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\title{Algebraic Geometry, PS 6}
\author{Bryden Cais}
\begin{document}
\maketitle

\section{Lecture 12}

\begin{enumerate}

\item We have seen that lines in $\P^3(K)$ are identified with points of $G(2,4)$.  Let $x\in \P^3(K)$ be any point.  By a projective
change of coordinates, we may assume without loss of generality that $x$ is the point $(1,0,0,0)=e_1$, where $\{e_i\}$ is the
standard basis of $K^4$.  We can write any element of $G(2,4)$ as $f_1\wedge f_2$ for $f_1,f_2\in K^4$.  The condition that
the line fiven by $f_1,f_2$ pass through $x$ is just the condition that
$$f_1\wedge f_2\wedge e_1=0.$$  Writing  $f_1\wedge f_2$ in Pl\"{u}cker coordinates as
$$f_1\wedge f_2=\sum_{i<j} p_{ij}e_i\wedge e_j$$ we see that the condition that $f_1\wedge f_2$ pass through $x$ is given by
$$p_{23}=p_{24}=p_{34}=0.$$  Since we have in addition the condition
$$p_{12}p_{34}-p_{13}p_{24}+p_{14}p_{23}=0,$$ we see that the set of lines passing through $x$, $\pi_x$, is a closed subset of $G(2,4)$ given
by the equations above.  Viewing $\pi_x$ via the Pl\"{u}cker embedding, we define the morphism
$$\varphi:\pi_x\rightarrow \P^2(K)$$ by $\varphi(\{p_{ij}\})=(p_{12},p_{13},p_{14})$.
From the defining equations (in Pl\"{u}cker coordinates) of $\pi_x$ it is clear that $\varphi$ is an isomorphism with inverse given by
$$\varphi^{-1}(x,y,z)=(x,y,z,0,0,0)=(p_{12},p_{13},p_{14},p_{23},p_{24},p_{34})\in G(2,4).$$

Now let $\pi_P$ be the set of lines contained in a given plane $P\subset \P^3(K)$.  Without loss of generality, as before,
we may suppose that $P$ is the span of $e_1,e_2,e_3$.  Then the condition that a line $f_1\wedge f_2\in G(2,4)$ in $\P^3(K)$
be contained in $P$ is just the condition
$$e_1\wedge e_2\wedge e_3\wedge f_1=e_1\wedge e_2\wedge e_3\wedge f_2=0,$$
and writing $f_i=\sum_j a_{ij}e_j$ we see that $f_1\wedge f_2\in \pi_P$ if and only if $a_{14}=a_{24}=0$.

\stepcounter{enumi}
\stepcounter{enumi}
\stepcounter{enumi}
\item \begin{enumerate}
    \item We first show that the set of flags
    $$F(k,l;n+1)=\{(U,W):U\subset W\subset K^{n+1}\}\subset G(k,n+1)\times G(l,n+1)$$ is a closed set.
    Write
    \begin{align*}
        U&=[v_1\wedge v_2\ldots\wedge v_k]\in G(k,n+1)\\
        W&=[w_1\wedge w_2\ldots\wedge w_l]\in G(l,n+1).
    \end{align*}
    Then the condition that $U\subset W$ is precisely the condition that $v_i\in W$ for all $i$, or equivalently, that
    $$v_i\wedge w_1\wedge w_2\ldots\wedge w_l=0$$ for $0\leq i\leq k$.  This gives a system of algebraic equations
    in pl\"{u}cker coordinates, thus showing that $F(k,l)$ is closed.

    Now let $$F(m_1,\ldots,m_r;n+1)\subset G(m_1,n+1)\times\cdots\times G(m_r,n+1)$$ be a flag of type $(m_1,\ldots,m_r)$ and for $i<j$ let
    $$\pi_{i,j}:F(m_1,\ldots,m_r;n+1)\rightarrow F(m_i,m_j;n+1)$$ be the natural projection map given by
    $\pi_{i,j}(V_0,V_1,\ldots,V_r)=(V_i,V_j)$.  Then it is clear that
    $$F(m_1,\ldots,m_r;n+1)=\bigcap_{i<j}\pi_{i,j}^{-1}(F(m_i,m_j;n+1)).$$  Obviously, $\pi_{i,j}$ is continuous for each $i<j$
    so that $\pi_{i,j}^{-1}(F(m_i,m_j;n+1))$ is closed since $F(m_i,m_j;n+1)$ is.  Thus, $F(m_1,\ldots,m_r;n+1)$ is a finite intersection
    of closed sets and hence closed.

