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\title{Algebraic Geometry, PS 4}
\author{Bryden Cais}
\begin{document}
\maketitle

\section{Lecture 7}
\exer{1}
    Let $X$ be a projective $d$ subspace in $\P^n_k$.  By definition, $X$ is given by $n-d$ linearly independent linear equations.
    By a projective change of coordinates, $X$ is isomorphic to the projective subspace of $\P^n_k$ given by the linear equations
    $$\{T_0=0,\ T_1=0,\ \ldots,\ T_{n-d}=0\}.$$  Then projection onto the last $d+1$ coordinates gives an isomorphism
    of $X$ with $\P_k^d$, as required.


\exer{2}
    Consider the {\em Segre} embedding $s(1,1)_K(\P^1_k(K)\times\P^1_k(K))\rightarrow \P^3_k(K)$ for any $K$.  It is defined by
    $$s(1,1)_K([X_0,X_1],[Y_0,Y_1])=[X_0Y_0,X_0Y_1,X_1Y_0,X_1Y_1]=[Z_0,Z_1,Z_2,Z_3].$$  Let $Q$ be the hypersurface in
    $\P^3_k$ given by
    \begin{align}
    Z_0Z_3-Z_1Z_2=0\label{defofQ}
    \end{align}
      It is clear that the image of $\P^1_k\times \P^1_k$ under $s(1,1)_K$ is contained
    in $Q$ since any point in the image satisfies \ref{defofQ}.  We show that the two are in fact isomorphic.  Define
    $f:Q\rightarrow \P_k^1\times\P_k^1$ by
    \begin{align*}
    f([Z_0,Z_1,Z_2,Z_3])&=\begin{cases} (1,Z_2/Z_0),(1,Z_1/Z_0) & \text{if $Z_0\neq 0$} \\
    (1,Z_3/Z_1), (Z_0/Z_1,1) & \text{if $Z_1\neq 0$} \\ (Z_0/Z_2,1),(1,Z_3/Z_2) & \text{if $Z_2\neq 0$} \\
    (Z_1/Z_3,1), (Z_2/Z_3,1) & \text{if $Z_3\neq 0$} \end{cases}.
    \end{align*}
    It is not difficult to check that $f$ is well defined: one need only observe that on the patches of $\P^3_k$
    where more than one $Z_i$ is nonzero the different definitions of $f$ all agree.  For example,
    if $Z_0,Z_1\neq 0$ we have
    \begin{align*}
    [1,Z_2/Z_0],[1,Z_1/Z_0]&=[Z_0,Z_2],[Z_0,Z_1]\\&=[Z_0Z_1,Z_2Z_1],[Z_0,Z_1]\\
    &=[Z_0Z_1,Z_0Z_3],[Z_0,Z_1]\\&=[Z_1,Z_3], [Z_0,Z_1]\\&=[1,Z_3/Z_1], [Z_0/Z_1,1].
    \end{align*}
    The other compatibility computations are identical, and we omit them.  Observe that
    $$s(1,1)_K\circ f=\id_f$$ and
    $$f\circ s(1,1)_K=\id_{\P^1_k\times \P^1_k}.$$  Since both $f,s(1,1)_K$ are morphisms, this shows that they are isomorphisms
    and $\P^1_k\times \P^1_k\simeq Q$.  Now suppose that $k$ is an algebraically closed field and let
    $$F(T_0,T_1,T_2,T_3)=(T_0,T_1,T_2,T_3)\begin{pmatrix}a_{00} & a_{01} & a_{02} & a_{03}\\
    a_{10} & a_{11} & a_{12} & a_{13}\\a_{20} & a_{21} & a_{22} & a_{23}\\a_{30} & a_{31} & a_{32} & a_{33}
    \end{pmatrix}\begin{pmatrix}T_0 \\T_1\\T_2\\T_3\end{pmatrix}$$
    be a symmetric, nondegenerate quadratic form (that is, $a_{ij}=a_{ji}$ and the matrix $(a_{ij})$ is nonsingular).  Since
    $k$ is algebraically closed and not of characteristic 2, we know that every symmetric, nonsingular matrix
    is equivalent (by elementary row and column operations) to the identity matrix.  Moreover, by a projective change of coordinates
    (projective automorphism) on the form $F$, we can perform any row and column operations on the matrix $(a_{ij})$.
    It follows that the hypersurface defined by $F$ is (via an appropriate projective change of coordinates)
    isomorphic to $Q$, which is given by the symmetric, nonsingular matrix
    $$\begin{pmatrix}0 & 0 & 0 & 1/2\\ 0 & 0 & -1/2 & 0\\ 0 & -1/2 & 0 & 0 \\1/2 & 0 & 0 & 0\end{pmatrix}.$$
    Since $Q$ is isomorphic to $\P^1_k\times \P^1_k$ by the above, we see that every hypersurface given by an equation of
    the form $F$ is isomorphic to $\P^1_k\times \P^1_k$.  The projection maps
    $p_i:Q\rightarrow \P^1_k$ for $i=1,2$ are given as the first and second components (respectively) of our function $f$ above.


