## I.2: Projective Varieties

> Note: in this section, $S \da \kxnz$, $I(Y)$ is the ideal generated by $\ts{f\in S \text{ homogeneous} \st f(Y) = 0 }$, and $S(Y) \da S/I(Y)$ is the homogeneous coordinate ring.

### 2.1   #to-work

Prove the "homogeneous Nullstellensatz," which says if $a \subseteq S$ is a homogeneous ideal, and if $f \in S$ is a homogeneous polynomial with deg $f>0$, such that $f(P)=0$ for all $P \in Z(a)$ in $\mathbf{P}^n$, then $f^u \in a$ for some $q>0$. [^hint_2.1]

[^hint_2.1]: 
Hint: Interpret the problem in terms of the affine $(n+1)$-space whose affine coordinate ring is $S$, and use the usual Nullstellensatz, (1.3A).

> [!solution]- Solution


### 2.2: The irrelevant ideal #completed

For a homogeneous ideal $a \subseteq S$, show that the following conditions are equivalent:

(i) $Z(a) = \varnothing$ (the empty set);
(ii) $\sqrt{a}=$ either $S$ or the ideal $S_{+}=\bigoplus_{d>0} S_d$;
(iii) $a \supseteq S_d$ for some $d>0$.

> [!solution]- Solution
> Some notational differences that are helpful: let $J$ be the ideal $a$ in the question, and write $V_p(J)$ for the projective vanishing locus and $V_a(J) = C(V_p(J))$ for the corresponding affine locus in the cone.
> 
> - $1\iff 2$: If $V_p(J) = \emptyset$ then $V_a(J) = \emptyset$ or $\vector 0\in \AA^{n+1}$ since $C(V_p(J)) = \ts{\vector 0}\union\ts{\tv{x_0,\cdots, x_n} \st \tv{x_0: \cdots : x_n}\in V_p(J)}$. Then $\sqrt{I} = I(V_a(J)) = I(\emptyset), I(\vector 0) = S, S_+$ since all elements of $S$ vanish on $\emptyset$ and any element in $S_+$ is divisible by a monomial. The same argument works in reverse.
> - $2\implies 3$: if $\sqrt{J} = S$ this is clear, since $1\in S \implies 1\in \sqrt{J} \implies 1\in J \implies J = S$ and consequently $J$ contains every graded piece $S_d$ of $S$.
> 	If $\sqrt{J} = S_+$ then $x_1,\cdots, x_n\in \sqrt{a}$ so $x_1^{N_1}, \cdots, x_n^{N_n}\in J$ for various integers $N_i$. Let $d = \lcm(N_i)$ so that $x_1^d,\cdots, x_n^d\in J$, then $S_d \subseteq J$.
> - $3\implies 1$: if $S_d \subseteq J$ then note that the functions $x_1^d,\cdots, x_n^d\in J$ have no common zero in $\PP^n$, so $V_p(a) = \emptyset$.
> 
> 
> 


### 2.3  #to-work

(a) If $T_1 \subseteq T_2$ are subsets of $S^h$. then $Z\left(T_1\right) \supseteq Z\left(T_2\right)$.
(b) If $Y_1 \subseteq Y_2$ are subsets of $\mathbf{P}^n$, then $I\left(Y_1\right) \supseteq I\left(Y_2\right)$.
(c) For any two subsets $Y_1, Y_2$ of $\mathbf{P}^n, I\left(Y_1 \cup Y_2\right)=I\left(Y_1\right) \cap I\left(Y_2\right)$.
(d) If $a \subseteq S$ is a homogeneous ideal with $Z(a) \neq \varnothing$. then $I(Z(a))=\sqrt a$.
(e) For any subset $Y \subseteq \mathbf{P}^n, Z(I(Y))=\bar{Y}$.

