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\begin{document}
\centerline{\sc Bryden Cais. \\ Algebraic Geometry, HW 1.\\   \today}

\medskip\noindent
1.  Let $X=\spec A$ with $P\in X$ and let $U$ be any open neighborhood of $P$.  We show that there is an affine open
neighborhood $V$ of $P$ contained in $U$.  Since $U$ is open, we have $U=X\setminus V(I)$ for some ideal $I\subset A$.
Pick any $f\in I$ such that $f\not\in P$.  This is possible since if $I\subset P$ then $P\in V(I)=\{Q:Q\supset I\}$ and
hence $P\not\in U$.  We know that $D(f)=X\setminus V(f)$ is an open affine neighborhood of $P$ since $P\not\in V(f)=\{Q:Q\supset (f)\}$
and $f\not\in P$, and therefore $P\in X\setminus V(f)=D(f)$.  Moreover, since $f\in I$, and prime ideal $Q$ containing $I$ also contains
$(f)$ so that
$$V(f)=\{Q:q\supset (f)\}\supset \{Q:Q\supset I\}=V(I).$$  It follows that
$X\setminus V(f)\subset X\setminus V(I)$ and hence $D(f)\subset U$.

\medskip\noindent
2.  If $X=\A-\{(0,0)\}$ were an open affine subvariety of $\A$, we would have a surjection $k[x,y]\twoheadrightarrow \O(X)$
given by restriction.  We claim that this surjection is also an injection.  For let $f,g\in k[x,y]$ and suppose that
$f=g$ on $X$.  Then in particular, $f-g$ is a polynomial vanishing on the line $x=0$ without the point $(0,0)$.  But
when restricted to this line, $f-g$ is a polynomial in one variable with infinitely many roots, so that $f-g$ is identically $0$.
Hence, we have an isomorphism $k[x,y]\simeq\O(X)$.  But this is impossible, as $X\not\simeq \A$, so that $X$ is not affine.

\medskip\noindent
3.  Let $A=k[t]/(t^2)$.  Observe that $A$ is a local ring with unique maximal ideal $(t)$.  Moreover, it is not hard to
see that $\spec A=\{(t)\}$ since prime ideals of the quotient $k[t]/(t^2)$ correspond to prime ideals of $k[t]$ containing $(t^2)$,
and it is clear that the only such ideal is $(t)$.  Since $A$ is {\em not} an integral domain, $(0)$ is not a prime ideal.
It follows that the closed sets of $Y=\spec A$ are just $\{\emptyset,Y\}$ and that these are also the open sets.

Let $X$ be any variety over $k$.  Then the associated scheme is $(\spec \mathcal{O}(X),\mathcal{O}_X)$
where $\mathcal{O}_X(U)=\mathcal{O}(U)$ for any open set $U\subset X$.  To define a morphism of schemes
$f:Y\rightarrow X$ we must give a continuous map of topological spaces $f:Y\rightarrow X$ together with a morphism
of sheaves $f^{\#}:\mathcal{O}_X(U)\rightarrow \mathcal{O}_Y(f^{-1}(U))$ for any open $u\subset X$.
Observe that to define $f:Y\rightarrow X$ (as a map of topological spaces) it is enough to pick a single point $x\in X$
as the image of $(t)\in Y$.  Clearly $f$ will be continuous no matter the choice of $x$, since $Y$ has the discrete topology.
We show that to give a map of sheaves $f^{\#}$ corresponds to giving a tangent vector $\phi\in T_x$.