    \item Consider the projection map
    $$\pi:F(m_1,\ldots,m_r;n+1)\rightarrow F(m_1,m_2,\ldots,m_{r-1};n+1).$$  The fibre over a point
    $$(V_1,V_2,\ldots,V_{r-1})\in F(m_1,m_2,\ldots,m_{r-1};n+1)$$ consists
    of all points $V_r\in G(m_r,n+1)$ with $V_{r-1}\subset V_r$.  Choosing such a $V_r$ corresponds
    to choosing a subspace $W$ of the orthogonal complement of $V_{r-1}$ in $V=K^{n+1}$ of dimension
    $m_r-m_{r-1}$, so that $V_r:=W\oplus V_{r-1}$.  But the set of all such spaces is identified with
    $$G(m_r-m_{r-1},n+1-m_{r-1}),$$ and this is therefore the fibre over any point.  Since our projection
    map is surjective and all fibres clearly have the same dimension, namely
    $$\dim G(m_r-m_{r-1},n+1-m_{r-1})=(m_r-m_{r-1})(n+1-m_r),$$
    it follows that we have
    $$\dim F(m_1,\ldots,m_r;n+1)=\dim F(m_1,m_2,\ldots,m_{r-1};n+1)+(m_r-m_{r-1})(n+1-m_r).$$
    Repeated application of this formula gives
    $$\dim F(m_1,\ldots,m_r;n+1)=\sum_{i=0}^{r-1}(m_{r-i}-m_{r-i-1})(n+1-m_{r-i}),$$
    where for convenience we set $m_0=0$.
\end{enumerate}

\stepcounter{enumi}
\stepcounter{enumi}
\stepcounter{enumi}
\stepcounter{enumi}

\item Set $k=k_0(t)$ where $k_0$ is an algebraically closed field and let $F\in k[T_0,\ldots,T_n]$ be a homogenous polynomial
of degree $d\leq n$.  We claim that $V(F)(k)\neq \emptyset.$  By clearing denominators, we may assume without loss of generality
that the coefficients of $F$ are all polynomials in $t$.  Let $m$ be the maximum of the degrees of these polynomials and for each $i$
write
$$T_i=\sum_{j=0}^{r} a_{ij} t^j$$ for unknowns $a_{ij}\in k_0$.  Substituting in the $T_i$ to the equation
$$F(T_0,T_1,\ldots,T_n)=0$$
we get a polynomial in $t$ all of whose coefficients must be equal to $0$.  Observe that the degree of this polynomial is
at most
$$m+rd,$$ and hence there are at most $m+rd+1$ equations.  However, there are $rn$ unknowns, and since $d\leq n$ we can choose $r$
large enough so that $1+m+rd < (r+1)(n+1)$ so that there are more unknowns than equations, whence there exists a nonzero solution, as claimed.