> [!solution]- Solution

### 2.4  #to-work

(a) There is a 1-1 inclusion-reversing correspondence between algebraic sets in $\mathbf{P}^n$. and homogeneous radical ideals of $S$ not equal to $S_{+}$ given by $Y \mapsto I(Y)$ and $a \mapsto Z(a)$. [^irrelevant_ideal]
(b) An algebraic set $Y \subseteq \mathbf{P}^n$ is irreducible if and only if $I\left(Y^{\prime}\right)$ is a prime ideal.
(c) Show that $\mathbf{P}^n$ itself is irreducible.


[^irrelevant_ideal]: 
Note: Since $S_{+}$ does not occur in this correspondence, it is sometimes called the **irrelevant maximal ideal of $S$**.

> [!solution]- Solution

### 2.5  #to-work

(a) $\mathbf{P}^n$ is a noetherian topological space.
(b) Every algebraic set in $\mathrm{P}^n$ can be written uniquely as a finite union of irreducible algebraic sets. no one containing another. These are called its irreducible components.

> [!solution]- Solution

### 2.6: Dimension of homogeneous coordinate ring  #completed

If $Y$ is a projective variety with homogeneous coordinate ring $S(Y)$, show that $\operatorname{dim} S(Y)=\operatorname{dim} Y+1$.[^hint_2.5]

[^hint_2.5]: 
Hint: Let $\varphi_i: U_i \rightarrow \mathbf{A}^n$ be the homeomorphism of (2.2), let $Y_i$ be the affine variety $\varphi_i\left(Y \cap U_i\right)$, and let $A\left(Y_i\right)$ be its affine coordinate ring.
Show that $A\left(Y_i\right)$ can be identified with the subring of elements of degree 0 of the localized ring $S(Y)_{x_i}$. 
Then show that $S(Y)_{x_i} \cong A\qty{Y_i}[x_i, x_i\inv]$. Now use (1.7), (1.8A), and (Ex 1.10), and look at transcendence degrees. Conclude also that $\operatorname{dim} Y=\operatorname{dim} Y_i$ whenever $Y_i$ is nonempty.

> Here $U_i \da \PP^n\sm H_i$ where $H_i = V(x_i)$ is the coordinate hyperplane, and the homeomorphism of 2.2 is $\phi_i: U_i \to \AA^n$ where $\tv{a_0: \cdots: a_n} \mapsto \tv{ {a_0\over a_i}, \cdots, \hat{a_i}, \cdots, {a_n\over a_i}}$.

> [!solution]- Solution
> Cover $Y$ by $Y_i$, then $\dim Y = \sup \dim Y_i$ and since there are only finitely many such charts, this sup is attained by wlog $Y_0$. Define a map
> $$\begin{align}
> A(Y_0) &\mapstofrom (S(Y)_{x_0})_0 \\
> f(x_0, x_1, \cdots, x_n) &\mapsto \alpha(f) \da f(1, x_1/x_0, \cdots, x_n/x_0) \\
> \beta(g) \da x_0^{\deg g} \cdot g(x_0, x_1,\cdots, x_n) &\mapsfrom g
> \end{align}$$
> Claim: this is an isomorphism of rings. This follows from checking that $\alpha(\beta(g)) = x_0^{\deg g} g(1, x_1/x_0, \cdots, x_n/x_0) = g(x_0, x_1,\cdots, x_n)$ and similarly $\beta(\alpha(f)) =f$. This shows the first part of the hint.
> Claim: $S(Y)_{x_0} \cong A(Y_0)[x_0, x_0\inv]$. This follows from $S(Y) \cong A(Y_0)[x_0]$ and $S(Y)_{x_0} \cong S(Y)[x_0\inv]$.
> Now use that $$[A(Y_0)[x_0][x_0\inv]: A(Y_0) ]_\trdeg = [A(Y_0)[x_0]: A(Y_0) ]_\trdeg = 1$$ since $x_0, x_0\inv$ satisfy the polynomial $f(u,v) = uv-1$, and so $$[A(Y_0)[x_0][x_0\inv]: k]_\trdeg = [A(Y_0): k]_\trdeg + 1.$$
> Using the isomorphism from the claim yields
> $$[S(Y)_{x_0} : k]_\trdeg = [A(Y_0): k]_\trdeg + 1.$$
> Finally use that $\dim Y_0 = [A(Y_0): k]_\trdeg$ and $\dim S(Y)_{x_0} = [S(Y)_{x_0} : k]_\trdeg$ to rewrite this as
> $$\dim S(Y)_{x_0} = \dim Y_0 + 1 = \dim Y + 1,$$
>  and so it STS $\dim S(Y)_{x_0} = \dim S(Y)$. This is true because $x_0\in S(Y)$ and $S(Y)_{x_0} \cong S(Y)[x_0\inv]$, so $x_0\inv$ is not algebraically independent since it satisfies a the polynomial $uv-1$.

