Chapter 1, Exercise 2.8: Hypersurfaces and dimension $n-1$

A projective variety $Y\subseteq\mathbb P^n$ 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 $\mathbb P^n$.

Let $Y=Z(f)$ be the zero set of one irreducible polynomial with $\deg f>0$. This has projective coordinate ring $S(Y)=S/(f)$. By the Hauptidealsatz $\operatorname{height}(f)=1$ (using $\deg f\neq 0$) and thus $\dim S(Y) = \dim S - \operatorname{height}(f)=n+1-1=n$. But $\dim Y=\dim S(Y)-1=n-1$.

Now suppose $\dim Y=n-1$, then $\dim S(Y)=n$ and further $I(Y)$ is prime. Also, $\operatorname{height}I(Y)=\dim S(Y)-\dim Y = 1$, so $I(Y)$ has a single nonconstant generator $g$ which can be taken to be homogeneous since $I(Y)$ is a homogeneous ideal. Then $I(Y)=(g)$ implies $Y=Z(g)$ as desired.

Chapter 1, Exercise 2.7: Basic properties of projective dimension

(a) $\dim\mathbb P^n=n$.

(b) If $Y\subseteq\mathbb P^n$ is a quasi-projective variety, then $\dim Y=\dim\bar Y$.
[Hint: Use (Ex. 2.6) to reduce to (1.10).]

(a) Note $I(\mathbb P^n)=(0)$, so $S(\mathbb P^n)=S$. But $\dim S=n+1$ and $\dim\mathbb P^n=\dim S(\mathbb P^n)-1=n$.

(b) Consider any nonempty component $Y_i=Y\cap U_i$: by 2.6 $\dim Y_i=\dim Y$. Further, $U_i$ is open, so $\bar Y_i=\bar Y\cap U_i$. By Proposition 1.10 and the fact that the $Y_i$ are Zariski homeomorphic to quasi-affine varieties, we find $\dim Y_i=\dim\bar Y_i$. But $\dim\bar Y_i=\dim\bar Y$ again by 2.6, so $\dim Y = \dim Y_i=\dim\bar Y_i=\dim\bar Y$.

Chapter 1, Exercise 2.6: Dimension of a projective coordinate ring

If $Y$ is a projective variety with homogeneous coordinate ring $S(Y)$, show that $\dim S(Y)=\dim Y+1$. [Hint: Let $\varphi_i:U_i\to\mathbb A^n$ be the homeomorphism of (2.2), let $Y_i$ be the affine variety $\varphi_i(Y\cap U_i)$, and let $A(Y_i)$ be its affine coordinate ring. Show that $A(Y_i)$ can be identified with the subring of elements of degree $0$ of the localized ring $S(Y)_{x_i}$. Then show that $S(Y)_{x_i}\cong A(Y_i)[x_i,x_i^{-1}]$. Now use (1.7), (1.8A), and (Ex 1.10), and look at transcendence degrees. Conclude also that $\dim Y=\dim Y_i$ whenever $Y_i$ is nonempty.]

$\newcommand{\trdeg}{\operatorname{tr.deg.}}$Our format for this exercise will be to list out the calculations that Hartshorne leaves for the reader in the hint sketch, completing them as we go along.

1. $\varphi_i(Y\cap U_i)$ is an affine variety. This is a direct consequence of Corollary 2.3.

2. $A(Y_i)\cong(S(Y)_{x_i})_0$. First, note that we can reduce to the case where $x_i\notin I(Y)$. This is because if $x_i\in I(Y)$, then $x_i=0$ in $S(Y)$, thus the localization $S(Y)_{x_i}$ is trivial. Similarly, we see $Y\cap U_i=\emptyset$, so $A(Y_i)$ is also trivial and the identification is just $0\mapsto 0$.

Now we assume $x_i\notin I(Y)$. Then elements of $S(Y)_{x_i}$ have the form $\frac f{x_i^e}$, where $f$ is homogeneous and $e$ is a positive integer. This has degree zero if $\deg f=e$, and we see that all elements of $(S(Y)_{x_i})_0$ have the form $$f\left(\frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},1,\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i}\right).$$ We can define $\alpha: (S(Y)_{x_i})_0\to A(Y_i)$ to be the dehomogenization map that takes such an $f$ to its image. Now, consider $\alpha^{-1}:A(Y_i)\to(S(Y)_{x_i})_0$ taking $f\mapsto x_i^{-\deg f}f$. By the proof of Proposition 2.2 these are inverses.

