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.

Chapter 1, Exercise 2.1: A homogeneous form of the Nullstellensatz

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

The first case of this problem (a trivial case) is when $Z(\mathfrak a)=\emptyset$ in $\mathbb P^n$ and the condition that $f$ vanishes on $Z(\mathfrak a)$ is vacuous. In this case, in $\mathbb A^{n+1}$ either $Z(\mathfrak a)=\emptyset$ or $Z(\mathfrak a)=\{(0,\dots,0)\}$. In the first case $f$ vanishes on $Z(\mathfrak a)$ vacuously and thus $f^q\in\mathfrak a$ for some $q$. In the second case, since $f$ is homogeneous with $\deg f>0$ we have $f(0,\dots,0)=0$, hence the same holds and we are done.

The nontrivial case occurs when $Z(\mathfrak a)\neq\emptyset$ in $\mathbb P^n$. Interpreting this in $\mathbb A^{n+1}$, if $\mathfrak a$ is homogeneous then we can interpret $Z(\mathfrak a)$ as an affine cone: a union of the lines through the origin representing points through the origin. To prove this, let $P=(a_0,\dots,a_n)\in Z(\mathfrak a)$ be a point. The ideal $\mathfrak a$ has generators that are homogeneous, and thus it suffices to show that any homogeneous polynomial vanishing at $P$ vanishes on such a line. Indeed, since it is homogeneous it vanishes on any multiple $\lambda P$, and thus on the entire line. Hence each line is contained in $Z(\mathfrak a)$. The other inclusion is trivial.

Now, since $f$ vanishes on $Z(\mathfrak a)$ and $f$ is homogeneous, it also vanishes on any affine lines through the origin and thus on the cone $Z(\mathfrak a)\subseteq\mathbb A^{n+1}$. By the ordinary Nullstellensatz, $f^q\in\mathfrak a$ for some $q>0$.

Chapter 1, Exercise 1.12: Algebraic geometry breaks down over a non-algebraically closed field

Give an example of an irreducible polynomial $f\in\mathbb R[x,y]$, whose zero set $Z(f)$ in $\mathbb A^2_{\mathbb R}$ is not irreducible (cf. 1.4.2).

This is as easy and pathological as it sounds - take $f(x,y)=x^2+y^2+1$, then $Z(f)=\emptyset$ in $\mathbb A^2_{\mathbb R}$ which is vacuously not irreducible.

Chapter 1, Exercise *1.11: A variety that is not a local complete intersection

Let $Y\subseteq\mathbb A^3$ be the curve given parametrically by $x=t^3,y=t^4,z=t^5$. Show that $I(Y)$ is a prime ideal of height $2$ in $k[x,y,z]$ which cannot be generated by $2$ elements. We say $Y$ is not a local complete intersection-- cf. (Ex. 2.17).

This is a starred exercise, however we will provide a solution here since it is quite fun and yields some enlightening computations in commutative algebra. The difficulty in this problem is that it is not immediately obvious as to how we should parameterize this curve - indeed, the obvious method would no longer be based on a polynomial system and would use radicals which are disallowed in algebraic geometry. Thus we must turn to a more subtle way of dealing with this problem. Note that we have $\operatorname{height}\mathfrak p = \operatorname{codim}Y$ where $Y$ is the associated variety, and thus we need to show that $Y$ has dimension one.

