7.30.

A conic over a field of characteristic not equal to 2 is an irreducible curve in $\mathbb{P}^2$ of degree 2. a. Using the coefficients of quadratic polynomials show that the set of all conics can be identified with an open subset $U$ of $\mathbb{P}^5$. (One says that $U$ is a moduli space for conics.) b. Given a point $p \in \mathbb{P}^2$ show that the subset of $U$ consisting of all conics passing through $p$ is the zero locus of a linear equation in the homogeneous coordinates of $U \subset \mathbb{P}^5$. c. Given 5 points in $\mathbb{P}^2$, no three of which lie on a line, show that there is a unique conic passing through all these points.

Lemma:

Solution. a. A quadratic polynomial on $\mathbb{P}^2$ is of the form $f=a_{00} x_0^2+a_{01} x_0 x_1+a_{02} x_0 x_2+a_{11} x_1^2+a_{12} x_1 x_2+a_{22} x_2^2 \in K\left[x_0, x_1, x_2\right] .$

Since its vanishing set $C=V(f) \subseteq \mathbb{P}^2$ does not change when scaling $f$, this vanishing set is uniquely determined by the vector $a=\left(a_{00}: a_{01}: a_{02}: a_{11}: a_{12}: a_{22}\right) \in \mathbb{P}^5$ describing the coefficients of $f=f_a$. Moreover, the vanishing set $C$ is an irreducible curve of degree 2 if and only if $f$ is irreducible (and in this case we have $I(C)=\langle f\rangle$, so $f$ up to scaling is uniquely determined by $C$ ). To conclude, we want to show that inside $\mathbb{P}^5$, the locus $U$ of points $a$ such that $f_a$ is irreducible is an open set. If $f$ had a factorization $f=f_1 \cdot f_2$, then necessarily $f_1, f_2$ would have to be homogeneous polynomials, and in order for the factorization to not be trivial, they would both have to be linear. Thus we see that the complement $\mathbb{P}^5 \backslash U$ is the image of the morphism $\begin{aligned} & \Phi: \mathbb{P}^2 \times \mathbb{P}^2 \rightarrow \mathbb{P}^5 \\ &(\underbrace{b_0 x_0+b_1 x_1+b_2 x_2}{f_1}, \underbrace{c_0 x_0+c_1 x_1+c_2 x_2}{f_2}) \mapsto f=f_1 \cdot f_2 . \end{aligned}$

Here again we identify $\mathbb{P}^2$ with the moduli space of linear polynomials (via the coefficient vectors $\left(b_0: b_1: b_2\right)$ and $\left(c_0: c_1: c_2\right)$ of $\left.f_1, f_2\right)$. One checks that $\Phi$ is a morphism, e.g. on an open cover of $\mathbb{P}^5$. Then since $\mathbb{P}^2$ is projective, also $\mathbb{P}^2 \times \mathbb{P}^2$ is projective, and thus complete. Hence the image of $\Phi$ is closed, and so its complement $U$ is open. b. For $p=\left(y_0: y_1: y_2\right) \in \mathbb{P}^2$ the condition that $p \in C=V\left(f_a\right)$ is just given by $a_{00} y_0^2+a_{01} y_0 y_1+a_{02} y_0 y_2+a_{11} y_1^2+a_{12} y_1 y_2+a_{22} y_2^2 .$

Since the $y_i$ are fixed numbers, this is just a linear condition on the coordinates $a_{i j}$ on $\mathbb{P}^5$. c. For each of the points $p_1, \ldots, p_5$, the condition that $p_i \in V\left(f_a\right)$ gives a linear subspace of $U$ of codimension 1. Intersecting these five subspaces, we expect to obtain a linear space of codimension 5, i.e. a single point. To make this precise: using lemma, we see that the condition of no three points lying on a line exactly means that no three representatives of the points are linearly dependent. Thus the points $p_i$ are in general position and we can find an isomorphism $f: \mathbb{P}^2 \rightarrow \mathbb{P}^2$ sending $p_1, p_2, p_3, p_4$ to $A_1=(1: 0: 0), A_2=(0: 1: 0)$, $A_3=(0: 0: 1)$ and $A_4=(1: 1: 1)$. Any such isomorphism sends a conic $C$ through the $p_i$ to a conic $f(C)$ through $A_1, A_2, A_3, A_4$ and $p=f\left(p_5\right)=\left(y_0: y_1: y_2\right)$, and this correspondence is a bijection.