3. $S(Y)_{x_i}\cong A(Y_i)[x_i,x_i^{-1}]$. This follows from the general obvious fact that $R_x\cong (R_x)_0[x_i,x_i^{-1}]$ if $R$ is a ring with $x\in R$. This isomorphism follows in turn from the definition of localization - we can divide or multiply by any power of $x_i$ in $S(Y)_{x_i}$.

4. The desired claim. First, note $$K((S(Y)_{x_i})_0)(x_i)\cong K(S(Y)_{x_i})\cong K(S(Y)).$$ Then $$\begin{align*}\dim S(Y)&=\trdeg K(S(Y))\\&=\trdeg K((S(Y)_{x_i})_0)(x_i)\\&=1+\trdeg K((S(Y)_{x_i})_0)\\&=1+\dim Y\cap U_i.\end{align*}$$ Since $Y\cap U_i$ open cover $Y$, we have $\dim Y\cap U_i=\dim Y$ for some $i$. But $\dim Y\cap U_i=\dim S(Y)-1$ for any $i$, so it is constant, thus $\dim Y\cap U_i=\dim Y$ for all $i$. Finally, $\dim Y = \dim Y\cap U_i=\dim S(Y)-1$ as desired.

5. $\dim Y=\dim Y_i$ if $Y_i\neq\emptyset$. This is immediate, since $\varphi_i$ preserves dimension.

Chapter 1, Exercise 2.5: Properties of $\mathbb P^n$

(a) $\mathbb P^n$ is an noetherian topological space.

(b) Every algebraic set in $\mathbb P^n$ can be written uniquely as a finite union of irreducible algebraic sets, no one containing another. These are called its irreducible components.

This exercise is essentially identical to the material covered in Section 1, with no technical difficulties arising from passing to projective space.

(a) If $Y_1\supseteq Y_2\supseteq\dots$ is a descending chain in $\mathbb P^n$, taking ideals gives an ascending chain $I(Y_1)\subseteq I(Y_2)\subseteq\dots$ in $S$, but $S$ is noetherian, hence this stabilizes, hence so does the original chain.

(b) Proposition 1.5 works in general even if $X=\mathbb P^n$ (by (a)), and we are done.

Chapter 1, Exercise 2.4: The projective fundamental correspondence (part 2)

(a) There is a 1-1 inclusion-reversing correspondence between algebraic sets in $\mathbb P^n$, and homogeneous radical ideals of $S$ not equal to $S_+$, given by $Y\mapsto I(Y)$ and $\mathfrak a\mapsto Z(\mathfrak a)$. Note: Since $S_+$ does not occur in this correspondence, it is sometimes called the irrelevant maximal ideal of $S$.

(b) An algebraic set $Y\subseteq\mathbb P^n$ is irreducible if and only if $I(Y)$ is a prime ideal.

(c) Show that $\mathbb P^n$ itself is irreducible.

(a) We know that both directions are inclusion-reversing by the preceding exercise. Thus we must show that both compositions of $I$ and $Z$ do nothing to their argument. For example, if $Y$ is algebraic (closed) then $Z(I(Y))=\bar Y = Y$. Similarly, if $\mathfrak a$ is a homogeneous radical ideal with $Z(\mathfrak a)\neq\emptyset$, then $I(Z(\mathfrak a))=\sqrt{\mathfrak a}=\mathfrak a$. Note that in the case $Z(\mathfrak a)=\emptyset$, since $\mathfrak a$ is radical we have either $\mathfrak a=S$ or $\mathfrak a=S_+$. The second case is omitted by hypothesis, and thus $S$ uniquely corresponds to $\emptyset$, completing the bijection.

(b) If $Y$ is irreducible and $f,g\in S^h,\,fg\in I(Y)$, then $(fg)\subseteq I(Y)$ so $Z(f)\cup Z(g)=Z(fg)\supseteq Y$. Decompose $Y$ as $Y=(Y\cap Z(f))\cup(Y\cap Z(g))$, both of which are closed, so $Y=Y\cap Z(f)$ or $Y=Y\cap Z(g)$. In the first case $Z(f)\supseteq Y$, so $f\in\sqrt{I(Y)}=I(Y)$. In the other case $g\in I(Y)$, so $I(Y)$ is prime.

