Exercise 7.15
Show by example that not every hypersurface $Y$ in a projective variety $X$ is of the form $V(f)$ for a homogeneous polynomial $f\in S(X)$. (One possibility is to consider the Segre embedding $X$ of $\mathbb{P}^1\times\mathbb{P}^1$ in $\mathbb{P}^3$ and $Y=\mathbb{P}^1\times \{0\} \subset \mathbb{P}^1\times\mathbb{P}^1$)
Proof
Consider Segre embedding $F : \mathbb{P}^1\times\mathbb{P}^1\to \mathbb{P}^3$, $Y=\mathbb{P}^1\times \{(0:1)\}$, $Y$ is irreducible since $\{(0:1)\}$ and $\mathbb{P}^1$ are irreducible. $\dim Y=1$ and $F(Y)=V(f),f=f(x_0y_0,x_0y_1,x_1y_0,x_1y_1)$ is homomorphism. $f=\sum\limits_{k_1,k_2}f_{k_1k_2}(x_0,x_1)y_0^{k_1}y_1^{k_2}$. Since $(x_0,x_1)\nsubseteq (f_{k_1,k_2}(x_0,x_1))$, $D(f_{k_1,k_2})\neq \{0\}$ $\implies$ $\exist (a: b)\in \mathop{\cap}\limits_{k_1,k_2}D(f_{k_1,k_2})$. Let $f'=f\vert_{x_0=a,x_1=b}, f^’(0,1)=0\implies$ $f'$ vanishes at $V(y_1)$ $\implies \sqrt{(f')}=(y_1)$ $\implies f'=ky_1^n$ for some $n>0$ $\implies f= f_{0,n}(x_0,x_1)y_1^n$ with $f_{0,1}(x_0,x_1)\neq 0$ a homomorphism $\implies Y\cong V(f)$
Exercise 7.28.
Proof
Let $N_d=\left\{\alpha \in \mathbb{N}^{n+1}: \sum_i \alpha_i=d\right\}$ be the index set of the coordinates of $\mathbb{P}^N$. Then we claim that the image $X=F\left(\mathbb{P}^n\right) \subseteq \mathbb{P}^N$ is cut out by the ideal $J=\left\langle z_\alpha \cdot z_\beta-z_\gamma \cdot z_\delta: \alpha, \beta, \gamma, \delta \in N_d \text { with } \alpha+\beta=\gamma+\delta\right\rangle$ in the coordinate ring $K\left[z_\alpha: \alpha \in N_d\right]$ of $\mathbb{P}^N$. From the formula of $F$, it's clear that $X \subseteq V(J)$ since $x^\alpha \cdot x^\beta=x^\gamma \cdot x^\delta$ by the assumption $\alpha+\beta=\gamma+\delta$. To show the other inclusion, note first that for $d e_i=(0: 0: \ldots: 0: d: 0: \ldots 0) \in N_d$ we have that $X$ is covered by the open sets $D\left(z_{d e_i}\right) \subseteq \mathbb{P}^N$. Indeed, if all $z_{d e_i}$ vanish on $f(x)$ this forces $x_i^d=0$ for all $i$, giving a contradiction. So let $z=\left(z_\alpha\right){\alpha \in N_d}$ be a point in $V(J)$ with $z_{d e_i} \neq 0$, so that by rescaling we can assume $z_{d e_i}=1$. Then we claim that $z=F(x)$ for $x=\left(z_{(d-1) e_i+e_j}\right){j=0, \ldots, n}$. We prove the equality $z_\alpha=F(x)\alpha=x^\alpha$ **for all $\alpha \in N_d$ by downward-induction on the $i$-th entry $\alpha_i$ of $\alpha$. The cases $\alpha_i=d, d-1$ follow immediately from the formulas (using $z_{d e_i}=x_i=1$ ). On the other hand, assume the statement is proven for $\alpha \in N_d$ with $\alpha_i>d^{\prime}$ and consider some $\alpha$ with $\alpha_i=d^{\prime}$. Then setting $\beta=d e_i$ we find $\gamma, \delta \in N_d$ with $\gamma_i, \delta_i>d^{\prime}$ and $\alpha+\beta=\gamma+\delta .{ }^1$ But then the defining equation of $J$ forces $z_\alpha=z_\gamma \cdot z_\delta$. Since we already showed $z_\gamma=x^\gamma, z_\delta=x^\delta$ this implies $z_\alpha=x^{\gamma+\delta}=x^{\alpha+\beta}=x^\alpha \cdot \underbrace{x^\beta}_{=x_i^d=1}=F(x)\alpha$ finishing the induction step.
We have seen in class that any projective variety $Y \subseteq \mathbb{P}^n$ can be written as $Y=$ $V\left(f_1, \ldots, f_m\right)$ for $f_i$ homogeneous of the same degree $d$. Let $f_i=\sum_{\alpha \in N_d} c_{i, \alpha} x^\alpha$. Then we claim that $Y \cong \underbrace{V\left(\sum_{\alpha \in N_d} c_{i, \alpha} z_\alpha: i=1, \ldots, n\right)}_L \cap F\left(\mathbb{P}^n\right) \subseteq \mathbb{P}^N .$
Indeed, we know that $F: \mathbb{P}^n \rightarrow F\left(\mathbb{P}^n\right)$ is an isomorphism, and the pull-back of the defining equations of $L$ are exactly the defining equations of $Y$. But $L$ is a linear subspace of $\mathbb{P}^N$, and thus isomorphic to some $\mathbb{P}^{N^{\prime}}$, and then the defining quadratic equations of $F\left(\mathbb{P}^n\right)$ found in the previous part of the exercise restrict to quadratic equations on $L$, cutting out $Y$.