### 2.7: Dimension of $\PP^n$ and closures of quasiprojectives  #completed

(a) $\operatorname{dim} \mathbf{P}^n=n$.
(b) If $Y \subseteq \mathbf{P}^n$ is a quasi-projective variety, then $\operatorname{dim} Y=\operatorname{dim} \bar{Y}$.[^hint_2.7b]

[^hint_2.7b]: 
Hint: Use (Ex. 2.6) to reduce to (1.10).

> [!solution]- Solution (Parts a and b)
> - Part(a): By 2.6 above, $\dim S(\PP^n) = \dim \PP^n + 1$, and $\dim S(\PP^n) = [\kxnz: k]_{\trdeg} = n+1$. Alternatively, take the standard open cover $\ts{U_i} \covers \PP^n$ and note $\dim U_i = \dim U_0 = \dim\ts{\tv{1: x_1/x_0: \cdots : x_n/x_0}} =\dim \AA^n = n$, and $\dim \PP^n = \sup \dim U_i = n$.
> - Part (b): Pass to $C(Y)$; then it STS $\dim C(Y) = \dim\bar{C(Y)}$ since $\dim(C(Y)) = \dim Y + 1$. This now follows immediately from proposition 1.10 since $C(Y)$ is quasi-affine.
> - How that proof goes: pick $(Z_i, <)_{0\leq i \leq n}$ witnessing $\dim Y = n$ and take closures to get $(\bar Z_i, <)_{0\leq i \leq n}$ to get $\dim Y \leq \dim \bar Y$. Then $Z_0 = \bar Z_0 = P$ is a point, corresponding to $m\in \mspec A(\bar Y)$ and the $\bar Z_i$ corresponding to primes $p_i\in \spec A(\bar Y)$ with $p_i \subset m$. Since the chain is length $n$ we have $\height m = n$. Now apply the height-quotient formula, 
>  $$
>  \height m + \dim A(\bar Y)/m = \dim A(\bar Y)\implies n + \dim k = \dim A(\bar Y) \implies \dim A(\bar Y)=n.
> $$

### 2.8: Codimension 1 iff hypersurface   #completed

A projective variety $Y \subseteq \mathbf{P}^n$ has dimension $n-1$ if and only if it is the zero set of a single irreducible homogeneous polynomial $f$ of positive degree. $Y$ is called a **hypersurface** in $\mathbf{P}^n$.

> [!solution]- Solution
> $\implies$: Note $\dim C(Y) = \dim Y + 1 = (n-1) + 1 = n$ in $\AA^{n+1}$, so $C(Y) = Z(f)$  for some irreducible $f \in \kxnz$. Letting $F = \beta(f)$ be the homogenization, we have $Y = Z(F)$ by the affine cone correspondence.
> $\impliedby$: Let $Y = Z(F)$ for an irreducible homogeneous polynomial. Then $\gens{F}$ corresponds to a minimal prime ideal of height 1 and 
> $$\height \gens{f} + \dim S/\gens{f} = \dim S \implies 1 + \dim A(Y) = n \implies \dim A(Y) = n-1.$$