 \exer{3}
    Recall that $\ver_1^n$ is the image of $\P_k^1$ under the map
    $$v_{1,n}([X_0,X_1])=[X_0^n,X_0^{n-1}X_1,\ldots,X_1^n]:=[Z_0,Z_1,\ldots,Z_n]\in \P_k^{n}.$$  Since $Z_i=X_0^{n-i}X_1^i,$
    we observe that any point of $\ver_1^n$ satisfies $Z_iZ_j=Z_kZ_l$ for $i+j=k+l$.  Moreover, since $v_{1,n}$ is a morphism,
    we know that the image $\ver_1^n$ is closed in $\P^n$.  Now let $C$ be the affine curve given by
    $$\left\{Z_i-Z_1^i=0\right\}_{i=2}^{n}.$$  Observe that for any point $(1,Z_1,Z_2,\ldots,Z_n)$ of $C$ we have
    $$Z_iZ_j=Z_1^{i+j},$$ which only depends on $i+j$.  Therefore, $C$ is contained (via the usual inclusion
    of affine space into projective space) in $\ver_1^n$ since the defining ideal of $\ver_1^n$ vanishes on
    $C$.  Observe that $C$ is not a closed subset of $\P^n_k.$  Indeed, any point of $C$ satisfied the equations
    $$Z_iZ_j=Z_kZ_l$$ for $i+j=k+l$.  However, $(0,0,\ldots,1)$ also satisfies these equations but is not a point of $C$.
    On the other hand, any point of $\ver_1^n(K)$ with $Z_0\neq 0$ has the form
    $$(1,t,t^2,\ldots,t^n)$$ for $t\in K$, while there is only one point with $Z_0=0,$ the point $(0,0,\ldots,1)$.
    It is clear that any affine point of $\ver_1^n(K)$ is also a point of $C$.  It follows that
    $\ver_1^n(K)=C\cup (0,0,\ldots,1)$.  Since $C$ is not closed in $\P^n_k$ and $\ver_1^n$ is, it follows that
    the projective closure of $C$ is $\ver_1^n$.


 \exer{6}
    \begin{enumerate}
        \item Let $M$ be the set of $(n+1)\times(n+1)$ rank 1 matrices, up to proportionality.  We know that the image $S$ of
        $\P^n_k(K)^{\prime\prime}\times \P^n_k(K)^{\prime\prime}$ under the Segre embedding is isomorphic to $\P^n_k(K)^{\prime\prime}\times
        \P^n_k(K)^{\prime\prime}$.  Any point of $S$ has the form
        $$\left(X_iY_j\right)_{i,j=0}^n$$ with $(X_0,\ldots,X_n),(Y_0,\ldots,Y_n)\in\P^n_k$.  Therefore, we define
        $$f:S\rightarrow M$$ by
        $$f(X_0Y_0,\ldots,X_nY_n)=(X_iY_j).$$  We must check that matrix $(X_iY_J)$ has rank 1.  Certainly, the rank
        of $(X_iY_j)$ is at least 1 since not all $X_iY_j$ are $0$ (this would imply that $X_i=0,Y_i$ for all $i$, which is impossible
        since $(X_0,\ldots,X_n),(Y_0,\ldots,Y_n)$ are points of projective space).  However, $(X_iY_j)$ has at most rank $1$ since the determinant of
        any $2\times 2$ minor is
        $$X_iY_j\cdot X_(i+r)Y_(j+s)-X_(i+r)Y_j\cdot X_iY_(j+s)=0,$$
        for any $1\leq r\leq n-i$ and $1\leq s\leq n-j$.  Therefore, the matrix $(X_iY_j)$ indeed has rank $1$.  Our map $f$
        is thus well defined.
        It is clear that $f$ is injective, for $X_iY_j=X_i^{\prime}Y_j^{\prime}$ for all $i,j$ implies that $X_i=X_i^{\prime},Y_j=Y_j^{\prime}$.
        Surjectivity is similarly clear: any $2\times 2$ minor of a rank 1 $(n+1)\times (n+1)$ matrix has determinant 0, so that
        the entries of our matrix (when considered as a lexographically ordered tuple) satisfy the defining equations of $S$ and hence
        gives a point of $S$.