Conversely, if $I(Y)$ is prime and $Y=Y_1\cup Y_2$, apply ideals to get $I(Y)=I(Y_1)\cap I(Y_2)$. Thus $I(Y)=I(Y_1)$ or $I(Y)=I(Y_2)$. In either case $Y=Y_i$ for one of the two $i$, so $Y$ is irreducible.

(c) We have $I(\mathbb P^n)=(0)$, which is prime, so $\mathbb P^n$ is irreducible.

Chapter 1, Exercise 2.3: The projective fundamental correspondence (part 1)

(a) If $T_1\subseteq T_2$ are subsets of $S^h$, then $Z(T_1)\supseteq Z(T_2)$.

(b) If $Y_1\subseteq Y_2$ are subsets of $\mathbb P^n$, then $I(Y_1)\supseteq I(Y_2)$.

(c) For any two subsets $Y_1,Y_2$ of $\mathbb P^n$, $I(Y_1\cup Y_2)=I(Y_1)\cap I(Y_2)$.

(d) If $\mathfrak a\subseteq S$ is a homogeneous ideal with $Z(\mathfrak a)\neq\emptyset$, then $I(Z(\mathfrak a))=\sqrt{\mathfrak a}$.

(e) For any subset $Y\subseteq\mathbb P^n$, $Z(I(Y))=\bar Y$.

(a) If $P\in Z(T_2)$ then all polynomials in $T_2$ vanish at $P$, which includes polynomials in $T_1$, so $P\in Z(T_1)$.

(b) If $f\in I(Y_2)$ then $f$ vanishes on $Y_2$, hence on $Y_1$, so $f\in I(Y_1)$.

(c) If $f\in I(Y_1\cup Y_2)$ then $f$ vanishes on all points of $Y_1$ and $Y_2$, hence $f\in I(Y_1)$ and $f\in I(Y_2)$. Thus $f\in I(Y_1)\cap I(Y_2)$. Now, if $f\in I(Y_1)$ and $f\in I(Y_2)$, then $f$ vanishes on points of $Y_1$ and $Y_2$, hence on $Y_1\cup Y_2$, hence $f\in I(Y_1\cap Y_2)$.

(d) The inclusion $\sqrt{\mathfrak a}\subseteq I(Z(\mathfrak a))$ is trivial. We know that any non-constant $f\in I(Z(\mathfrak a))$ lies in $\sqrt{\mathfrak a}$ by Exercise 2.1. If $f$ is constant, then either $f\equiv 0$ or $f\not\equiv 0$. In the first case $f\in\sqrt{\mathfrak a}$ vacuously (since $\sqrt{\mathfrak a}$ is an ideal). In the second, we find $Z(\mathfrak a)=\emptyset$ which is disallowed by hypothesis and thus unimportant.

(e) The same argument works in this case as in the proof of Proposition 1.2 in the book, and thus we need not repeat it here.

Chapter 1, Exercise 2.2: Homogeneous ideals with no zero set

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

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

$(i)\implies(ii)$: Suppose $Z(\mathfrak a)=\emptyset$. From the previous proof we see that, in $\mathbb A^{n+1}$, either $Z(\mathfrak a)=\emptyset$ or $Z(\mathfrak a)=\{0\}$. In the first case, taking affine ideals yields $\mathfrak a=S$, so $\sqrt{\mathfrak a}=S$. In the second case, we find $\mathfrak a=I(\{0\})=S_+$, and $\sqrt{\mathfrak a}=S_+$.

$(ii)\implies(iii)$: If $\sqrt{\mathfrak a}=S$ then $1\in\sqrt{\mathfrak a}$, so $1\in\mathfrak a$ and $\mathfrak a=S$ so $\mathfrak a\supseteq S_d$ for any $d$. If $\sqrt{\mathfrak a}=S_+$ then we can find some $r>0$ such that $x_i^r\in\mathfrak a$ for all $i$. By the pigeonhole principle, each polynomial in $S_{r(n+1)}$ has some $x_i^r$ in every term, so $\mathfrak a\supseteq S_{r(n+1)}$.

$(iii)\implies(i)$: Suppose $\mathfrak a\supseteq S_d$ and $P\in Z(\mathfrak a)$. Then $P\in Z(S_d)$ and thus every homogeneous polynomial of degree $d>0$ vanishes at $P$. Since $P\neq(0,\dots,0)$ this is absurd.