### 2.9 Projective Closure of an Affine Variety (Twisted Cubic)   #completed

If $Y \subseteq \mathbf{A}^n$ is an affine variety, we identify $\mathbf{A}^n$ with an open set $U_0 \subseteq \mathbf{P}^n$ by the homeomorphism $\varphi_0$. Then we can speak of $\bar{Y}$, the closure of $Y$ in $\mathbf{P}^n$, which is called the **projective closure** of $Y$.

a. Show that $I(\bar{Y})$ is the ideal generated by $\beta(I(Y))$, using the notation of the proof of $(2.2)$.

b. Let $Y \subseteq \mathbf{A}^3$ be the twisted cubic of (Ex. 1.2). Its projective closure $\bar{Y} \subseteq \mathbf{P}^3$ is called the **twisted cubic curve** in $\mathbf{P}^3$. 
Find generators for $I(Y)$ and $I(\bar{Y})$, and use this example to show that if $f_1, \ldots, f_r$ generate $I(Y)$, then $\beta\left(f_1\right), \ldots, \beta\left(f_r\right)$ do not necessarily generate $I(\bar{Y})$.

> [!hint] Hint
> $S = \kxnz$.
> $\alpha: S^h \to k[y_1,\cdots, y_n]$ is $f(x_0,\cdots, x_n) \mapsto g(y_1, \cdots, y_n) \da f(1, y_1,\cdots, y_n)$ and $\beta: k[y_1,\cdots, y_n]\to S^h$ is $g \mapsto x_0^{\deg g} g(x_1/x_0,\cdots, x_n/x_0)$.
> Many posted solutions seem to use $\alpha$ instead of $\beta$....
> The twisted cubic is $Y = \ts{(t, t^2, t^3)\st t\in k} \subseteq \AA^3$.


> [!solution]- Solution (Part a)
> WTS: $I(\bar Y) = \gens{\beta(I(Y))}$.
> Write $H_i = V(x_i)\subseteq \PP^n$ and embed $\AA^n\injects \PP^n$ by inclusion along $U_0\da H_0^c = \ts{x_0 = 1}$, so $\tv{x_1,\cdots, x_n} \mapsto \tv{x_0: x_1:\cdots: x_n}$.
> 
> $I(\bar Y) \subseteq \gens{\beta(I(Y))}$: 
> Let $g\in I(\bar Y)$ so $g(\vector a) \da g(a_0, \cdots, a_n) = 0$ when $\tv{a_0: \cdots : a_n}\in \bar Y \subseteq \PP^n$, and we can assume $g$ is homogeneous. In particular $g$ vanishes on $Y_0\da \bar Y \intersect U_0 = Y$, where $x_0\neq 0$, so we can write $\vector a = \tv{1: b_1, \cdots, b_n}$ where $b_i = a_i/a_0$ and $g(1, b_1,\cdots b_n) = 0$. Note that $\vector b$ defines a point in $Y$. We want to write $g = \beta(f)$ for some $f\in I(Y)$, so take $f(y_1,\cdots, y_n) \da g(1, y_1,\cdots, y_n)$, then $$\beta(f)(x_0, \cdots, x_n) \da x_0^{\deg f} f(x_1/x_0, \cdots, x_n/x_0) \da x_0^{\deg g} g(1, x_1/x_0, \cdots, x_n/x_0) = g(x_0, x_1,\cdots, x_n),$$
>and moreover $f$ vanishes on $Y$, so $g\in \gens{\beta(I(Y))}$.
>Note that $g$ need not literally be in the image of $\beta$.
>
>$\gens{\beta(I(Y))} \subseteq I(\bar Y)$:
>Let $f\in \beta(I(Y))$ so $f = \beta(g)$ where $g$ vanishes on $Y$. Then $f(x_0, \cdots, x_n) \da x_0^{\deg g}g(x_1/x_0,\cdots, x_n/x_0)$. We want to show $f$ vanishes on $\bar{Y}$. We know $f$ vanishes on $Y_0$, since $f(1, a_1,\cdots, a_n) \da 1^{\deg f}g(a_1/1,\cdots, a_n/n) = g(a_1,\cdots, a_n) = 0$ since $\vector a \in Y$. So $f$ vanishes on $Y$, and $\beta(f) = g$ vanishes on $\bar Y$, yielding $g\in I(\bar Y)$.

