that $g$ must vanish on some divisor $Z\subset X$.  Then $f$ vanishes on $\pi^{-1}(Z-P)^-\neq E$
so that $(f)\neq nE$.  It follows that we have an exact sequence
$$\begin{CD}0 @>>> \Z @>>> \cl(\tilde{X}) @>\pi_*>> \cl(X)@>>> 0\end{CD}.$$

\medskip\noindent
3.  Let $X\subset \P^3$ be a smooth cubic surface.  Then $X$ contains a line $l$.  Recall that $\cl(\P^3)$ is generated by any hyperplane $H$.
Without loss of generality, we may suppose that $l\subset H$.  We show that the induced map $\cl(\P^3)\rightarrow \cl(X)$
given by $Y\mapsto H\cap X$ for any prime divisor $Y$ of $\P^3$ is not surjective.    We claim that for every integer $n$, $n(H\cap X)$ is not linearly
equivalent to $l$.  Since $X$ is a cubic surface, it is given by a degree 3 homogenous polynomial.  Moreover, since we have taken
$H$ to contain $l$, we see that $H\cap X$ consists of $l$ together with a conic $Q$.  We thus have $H\cap X=l+Q$.
Suppose that $nH\cap X\sim l$ and let $H'$ be any hyperplane in $\P^3$ intersecting $H$ in a line.  Then
$n(H\cap H'\cap X)=n(H'\cap l+H'\cap Q)$ is a divisor on the irreducible cubic curve $X\cap H'$ of degree $3n$ (since $H'\cap Q$ is a divisor of degree 2).
On the other hand, $nH\cap H'\cap X\sim l\cap H'$ has degree 1 since $l\cap H'$ consists of a single point.  This is a contradiction since there
is no integer $n$ satisfying $3n=1$, so that $l$ is not in the image of $\cl(\P^3)\rightarrow \cl(X)$, as asserted.

\medskip\noindent
5.  Let $X$ be a smooth curve of genus 1 over a field $k$ of characteristic not equal to 2.  Then there
exists a finite map $\varphi: X\rightarrow \P^1$ of degree 2, ramified over 4 points.  Let $U\subset X$
be some affine open subset mapping to $\A_t^1\subset \P^1$.  Then we have seen that
$k(U)=k(X)$ and that $k[U]$ is the integral closure of $k[t]$ in $k(X)$ (here we use $\varphi$ to realize
$k(t)$ as a subfield of $k(X)$).  Since $f$ is of degree 2, the extension $k(X)/k(t)$ is a degree 2 field extension.
Let $f\in k[U]-k[t]$.  Then since $f\not\in k[t]$ and $k[U]$ is the integral closure of $k[t]$ in $k(X)$, we have that
$\{1,f\}$is a basis for $k[U]$ as a $k[t]$ vector space.  Thus, there exist $\alpha,\beta\in k[t]$ with
$f^2=\alpha f +\beta$.  Since $\Char(k)\neq 2$, it follows that we have $(f-\alpha/2)^2=\beta+\alpha^2/4$
so setting $g=f-\alpha/2$ we see that $g\not\in k[t]$ and that $k(X)=k(t)(g)$.  On the other hand, we clearly have $g^2\in k[t]$,
so write $g^2=\prod (t-e_i)$ for $e_i\in \bar{k}$.  Then since $U$ is affine, we see that
$k[U]=k[t,y]/(y^2-\prod (t-e_i))$ and that $\varphi$ is given by $(t,y)\mapsto t$ (that $\varphi$ takes this form comes
from the fact that we have been viewing $k(t)$ as a subfield of $k(X)$ rather than imbedded via $\phi_*$ in $k(X)$.
This is just for notational convenience and clearly does not change anything).  Over every point $t\in \A^1$ there are
two points (counted with multiplicity) $(t,\pm y)$ satisfying $y^2-\prod (t-e_i)=0$.  These points are distinct unless
$t=e_i$, over which points the map $\varphi$ is ramified.  Since $\varphi$ is ramified at exactly 4 points, we see that $g^2$
has precisely 4 roots.  Thus $k(X)=k(t)(g)$ where $g^2\in k[t]$ is a polynomial of degree 4.



\end{document}