So it suffices to solve the problem for the points $A_1, A_2, A_3, A_4, p$. Looking at the equation above, the conditions $A_1, A_2, A_3 \in V\left(f_a\right)$ mean that $a_{00}=a_{11}=a_{22}=0$. Moreover, the condition $A_4 \in V\left(f_a\right)$ implies that $a_{01}+a_{02}+a_{12}=0$. Solving this last equation for $a_{12}=-a_{01}-a_{02}$ the set of quadratic equations whose vanishing set locus contains $A_1, \ldots, A_4$ is given by $f$ of the form $f=a_{01}\left(x_0 x_1-x_1 x_2\right)+a_{02}\left(x_0 x_2-x_1 x_2\right)$

We claim that the additional condition $f(p)=f\left(y_0, y_1, y_2\right)=0$ gives one more linear condition on the coefficients $a_{01}, a_{02}$ and thus we obtain a unique solution for $f$ up to scaling, so that there is a unique quadratic equation $f$ satisfied by the five points. If $f(p)=0$ was not an additional condition, then $f(p)$ would have to vanish as a polynomial in $a_{01}, a_{02}$ and thus we would have $0=y_0 y_1-y_1 y_2=\left(y_0-y_2\right) y_1 \text { and } 0=y_0 y_2-y_1 y_2=\left(y_0-y_1\right) y_2 .$

Going through the cases, these two equations can only be satisfied if $p=\left(y_0: y_1 : y_2\right) \in\left\{A_1, \ldots, A_4\right\}$, which gives a contradiction. We have thus proved that there is a unique quadratic equation $f$ (up to scaling) satisfied by the five points $A_1, A_2, A_3, A_4, p$. If $f$ was not irreducible, it would decompose as the product of two linear polynomials $f=f_1 f_2$. But then the five points are contained in the union $L_1 \cup L_2=V\left(f_1\right) \cup V\left(f_2\right)$ of lines, so one of the two lines must contain at least three points. This gives a contradiction to the assumption.

8.22

We denote the Plücker coordinates of the Grassmannian $G(2,4)$ in $\mathbb{P}^5$ by $x_{i j}$ for $1 \leq i<j \leq 4$. a. Show that $G(2,4)=V\left(x_{12} x_{34}-x_{13} x_{24}+x_{14} x_{23}\right)$. b. Let $L \subseteq \mathbb{P}^3$ be an arbitrary line. Show that the set of lines in $\mathbb{P}^3$ that intersect $L$, considered as a subset of $G(2,4) \subseteq \mathbb{P}^5$, is the zero locus of a homogeneous linear polynomial.

How many lines in $\mathbb{P}^3$ would you expect to intersect four general given lines? Solution. a. The Grassmannian $G(2,4)$ is cut out by the $3 \times 3$ minors of the matrix $M=\left(\begin{array}{cccc} x_{23} & -x_{13} & x_{12} & 0 \\ x_{24} & -x_{14} & 0 & x_{12} \\ x_{34} & 0 & -x_{14} & x_{13} \\ 0 & x_{34} & -x_{24} & x_{23} \end{array}\right)$

As an example, the upper left minor is given by $x_{23} x_{14}^2+x_{13} x_{24}\left(-x_{14}\right)+x_{12} x_{34} x_{14}=x_{14}\left(x_{12} x_{34}-x_{13} x_{24}+x_{14} x_{23}\right)$

Thus we have $G(2,4) \subseteq V\left(x_{14}\right) \cup V\left(x_{12} x_{34}-x_{13} x_{24}+x_{14} x_{23}\right) .$

On the other hand, we know that $G(2,4)$ is irreducible of dimension 4 , and from Krull's principal ideal theorem one checks that the same is true for $V\left(x_{14}\right)$ and $V\left(x_{12} x_{34}-x_{13} x_{24}+x_{14} x_{23}\right)$. Since $G(2,4)$ is not contained in $V\left(x_{14}\right)$, it must be contained in $V\left(x_{12} x_{34}-x_{13} x_{24}+x_{14} x_{23}\right)$, and for dimension reasons this containment must be an equality. b. Assume $L$ corresponds to the span $\operatorname{Lin}(a, b)$ of two vectors $a=\left(a_1, a_2, a_3, a_4\right)$ and $b=\left(b_1, b_2, b_3, b_4\right)$. Let $L^{\prime}=\operatorname{Lin}(c, d) \in G(2,4)$ be another line. Then $L, L^{\prime}$ meet in a point if and only if $\operatorname{Lin}(a, b) \cap \operatorname{Lin}(c, d)$ has dimension at least 1 (and in this case, any one-dimensional sub-vectorspace contained in this intersection gives a point of $\mathbb{P}^3$ where $L, L^{\prime}$ meet.

Having such a positive-dimensional intersection is equivalent to the vectors $a, b, c, d$ not being linearly independent, and thus equivalent to the vanishing of the determinant of the matrix $N=\left(\begin{array}{llll} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \end{array}\right) .$

By expanding this determinant along the first two columns (applying Laplace expansion twice), we see that for fixed $a_i, b_i$, the determinant $\operatorname{det}(N)$ is a linear polynomial in the $2 \times 2$-minors of the last two columns. But these are exactly the Plücker coordinates of the line $L^{\prime}$. Thus the condition $L \cap L^{\prime} \neq \emptyset$ is described by a linear equation in the Plücker coordinates of $L^{\prime}$.