> [!solution] Solution (Part b)
> $Y$ is codimension 2, so we need two equations. Take $I(Y) = \gens{f_1 = x^2-y, f_2 = x^3-z}$, then $\beta(f_1) =w^2 \qty{ \qty{x\over 2}^2 + {y\over w}} = x^2 - yw \in k[w,x,y,z]$ and $\beta(f_2) = w^3\qty{\qty{x\over w }^3 - {z\over w} }= x^3 - w^2z$.
> Thus $J \da \gens{\beta(f_1), \beta(f_2)} = \gens{ wy-x^2, x^3-w^2z}$.
> 
> One can show that $\bar Y$ is the image of $[s: t]\mapsto [s^3: s^2 t: st^2 : t^3]$ and has defining ideal $I(\bar Y) = \gens{wy-x^2, wz-xy, xz-y^2}\in k[w,x,y,z]$ by homogenizing a Grobner basis for $I(Y)$. Inspection shows that, for example, that $xz-y^2\not\in J$, since every element in $J$ has $x\dash$adic valuation either 0 or at least 2.

### 2.10 The Cone Over a Projective Variety (Fig. 1)   #completed

Let $Y \subseteq \mathbf{P}^n$ be a nonempty algebraic set, and let $\theta: \mathbf{A}^{n+1}-\{(0, \ldots, 0)\} \rightarrow \mathbf{P}^n$ be the map which sends the point with affine coordinates $\left(a_0, \ldots, a_n\right)$ to the point with homogeneous coordinates $\left(a_0, \ldots, a_n\right)$. We define the affine cone over $Y$ to be
$$
C(Y)=\theta^{-1}(Y) \cup\{(0, \ldots, 0)\} .
$$

(a) Show that $C(Y)$ is an algebraic set in $\mathbf{A}^{n+1}$, whose ideal is equal to $I(Y)$, considered as an ordinary ideal in $k\left[x_0, \ldots, x_n\right]$.
(b) $C(Y)$ is irreducible if and only if $Y$ is.
(c) $\operatorname{dim} C(Y)=\operatorname{dim} Y+1$.

Sometimes we consider the projective closure $\overline{C(Y)}$ of $C(Y)$ in $\mathbf{P}^{n+1}$. This is called the **projective cone** over $Y$.

![](attachments/2022-09-17_22-10-43.png)

> [!solution]- Solution (Part a)
> This will clearly be an algebraic set in $\AA^{n+1}$ since it will be of the form $C(Y) = V(I(Y))$. That this is the same ideal is clear, since functions vanishing on $C(Y)$ are those $f$ for which $f(\vector a) = 0$ when $\vector a\neq 0$ regarded as coordinates in $\AA^{n+1}$, but then regarding $\vector a$ as coordinates on $\PP^n$ we still have $f(\vector a)=0$. Conversely, if $f\in I(Y)$ then $f(\lambda \vector a) = \lambda^{\deg f} f(\vector a) = 0$  so $f$ vanishes on $C(Y)\sm\ts{\vector 0}$. Finally, any homogeneous polynomial vanishes at $\vector 0$.

> [!solution]- Solution (Part b)
> $Y$ is an irreducible variety iff $I(Y)$ is prime, but $I(Y) = I(C(Y))$.

> [!solution]- Solution (Part c)
>  Check that $\dim C(Y) = \dim S(C(Y)) = \dim S(Y) = \dim Y + 1$ by 2.6.