To show $\dim Y=1$ (and that $I(Y)$ is prime in the process), we will construct a topological homeomorphism $\mathbb A^1\to Y$. Dimension and irreducibility are both preserved under homeomorphism and our desired conclusion would follow. The homeomorphism, as with most things in algebraic geometry, is canonical: by definition $Y$ is the image of $\varphi:\mathbb A^1\to\mathbb A^3$ taking $t\mapsto(t^3,t^4,t^5)$. To see that $\varphi$ is injective, suppose $\varphi(s)=\varphi(t)$ which simply means $$s^3=t^3\qquad s^4=t^4\qquad s^5=t^5.$$ If either variable is zero we are done (since we are working in an integral domain). If they are both nonzero, divide equation two by equation one to find $s=t$ again. Hence $\varphi$ is a bijection and it remains to show it is continuous. This means that if $W$ is any algebraic set in $\mathbb A^3$, then $\varphi^{-1}(W)$ is algebraic in $\mathbb A^1$. In other words, it is either finite or all of $\mathbb A^1$. Suppose that $\varphi^{-1}(W)$ is infinite, that is, there are infinitely many $t\in\mathbb A^1$ such that $(t^3,t^4,t^5)\in W$. If $W=Z(f_1,\dots,f_r)$ then each $f_i$ must vanish on infinitely many points, thus $f_i\equiv 0$ and $W=\mathbb A^3$. Consequently, $\varphi^{-1}(W)=\mathbb A^1$ which is indeed closed. Hence $\varphi$ is a homeomorphism and the first half of the exercise follows.

To speak about generators of $I(Y)$, we should give a concrete description of this ideal. Suppose $f\in A=k[x,y,z]$ such that $f(t^3,t^4,t^5)=0$ for all $t\in k$. In general we can write $$f(x,y,z)=\sum_{a_1,a_2,a_3}c_{a_1a_2a_3}x^{a_1}y^{a_2}z^{a_3}.$$ Plugging in $x=t^3,y=t^4,z=t^5$, we find $$f(t^3,t^4,t^5)=\sum_{a_1,a_2,a_3}c_{a_1,a_2,a_3}t^{3a_1+4a_2+5a_3}\stackrel{?}{=}0\quad\forall t\in k.$$ For the last condition to be true we must have that for any given value of $3a_1+4a_2+5a_3$ all valid $c_{a_1a_2a_3}$ sum to zero, in other words $$\tag{*}\sum_{3a_1+4a_2+5a_3=s}c_{a_1a_2a_3}=0$$ for any $s$. These polynomials make up $I(Y)$.

The next step is an extraordinarily tedious calculation, in which we work out the meaning of condition $(*)$ for $s\in\{0,1,2,\dots,10\}$. These eleven calculations are all extremely trivial: we will work out $s=5$ and $s=10$. For $s=5$, the only valid set of $a_i$ is $(0,0,1)$, and the sum collapses to a single term giving $c_{001}=0$. In other words, $f$ must have no term proportional to $z$. For $s=10$, the two valid combinations are $(2,1,0)$ and $(0,0,2)$ and thus $c_{210}+c_{002}=0$. In other words, the $x^2y$ and $z^2$ terms must have equal and opposite coefficients. In the end, we rule out the following terms: $$\text{constant},x,y,z,x^2,xy.$$ The following pairs must have equal and opposite coefficients: $$xz,y^2\qquad x^2,yz\qquad x^2y,z^2.$$ Thus, the polynomials $y^2-xz,x^2-yz,x^2y-z^2$ lie in $I(Y)$, and further generate it. They are minimal - if we were to remove one polynomial then the associated condition would no longer be assured. Thus $I(Y)$ cannot have fewer than three generators.

Chapter 1, Exercise 1.10: Topological properties of dimension

(a) If $Y$ is any subset of a topological space $X$, then $\dim Y\leq\dim X$.

(b) If $X$ is a topological space which is covered by a family of open sets $\{U_i\}$, then $\dim X=\sup\dim U_i$.

(c) Give an example of a topological space $X$ and a dense open subset $U$ with $\dim U<\dim X$.

(d) If $Y$ is a closed subset of an irreducible finite-dimensional topological space $X$, and if $\dim Y=\dim X$, then $Y=X$.

(e) Give an example of a noetherian topological space of infinite dimension.

(a) This is quite trivial: any closed irreducible chain in $Y$ is also such a chain in $X$, and as such cannot be longer than $\dim X$.