Let $U\subset X$ be open.  If $x\not\in U$ then $f^{-1}(U)=\emptyset$, and hence, by definition,
$\mathcal{O}_Y(f^{-1}(U))=\mathcal{O}_Y(\emptyset)=0$ so that for any $U$ not containing $x$, we define
$f^{\#}(\mathcal{O}_X)(U)=0$.  On the other hand, if $x\in U$, then $f^{-1}(U)=Y$ so for such $U$ we must define
$f^{\#}:\mathcal{O}_X(U)\rightarrow \mathcal{O}_{Y}(Y)=A$.  Now giving such a map $f^{\#}$ for every open $U$ containing $x$
is equivalent to giving a map of stalks
$$\tilde{f}:\mathcal{O}_{X,x}\rightarrow A,$$
with $\tilde{f}$ a local homomorphism of local rings.  However, giving such a map $\tilde{f}$ is equivalent to giving a map
$\bar{f}:\m_{X,x}\rightarrow (t)$ (since for $\varphi\in\O_{X,x}$ we then define $\tilde{f}(\varphi)=\bar{f}(\varphi-\varphi(x))+\varphi(x)$, which
is then a local homomorphism of local rings), and since $(t)=\{at:a\in k\}$, we see that this is equivalent to giving a map
$\phi:\m_{X,x}\rightarrow k$ and then defining $\bar{f}(\varphi)=\phi(\varphi)t$.  However, observe that since $\tilde{f}$
must be a local homomorphism of local rings, we must have $\bar{f}(\varphi^2)=\bar{f}(\varphi)^2=(\phi(\varphi)t)^2=0$ in $A$.
Thus, giving a local homomorphism of local rings $\tilde{f}:\mathcal{O}_{X,x}\rightarrow A$ is equivalent to giving a homomorphism
$\phi:\m_{X,x}\rightarrow k$ that kills $\m_{X,x}^2$, or in other words, a homomorphism $\phi:\m_{X,x}/\m_{X,x}^2\rightarrow k$.
But this is precisely an element of $T_x=(\m_{X,x}/\m_{X,x}^2)^{\ast}$.  Thus we see that a morphism of schemes
$f:Y\rightarrow X$ is given by a point $x\in X$ (to be the image of $(t)$) and a tangent vector $\phi\in T_x$
(which gives the map $f^{\#}$ of sheaves when $f^{-1}(U)\neq \emptyset$).  Conversely, from our discussion it is clear that any morphism $f$
of schemes gives a point $f((t))$ of $X$ and a tangent vector induced by the homomorphism of local rings given
by $f^{\#}$.

\medskip\noindent
4.  Let $\m$ be a maximal ideal of $A$, i.e. $\m$ is a closed point of $X$.  We show that the maximal ideals of $A'$ containing $\m$
lie in the same $\Z/2\Z$ orbit on $X'$.  Let $\m_1,\m_2$ be distinct maximal ideals of $A'$ containing $\m$ and suppose that
$\s$ is a generator for $\Z/2\Z$.  Suppose that $m_1\neq \s\m_2$.  Then $\m_1\nsubseteq \s\m_2$ since $\m_1$ is maximal.
Then there exists $x\in \m_1$ with $x\not\in \s\m_2$.  Put $y=x\s(x)$.  By construction, $y\in \m_1$ since $\m_1$ is an ideal.
Clearly, $y$ lies in the field of fractions of $A$ since
it is $\Z/2\Z$ invariant, and since $A$ is integrally closed and $x\in A'$ is integral over $A$, we have $y\in A\cap \m_1=\m$
(since $A\cap\m_1$ is an ideal of $A$ containing $\m$ and $\m$ is maximal).  But $\m$ is contained in $\m_2$ so that $y\in\m_2$.
Since $\m_2$ is maximal, and hence prime, and $x\not\in\m_2$, we have $\s(x)\in\m_2$, so that, since $\s$ has order 2,
$x\in\s(\m_2)$.  Since $x\in \m_1$ was arbitrary, it follows that $\m_1\subseteq \s\m_2$ so that, since $\m_1$ is maximal,
$\m_1=\s\m_2$, and this is a contradiction.  It follows that $\m_1=\s\m_2$.
Conversely, if $\m_1=\s\m_2$ then $A\cap\m_1=A\cap\m_2$ is a maximal ideal of $A$ so that the closed points
of $X$ correspond to the $\Z/2\Z$ orbits on the closed points of $X'$.



\end{document}