        \item We know that any point of $\ver_n^2(K)^{\prime\prime}$ can be written
        $$(Z_{ij}):=(X_iX_j)$$ for $i,j=0\ldots n$ and $(X_0,\ldots, X_n)\in \P^n_k(K)^{\prime\prime}$.
        We therefore define
        $$f:\ver_n^2(K)^{\prime\prime}\rightarrow M$$ (where $M$ is the set of rank 1 symmetric square matrices with coefficients in $K$
        of size $m=\binom{n+2}{2}$ up to proportionality) by
        $$f(Z_{ij})=(Z_{ij})$$ (where we clearly mean the matrix with entries $Z_{ij}$ on the right and the point $Z_{ij}$ on the left).
        This is well defined since both the point $(Z_{ij})$ and the matrix $(Z_{ij})$ are defined up to proportionality and the symmetric
        matrix $(Z_{ij})$ has rank 1 since it certainly does not have rank $0$ (since not all entries are $0$) but we have
        $$Z_{ij}Z_{(i+r)(j+s)}-Z_{(i+r)j}Z_{i(j+s)}=X_iX_jX_{i+r}X_{j+s}-X_{i+r}X_jX_iX_{j+s}=0$$
        so that every $2\times 2$ minor has determinant 0.  As in part a), it is clear that $f$ is injective.  Surjectivity follows similarly
        since any $2\times 2$ minor of a  rank 1 symmetric matrix must have determinant 0, that is, any such matrix, when considered
        as a lexographically ordered tuple (that is, $(a_{ij})\mapsto (a_ij)$ with $0\leq i\leq j\leq n$) satisfies the defining equations
        of $\ver_n^2(K)^{\prime\prime}$.
    \end{enumerate}


\section{Lecture 8}
 \exer{1}
    Let $S=\{(a,b,c)\in \P^2(K):a\neq 0,b\neq 0\}\cup \{(1,0,0)\}$.  If $S=Z_1\setminus Z_2$ for some projective algebraic
    sets $Z_1,Z_2$, then for any copy of affine space $\A^2(K)\subset\P^2(K)$ we have $S\cap \A^2(K)=Z_1\cap \A^2(K)\setminus Z_2\cap \A^2(K)$,
    and $Z_i\cap \A^2(K)$ is a closed (in the subspace topology) set in $\A^2(K)$.  Therefore, it is enough to show that $S\cap \A^2(K)$
    is not of the form $Z_1\setminus Z_2$ for affine algebraic sets $Z_1,Z_2$.  We choose the copy of $\A^2(K)$ corresponding
    to $a=1$.  Then $T:=S\cap \A^2(K)=\{(b,c)\in \A^2(K):b\neq 0\}\cup (0,0)$.  Since (by a previous problem set) $T$ is dense in $\A^2(K)$,
    we must have $Z_1=\A^2(K)$.  But this requires that $Z_2=\{(b,c)\in \A^2(K):b=0\}\setminus (0,0)$, which is not closed since
    any polynomial vanishing on the line $b=0$ must also vanish at the origin (a polynomial restricted to a line is a polynomial in
    one variable and hence has finitely many roots unless it is identically 0 on the line).  Thus, $T$ is not locally closed, and it
    follows that $S$ is not quasi projective.


  \exer{4}
    The union of quasi projective algebraic sets is not necessarily quasi projective.  For example, the sets
    $S_1=\{(a,b,c)\in\P^2(K):a\neq 0,b\neq 0\}$ and $S_2=(1,0,0)$ are both quasi projective, but their union is,
    by problem 1, not quasi projective.  Any finite intersection of quasi projective sets is again quasi projective.
    For if $S_1,S_2$ are quasi projective, we may write $S_i=Z_{1i}\setminus Z_{2i}$ for closed sets $Z_{ij}$ and
    $i,j=1,2$.  It follows that
    $$S_1\cap S_2=\{Z_{11}\setminus Z_{21}\}\cap\{Z_{12}\setminus Z_{22}\}=(Z_{11}\cap Z_{21})\setminus (Z_{21}\cup Z_{22})$$
    is quasi projective since the intersection and union of closed (algebraic) sets is again closed.  However, an arbitrary intersection
    of quasi projective varieties is not necessarily quasi projective.  For example,
    $$\bigcap_{x\in \A^1(K)\setminus 0} \{(b,c)\in\A^2(K):c\neq x\}$$
    is an arbitrary intersection of quasi projective sets that is (by problem 1) not quasi projective.