8.23

Show that the following sets are projective varieties: a. the incidence correspondence $\left\{(L, a) \in G(k, n) \times \mathbb{P}^{n-1}: L \subseteq \mathbb{P}^{n-1} \text { a }(k-1) \text {-dimensional linear subspace and } a \in L\right\} \text {; }$ b) the join of two disjoint varieties $X, Y \subseteq \mathbb{P}^n$, i.e., the union of all lines in $\mathbb{P}^n$ intersecting both $X$ and $Y$.

Solution. a) Assume that $L=\operatorname{Lin}\left(b_1, \ldots, b_k\right)$ is the span of $k$ vectors $b_i$, then the containment $a \in L$ is equivalent to the matrix $M=\left(\begin{array}{cccc} a_1 & b_{11} & \ldots & b_{k 1} \\ a_2 & b_{12} & \ldots & b_{k 2} \\ \vdots & & & \vdots \\ a_n & b_{1 n} & \ldots & b_{k n} \end{array}\right)$ having rank exactly $k$ (so that the first column $a$ must be linearly dependent on the last $k$ columns $\left.b_1, \ldots, b_k\right)$. Since the last $k$ columns are already linearly independent (by the assumption that $L$ is a point of $G(k, n))$, this is equivalent to all $(k+1) \times(k+1)$ -minors of $M$ vanishing. Expanding such a minor by the first column using the Laplace rule, we see that it is a linear combination of products of $a_i$ with maximal minors of the matrix $B=\left(b_1, \ldots, b_k\right)$. These minors are the Plücker coordinates

of $L$, whereas the $a_i$ are the homogeneous coordinates of $\mathbb{P}^{n-1}$. On the standard open charts of $G(k, n)$ and $\mathbb{P}^{n-1}$ these become polynomial equations in the affine coordinates, so that the incidence correspondence is indeed a Zariski closed subset of $G(k, n) \times \mathbb{P}^{n-1}$. We have seen that both of these are projective varieties, and that the product of projective varieties is still projective. Thus the closed subset formed by the incidence correspondence is also projective. b) Our first goal is to show that the variety $Z=\{L \in G(2, n+1): L \text { intersects both } X \text { and } Y\} \subseteq G(2, n+1)$ of all lines appearing in the definition of the join is projective. To see that, let $I=\{(L, a): a \in L\}$ be the incidence correspondence from part a) applied to the case $G(2, n+1)$. Then $I$ is projective and so the product $I^2=I \times I=\left\{\left(L_1, a_1, L_2, a_2\right): a_1 \in L_1, a_2 \in L_2\right\} \subseteq G(2, n+1) \times \mathbb{P}^n \times G(2, n+1) \times \mathbb{P}^n$ is projective as well. We have two projection morphisms $\pi_1: I^2 \rightarrow G(2, n+1) \times G(2, n+1) \text { and } \pi_2: I^2 \rightarrow \mathbb{P}^n \times \mathbb{P}^n$ which remember the pairs $\left(L_1, L_2\right)$ and $\left(a_1, a_2\right)$ respectively. Let $\Delta \subseteq G(2, n+1) \times$ $G(2, n+1)$ be the diagonal and $X \times Y \subseteq \mathbb{P}^n \times \mathbb{P}^n$ be the product of $X, Y$, which are both closed subsets (here we use that $G(2, n)$ is a variety, and that the product of two closed embeddings is a closed embedding). Then we have that $\pi_1^{-1}(\Delta) \cap \pi_2^{-1}(X \times Y)=\left\{\left(L, a_1, L, a_2\right): a_1 \in L \cap X, a_2 \in L \cap Y\right\}$ is closed inside $I^2$ and thus projective as well. Our original variety $Z$ at the beginning is the image of this projective variety under the projection to $G(2, n+1)$ (e.g. to the first factor). Since the image of a projective variety under a map to another variety is still projective, we conclude that $Z$ is projective. Finally, to obtain the join consider the diagram $G(2, n+1) \stackrel{p_1}{\rightleftarrows} I \xrightarrow{p_2} \mathbb{P}^n$ with $p_1, p_2$ the projections to the two factors. Then the join $J(X, Y)$ is given by $J(X, Y)=p_2\left(p_1^{-1}(Z)\right)=p_2(\{(L, a): L \text { intersects both } X \text { and } Y, a \in L\} .$

Again since $Z \subseteq G(2, n+1)$ is closed, the closed subset $p_1^{-1}(Z) \subseteq I$ is projective and so its image $J(X, Y)$ under $p_2$ is projective as well.