### 2.11 Linear Varieties in $\mathbf{P}^n$   #to-work

A hypersurface defined by a linear polynomial is called a hyperplane.

(a) Show that the following two conditions are equivalent for a variety $Y$ in $\mathbf{P}^n$ :
    (i) $I(Y)$ can be generated by linear polynomials.
    (ii) $Y$ can be written as an intersection of hyperplanes.

    In this case we say that $Y$ is a **linear variety** in $\mathbf{P}^n$.

(b) If $Y$ is a linear variety of dimension $r$ in $\mathbf{P}^n$, show that $I(Y)$ is minimally generated by $n-r$ linear polynomials.
(c) Let $Y, Z$ be linear varieties in $\mathbf{P}^n$, with $\operatorname{dim} Y=i, \operatorname{dim} Z=$ s. If $r+s-n \geqslant 0$, then $Y \cap Z \neq \varnothing$. Furthermore, if $Y \cap Z \neq \varnothing$, then $Y \cap Z$ is a linear variety of dimension $\geqslant r+s-n$. [^hint_2.11b]

[^hint_2.11b]: 
Think of $\mathbf{A}^{n+1}$ as a vector space over $k$, and work with its subspaces.

> [!solution]- Solution

### 2.12 The $d$-uple Embedding   #to-work

For given $n, d>0$, let $M_0, M_1, \ldots, M_N$ be all the monomials of degree $d$ in the $n+1$ variables $x_0, \ldots, x_n$, where $N={n+d\choose n}-1$. We define a mapping $\rho_d: \mathbf{P}^n \rightarrow \mathbf{P}^{N}$ by sending the point $P=\left(a_0, \ldots, a_n\right)$ to the point $\rho_d(P)=\left(M_0(a), \ldots, M_N(a)\right)$ obtained by substituting the $a_t$ in the monomials $M_J$. This is called the $d$-uple embedding of $\mathbf{P}^n$ in $\mathbf{P}^N$. For example, if $n=1, d=2$, then $N=2$, and the image $Y$ of the 2-uple embedding of $\mathbf{P}^1$ in $\mathbf{P}^2$ is a conic.

(a) Let $\theta: k\left[y_0, \ldots, y_v\right] \rightarrow k\left[x_0, \ldots, x_n\right]$ be the homomorphism defined by sending $y_i$ to $M_i$, and let a be the kernel of $\theta$. Then $a$ is a homogeneous prime ideal, and so $Z$ (a) is a projective variety in $\mathbf{P}^{N}$.
(b) Show that the image of $\rho_d$ is exactly $Z(a)$. [^rmk-2.12c]
(c) Now show that $\rho_d$ is a homeomorphism of $\mathbf{P}^n$ onto the projective variety $Z$ (a).
(d) Show that the twisted cubic curve in $\mathbf{P}^3$ (Ex. 2.9) is equal to the 3-uple embedding of $\mathbf{P}^1$ in $\mathbf{P}^3$, for suitable choice of coordinates.


[^rmk-2.12c]: 
One inclusion is easy. The other will require some calculation.

> [!solution]- Solution

### 2.13   #to-work 

Let $Y$ be the image of the 2-uple embedding of $\mathbf{P}^2$ in $\mathbf{P}^5$. This is the Veronese surface. If $Z \subseteq Y$ is a closed curve (a **curve** is a variety of dimension 1), show that there exists a hypersurface $V \subseteq \mathbf{P}^5$ such that $V \cap Y = Z$.