\exer{5}
    Let us find the irreducible components of the projective subset $S$ of $\P^3(K)$given by $T_2T_0-T_1^2=0,\ T_1T_3-T_2^2=0$.
    Suppose first that $T_0\neq 0$.  Then if $T_1=0,$ we must also have $T_2=0$, while there are no restrictions on $T_3$.
    This gives the set of points $(1,0,0,t)$ for $t\in K$.  If $T_1\neq 0$, then we have $T_2=T_1^2$ and $T_1T_3=T_1^4$,
    so we get the points $(1,t,t^2,t^3)$ for $t\in K$.  If $T_0=0$, we get the point $(0,0,0,1)$.  It is clear from this
    that $S$ is the union of two irreducible curves given by
    \begin{align*}
        C_1:&\{(1,0,0,t):t\in K\}\cup \{(0,0,0,1)\}\\
        C_2:&\{(1,t,t^2,t^3):t\in K\}\cup \{(0,0,0,1)\},
    \end{align*}
    or given as projective curves by the equations
    \begin{align*}
        &\{Z_1=0,\ Z_2=0\}\\
        &\{Z_1^2-Z_2Z_0=0,\ Z_1^3-Z_3Z_0^2=0,\ Z_2^3-Z_0Z_3^2\}.
    \end{align*}


  \exer{8}
    We first show that a regular map is continuous.  To do this, we must show that the inverse image of a closed set is closed.
    We know that a regular map is given locally by polynomials.  Let $f:X\rightarrow Y$ be a regular map of quasi projective
    varieties and let $C$ be a closed set in $Y$ given by the polynomials $g_s$.  Let $U_i$ be a finite covering of $X$ by open sets.
    Then $f\arrowvert_{U_i}$ is just a polynomial map, for each $i$, so set $f\arrowvert_{U_i}=f_i$.  Set $V_i=f^{-1}(C)\cap U_i$.
    Then it is clear that $V_i$ is the quasi projective algebraic set defined by $g_s\circ f_i$.  For these polynomials
    certainly vanish on $V_i$.  On the other hand, if they vanished at some point not in $V_i$ then $g_s$ would vanish
    at a point not in $C$, a contradiction.  This shows that $V_i$ is closed.  The union of all $V_i$ is $f^{-1}(C)$, and since this is a finite
    union, we see that $f$ is continuous.

    Now let us show that the composition of regular maps is regular.  Let $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ be regular
    maps of quasi projective varieties.  Let $U_i,W_j$ be finite coverings of $X,Y$ by open sets such that
    $f\arrowvert_{U_i}:=f_i,\ g\arrowvert_{W_j}:=g_j$ are polynomial maps.  Set
    $$V_{ij}=f^{-1}(W_j)\cap U_i.$$  Since $f$ is continuous and $W_j$ is open, we see that $V_{ij}$ gives an open cover of $X$.
    Moreover, $f\arrowvert_{V_{ij}}=f_i$ and $g\arrowvert_{W_j\cap f_i(U_i)}=g_j$.  It follows that
    $$g\circ f\arrowvert_{V_ij}=f_i\circ g_j$$ is the composition of two polynomial maps and hence given by polynomials.  It follows that
    $g\circ f$ is regular.