Exercise 7.31.
Let $X \subset \mathbb{P}^3$ be the degree-3 Veronese embedding of $\mathbb{P}^1$, i. e. the image of the morphism
$$ \mathbb{P}^1 \rightarrow \mathbb{P}^3,\left(x_0: x_1\right) \mapsto\left(y_0: y_1: y_2: y_3\right)=\left(x_0^3: x_0^2 x_1: x_0 x_1^2: x_1^3\right) . $$
Moreover, let $a=(0: 0: 1: 0) \in \mathbb{P}^3$ and $L=V\left(y_2\right) \subset \mathbb{P}^3$, and consider the projection $f$ from $a$ to $L$ as in Example 7.5 (b). (a) Determine an equation of the curve $f(X)$ in $L \cong \mathbb{P}^2$. (b) Is $f: X \rightarrow f(X)$ an isomorphism onto its image?
Proof
(a) First, we claim that the projection $f: \mathbb{P}^3 \longrightarrow \mathbb{P}^2$ is given by $f\left(y_0: y_1: y_2: y_3\right)=$ $\left(y_0: y_1: y_3\right)$. Indeed, the unique line through $a$ and $\left(y_0: y_1: y_2: y_3\right)$ is given by $\left\{\left(s y_0: s y_1: s y_2+t: s y_3\right):(s: t) \in \mathbb{P}^1\right\}$ intersecting $L=V\left(y_2\right)$ at the point $\left(y_0: y_1: 0: y_3\right)$ for $(s: t)=\left(1:-y_2\right)$. Identifying $L$ with $\mathbb{P}^2$ by sending $\left(y_0: y_1: 0: y_3\right) \in L$ to $\left(y_0: y_1: y_3\right) \in \mathbb{P}^2$ gives the claimed form of $f$. From the form of the morphism $g: \mathbb{P}^1 \rightarrow \mathbb{P}^3$ given in the exercise, it's clear that $a$ is not contained in the image $g\left(\mathbb{P}^1\right)=X$. Since $a$ is the only point where $f$ is not defined, the image $Y=f(X)=(f \circ g)\left(\mathbb{P}^1\right)$ is just the image of the morphism $f \circ g: \mathbb{P}^1 \rightarrow \mathbb{P}^2,\left(x_0: x_1\right) \mapsto\left(x_0^3: x_0^2 x_1: x_1^3\right) .$
Denote the coordinates of $\mathbb{P}^2$ by $\left(z_0: z_1: z_2\right)$ to avoid confusion. Then from the formula of $f \circ g$ we see that $Y$ is contained in the vanishing locus $V\left(z_0^2 z_2-z_1^3\right)$. In fact we claim that $f \circ g: \mathbb{P}^1 \rightarrow V\left(z_0^2 z_2-z_1^3\right)$ is a bijection, so that $Y=V\left(z_0^2 z_2-z_1^3\right)$. To prove this, let $U_2=\left\{\left(z_0: z_1: z_2\right) \in \mathbb{P}^2: z_2 \neq 0\right\}$, then $(f \circ g)^{-1}\left(U_2\right)=\mathbb{P}^1 \backslash\{(1: 0)\} \cong \mathbb{A}^1 .$
Then the unique preimage of $\left(z_0: z_1: 1\right) \in V\left(z_0^2 z_2-z_1^3\right) \cap U_2$ is given by $\left(x_0: 1\right)=$ $\left(z_0 / z_1: 1\right)$ if $z_1 \neq 0$ and $(0: 1)$ if $z_1=0$ (which forces $\left.z_0=0\right)$. On the other hand, the unique preimage of $V\left(z_0^2 z_2-z_1^3\right) \cap V\left(z_2\right)=\{(1: 0: 0)\}$ under $f \circ g$ is given by $(1: 0)$. This shows the claimed bijection $f \circ g: \mathbb{P}^1 \rightarrow V\left(z_0^2 z_2-z_1^3\right)$.
(b) $g: \mathbb{P}^1 \rightarrow \mathbb{P}^3$ is an isomorphism onto its image $X$. Thus $f: X \rightarrow Y$ is an isomorphism if and only if $f \circ g: \mathbb{P}^1 \rightarrow Y$ is an isomorphism. But by the analysis above, on the open subset $U_2 \cong \mathbb{A}^2 \subseteq \mathbb{P}^2$ the map $f \circ g$ is given by $\mathbb{A}^1 \rightarrow V\left(z_0^2-z_1^3\right), x_0 \mapsto\left(x_0^3, x_0^2\right)$ in affine coordinates (setting $x_1=1$ and $z_2=1$ ). This map is not an isomorphism, and since an isomorphism must restrict to an isomorphism over any open subset of its image, this gives a contradiction.