### 2.14 The Segre Embedding  #to-work

Let $\psi: \mathbf{P}^r \times \mathbf{P}^{s} \rightarrow \mathbf{P}^{N}$ be the map defined by sending the ordered pair $\left(a_0, \ldots, a_r\right) \times\left(b_0, \ldots, b_s\right)$ to $\left(\ldots, a_i b_j, \ldots\right)$ in lexicographic order. where $N=r s+r+s$. Note that $\psi$ is well-defined and injective. It is called the Segre embedding. Show that the image of $\psi$ is a subvariety of $\mathbf{P}^N$. [^hint_2.14]

[^hint_2.14]: 
Hint: Let the homogeneous coordinates of $\mathbf{P}^N$ be $\ts{z_{ij} \st 0\leq i, j\leq r}$ and let $a$ be the kernel of the homomorphism $k[\ts{z_{ij}}] \to k[x_0,\cdots, x_r, y_0, \cdots, y_s]$, which sends $z_{ij}$ to $x_i y_j$. Then show that $\im \psi = Z(a)$.

> [!solution]- Solution

### 2.15 The Quadric Surface in $\mathbf{P}^3$ (Fig. 2)   #to-work

Consider the surface $Q$ (a surface is a variety of dimension 2) in $\mathbf{P}^3$ defined by the equation $x y-zw =0$.

(a) Show that $Q$ is equal to the Segre embedding of $\mathbf{P}^1 \times \mathbf{P}^1$ in $\mathbf{P}^3$. for suitable choice of coordinates.
(b) Show that $Q$ contains two families of lines (a line is a linear variety of dimension 1) $\left\{L_t\right\},\ts{M_t}$, each parametrized by $t \in \mathbf{P}^1$. with the properties that if $L_t \neq L_u$. then $L_t \cap L_u=\varnothing$ : if $M_t \neq M_u, M_t \cap M_u=\varnothing$, and for all $t,u$, $L_t \cap M_u=$ one point.
(c) Show that $Q$ contains other curves besides these lines, and deduce that the Zariski topology on $Q$ is not homeomorphic via $\psi$ to the product topology on $\mathbf{P}^1 \times \mathbf{P}^1$ (where each $\mathbf{P}^1$ has its Zariski topology).


![](attachments/2022-09-17_22-24-16.png)

> [!solution]- Solution

### 2.16  #to-work

(a) The intersection of two varieties need not be a variety. For example, let $Q_1$ and $Q_2$ be the quadric surfaces in $\mathbf{P}^3$ given by the equations $x^2-y w=0$ and $x y-z w=0$, respectively. Show that $Q_1 \cap Q_2$ is the union of a twisted cubic curve and a line.
(b) Even if the intersection of two varieties is a variety, the ideal of the intersection may not be the sum of the ideals. For example, let $C$ be the conic in $\mathbf{P}^2$ given by the equation $xy-zw=0$. Let $L$ be the line given by $y=0$. Show that $C \cap L$ consists of one point $P$, but that $I(C)+I(L) \neq I(P)$.

> [!solution]- Solution

### 2.17 Complete intersections   #to-work

A variety $Y$ of dimension $r$ in $\mathbf{P}^n$ is a (strict) complete intersection if $I(Y)$ can be generated by $n-r$ elements. $Y$ is a set-theoretic complete intersection if $Y$ can be written as the intersection of $n-r$ hypersurfaces.

(a) Let $Y$ be a variety in $\mathbf{P}^n$, let $Y=Z(a)$; and suppose that a can be generated by $q$ elements. Then show that $\operatorname{dim} Y \geqslant n-q$.
(b) Show that a strict complete intersection is a set-theoretic complete intersection.
(c) \* The converse of (b) is false. For example let $Y$ be the twisted cubic curve in $\mathbf{P}^3$ (Ex. 2.9). Show that $I(Y)$ cannot be generated by two elements. On the other hand, find hypersurfaces $\mathrm{H}_1, \mathrm{H}_2$ of degrees 2,3 respectively, such that $Y=H_1 \cap H_2$.
(d) \*\* It is an unsolved problem whether every closed irreducible curve in $\mathbf{P}^3$ is a set-theoretic intersection of two surfaces. See Hartshorne $[1]$ and Hartshorne $[5.III, \text{section } 5]$ for commentary.

> [!solution]- Solution