\section{Lecture 9}
  \exer{1}
        We show that the set of all degree $m$ homogenous polynomials that split into the product of a degree $k$ and a degree
        $m-k$ homogenous polynomial is closed.  It will then follow that the set of degree $m$ reducible homogenous polynomials
        is closed since it is a finite union of closed sets.  Under the Veronese embedding, $v_n^m$ hypersurfaces of degree $m$
        in $\P^n$ correspond to points of $\P^{N_m}$, where $N_m=\binom{n+m}{m}-1$,  since such a hypersurface is just a linear subspace
        (two homogenous degree $m$ polynomials give the same hypersurface if and only if they are proportional) of $\A^{N_m}$.
        Multiplication of polynomials gives a regular map (given by appropriate equations in the coefficients)
        $$f:\P^{N_k}\times \P^{N_{m-k}}\rightarrow \P^{N_m}$$ where $N_i=\binom{n+i}{i}-1$ as above.
        Clearly, the set of reducible homogenous degree $m$ polynomials that split as a product of a degree $m-k$ and a degree $k$
        homogenous polynomial is the image of $f$.  By the Segre embedding, we have seen that $\P^{N_k}\times \P^{N_{m-k}}$
        is a projective variety.  By Theorem 1 of Lecture 9, the image of $f$ is closed since $f$ is regular
        and $\P^{N_k}\times \P^{N_{m-k}}$ is a projective variety and we are done.

        Now let $n=d=2.$  We show that a degree $2$ homogenous polynomial
        $$F=\sum a_{ij}T_iT_j$$
        in three variables $T_0,T_1,T_2$ is reducible if and only if $\det(a_{ij})=0$.  Since we are working over an
        algebraically closed field, we may apply a projective change of coordinates to put $F$ in ``diagonal form''.
        It is not difficult to see that $F$ is irreducible if and only if the new, diagonalized form is irreducible.
        But there are only three possibilities for this form: $T_i^2,\ T_i^2+T_j^2,\ T_0^2+T_1^2+T_2^2$.
        Observe that since we are working over an algebraically closed field, the first two forms are reducible.
        Indeed, this shows that a quadratic form $F$ is reducible if and only if it is equivalent to one of the first two
        forms above (since $T_0^2+T_1^2+T_2^2$ is clearly not reducible).  But $F$ is equivalent to one of these forms
        if and only if the matrix $(a_{ij})$ does not have maximal rank; that is, if and only if the determinant of $(a_{ij})$
        is 0.  It follows that the projective variety of degree two reducible polynomials in three variables is given by
        $$\begin{vmatrix}a_{00} & a_{01} & a_{02}\\ a_{10} & a_{11} & a_{12}\\ a_{20} & a_{21} & a_{22}\\\end{vmatrix}=0.$$



 \exer{2}
    Let $x\in K^2$.  If $K^2\setminus\{x\}$ were an affine algebraic set, it would be given as the zero set of a system of equations,
    say $F_s=0$.  But any
    polynomial that vanishes on infinitely many points of a line vanishes on the entire line.  We thus consider the restriction
    of  the $F_s$ to any line $L$ passing through $x$.  This shows that all the $F_s$ vanish at $x$, a contradiction.
    If $K^2\setminus\{x\}$
    were a projective algebraic set then it's image under a regular map would be closed.  But the inclusion
    $K^2\setminus \{x\}\hookrightarrow \A^2(K)$ is certainly regular, but since $\A^2$ is irreducible,
    $K^2\setminus \{x\}$ is not closed, a contradiction.
    Now let $y\in \P^n(K)$.  Similarly, if $S=\P^n(K)\setminus \{y\}$ were a projective algebraic set, it would be closed.  It would follow that
    $\P^n(K)=S\cup \{y\}$ is the union of two closed sets, a contradiction since we know that $\P^n(K)$ is irreducible.  It is false that
    $\P^n(K)\setminus\{y\}$ is not an affine set.  For example, let $n=1,y=(0,1)$.  Then $\P^1(K)\setminus\{y\}\simeq \A^1(K)$.



  \exer{4}
    Let $X\subset \P^n$ be a connected projective algebraic set and let $Y$ be a projective set defined by $F=0$, where
    $F$ is a homogenous polynomial of degree $d$.  Suppose that $X\cap Y=\emptyset$ and let $v_n^d$ be the Veronese map.
    Then $v_n^d(X)\cap v_n^d(Y)=\emptyset$ since $v_n^d$ is injective (and moreover $v_n^d(X)$ is isomorphic
    to $X$).  Since $F$ is of degree $d$, we see that $v_n^d(Y)$ is given by a linear equation.  By a projective change of
    coordinates, we may assume that $v_n^d(Y)$ is given by $T_0=0$ and that $v_n^d(X)\cap v_n^d(Y)=\emptyset$.  But then
    it follows that $v_n^d(X)$ is contained in an affine subset of $\P^N$.  By Corollary 2 of Lecture 9, we conclude that
    $v_n^d(X)$ is a single point, and hence, since $v_n^d$ is injective, that $X$ is a single point.









\end{document}