\end{enumerate}

\section{Lecture 13}

\begin{enumerate}
\item Let $k$ be an algebraically closed field of characteristic 0 and let $V$ be an irreducible plane cubic.  We know that $V(k)$ has
an inflection point, and without loss of generality (by a projective change of coordinates) we may suppose that the point
$(1,0,0)\in\P^2(k)$ is an inflection point.  It follows that this is the only point in $V(k)$ on the line at infinity (since it
is a triple point and bezout's theorem tells us that the line at infinity has exactly $3$ intersection points with $V(k)$, counting
multiplicity).  We then see that $X$ is given by an equation of the form
$$a_0X^3+a_1X^2 Z+a_2 X Z^2+a_3 Z^3 + b_1 Y^2 Z+b_2 YZ^2+c_0 XYZ=0.$$
Observe that we must have $a_0\neq 0$ (since otherwise our cubic is reducible).  Moreover, since the only point on the line at infinity
is smooth, if $V(k)$ has a singular point, it lies in the affine space $Z=1$.  Therefore, scaling $X$ and dehomogenizing, our cubic
takes the form
$$X^3+a_1X^2 +a_2 X +a_3  + b_1 Y^2 +b_2 Y+c_0 XY=0.$$
Replacing $X$ by $X-a_1/3$  brings this to the form
$$X^3 +a X +b  + c Y^2 +d Y+e XY=0.$$
If $c=0$ and $e=0$ then our cubic is obviously smooth since it is of the form $Y=X^3+aX+b$ (or possibly $X^3+aX+b=0$).  So
we may assume that $e\neq 0$, and then we change $Y$ to $(Y-a)/e$ to bring our cubic to the form
$$X^3 + XY +u  +w Y =0.$$  Then change $X$ to $X-w$ to obtain
$$(X-w)^3+XY+u=0$$  Any singular point $(x,y)$ of a cubic of this form must satisfy
$$3(x-w)^2+y=x=(x-w)^3+xy+u=0,$$ so that we must have $x=0$, $w^2=-y$ and $w^3=u$.  However, if $w^3=u$ then our cubic has the form
$$(X-w)^3+w^3+XY,$$ which is reducible.  Thus we conclude that in this case, there can be no singular cubics, so that any singular
cubic has $c\neq 0$ above.  Thus we can scale $Y$ and bring the cubic to the form
$$X^3 +a X +b  +  Y^2 +u Y+v XY=0.$$  Now change $Y$ to $Y-u/2-vX/2$ to bring it to the form
$$Y^2=X^3+A X^2+B X+C,$$ and change $X$ to $X-A/3$ to finally get our cubic in the form
$$Y^2=X^3+pX+q.$$  A point $(x,y)$ of this cubic is singular if and only if
\begin{align*}
    y^2&=x^3+px+q\\
    y&=0\\
    3x^2&=-p.
\end{align*}
These equations imply that we must have
$$2px+3q=0,$$ and hence that
$$27q^2+4p^3=0.$$  There are then two possibilities: either $p=q=0$ and we obtain the cubic
$$Y^2=X^3$$ which has a cusp at the point $(0,0)$, or $p,q\neq 0$ but $27q^2+4p^3=0$ so that the cubic
$X^3+pX+q$ has a double (but not a triple) root.  Again by a projective change of coordinates we may assume that this
double root is $X=0$, whence (after scaling $X,Y$) we get the cubic
$$Y^2=X^3+X^2,$$ which has a node at $(0,0)$.  Our analysis shows that up to projective changes of coordinates, these are the only
two singular cubics in $\P^2(k)$.
\stepcounter{enumi}
\stepcounter{enumi}
\stepcounter{enumi}
\item Any singular point of the quartic $V(T_0^2T_1^2+T_1^2T_2^2+T_0^2T_2^2-T_0T_1T_2T_3)$ in $\P^3(K)$
must satisfy
\begin{align}
T_0^2T_1^2+T_1^2T_2^2+T_0^2T_2^2-T_0T_1T_2T_3&=0 \label{pt}\\
2T_0(T_1^2+T_2^2)-T_1T_2T_3&=0\label{dif0}\\
2T_1(T_0^2+T_2^2)-T_0T_2T_3&=0\label{dif1}\\
2T_2(T_0^2+T_1^2)-T_0T_1T_3&=0\label{dif2}\\
T_0T_1T_2&=0\label{dif3}.
\end{align}
The last equation gives us three possibilities:
\begin{enumerate}[{Case} 1:]
    \item $T_0=0$.  Then by (\ref{dif2}) we have $T_1T_2=0$ (and it is seen that (\ref{dif1}),(\ref{dif0}),(\ref{pt}) give no further information).
    It is therefore easily checked that the locus of points
    $$(0,0,1,t),(0,1,0,t),(0,0,0,1)$$ for any $t\in K$ are singular points.

    \item $T_1=0$. Clearly, the equation under consideration is invariant under any permutation of $T_0,T_1,T_2$.  We can thus
    immediately write down the locus of singular points in this case:
    $$(0,0,1,t),(1,0,0,t),(0,0,0,1)$$ for any $t\in K$.

    \item $T_2=0$.  Again, we find that the points
    $$(1,0,0,t),(0,1,0,t),(0,0,0,1)$$ for any $t\in K$ are singular points.
\end{enumerate}

In summary, we have found all singular points:
$$(0,0,1,t),(0,1,0,t),(1,0,0,t),(0,0,0,1)$$ for any $t\in K$.

\item Suppose that $X$ is a surface in $\P^3(K)$ that contains three non-coplanar lines passing through a point $x$.  Then clearly,
we have
$$\dim T(X)_x\geq 3.$$  On the other hand, since $X$ is a surface in $\P^3(K)$, it has dimension $2$ so that
$$\dim T(X)_x> \dim X.$$  Since one of the characterizations of a smooth point is
$$\dim T(X)_x = \dim X,$$ we see that $x$ must be singular.

\end{enumerate}
\end{document}