(b) By (a), each $\dim U_i$ is $\leq\dim X$ and thus $\sup\dim U_i\leq X$. Consider some chain $X_0\subset\dots\subset X_n$ in $X$, where $n=\dim X$. This has maximal length, thus $X_0$ must be a minimal closed irreducible subset of $X$. Thus $X_0$ is a point (more precisely, a singleton). We have $X_0\in U_i$ for some $i$. Then $X_j\cap U_i\neq\emptyset$ for each $j\leq n$, since each $X_j$ contains $X_0$, as does $U_i$. Further, each $U_i\cap X_j\subseteq X_j$ is irreducible and dense in $X_j$ by Exercise 1.6, and we see that $X_j\neq X_{j-1}$. Thus we can construct a chain $U_i\cap X_0\subset\dots\subset U_i\cap X_n$ of closed distinct irreducible sets in $U_i$, but this chain must have length $\leq\sup\dim U_i$.

(c) When looking for strange counterexamples involving "small" dense sets, the usual place to look is the Sierpinski space $X=\{0,1\}$ with open sets $\emptyset,\{1\},\{0,1\}$. Note that the longest applicable chain in $X$ is $\{1\}\subset\{0,1\}$, so $\dim X=2$, but $\dim\{1\}=1$ in $X$. Now, the only closed set containing $\{1\}$ is $X$ itself, so $\{1\}$ is dense but $\dim\{1\}<\dim\{0,1\}$.

(d) Suppose $Y\neq X$, and $Y_0\subset\dots\subset Y_n$ is a maximal chain in $Y$ (so $n=\dim Y$). Then $Y_0\subset\dots\subset Y_n\subset X$ is a longer chain in $X$ satisfying the hypotheses since $X$ is irreducible.

(e) Take any well-ordered set $S$ and give it the topology with closed sets being all initial segments. For example, choose $S=\mathbb N=\{0,1,2,\dots\}$. This satisfies a descending chain condition, since every chain will eventually reach $\{0\}$. However, it is infinite-dimensional since $\{0\}\subset\{0,1\}\subset\dots$ is a valid infinite ascending chain.

Chapter 1, Exercise 1.9: A lower bound on the dimension of irreducible components

Let $\mathfrak a\subseteq A=k[x_1,\dots,x_n]$ be an ideal which can be generated by $r$ elements. Then every irreducible component of $Z(\mathfrak a)$ has dimension $\geq n-r$.

This is an exercise that looks somewhat nontrivial, but ultimately reduces to a geometric realization of a lemma from commutative algebra generalizing Krull's Hauptidealsatz. For this reduction, we use the following consequence of Theorem 1.8Ab): if $Y$ is any affine variety in $\mathbb A^n$, then $$\dim Y = \dim A/I(Y) = n - \operatorname{height}I(Y).$$ In our case, if $Z(\mathfrak a)=\bigcup G_i$, then we need only prove that $\operatorname{height}I(G_i)\leq r$. In particular, the ideals $I(G_i)$ are minimal prime ideals over $\mathfrak a=I(Y)$ (where $Z(\mathfrak a)=Y$), so that the following lemma from commutative algebra (Krull's height lemma) will suffice.

Lemma 1. If $A$ is a noetherian ring, $\mathfrak a$ is a prime ideal generated by $r$ elements, and $\mathfrak p$ is a minimal prime ideal over $\mathfrak a$, then $\operatorname{height}\mathfrak p\leq r$.

This is true, and is sufficient.

Chapter 1, Exercise 1.8: Intersecting with a hypersurface decreases dimension by one

Let $Y$ be an affine variety of dimension $r$ in $\mathbb A^n$. Let $H$ be a hypersurface in $\mathbb A^n$, and assume that $Y\not\subseteq H$. Then every irreducible component of $Y\cap H$ has dimension $r-1$. (See (7.1) for a generalization.)

Note: It is implicit, but not stated, in this problem that $Y\cap H$ is nonempty. In the case $Y\cap H=\emptyset$ it is not even meaningful to talk about irreducible components since $\emptyset$ is not irreducible and as such has no decomposition.

This problem, probably the first truly nontrivial result given in an exercise thus far, works out most naturally algebraically because of the quotient decomposition of dimension. It is notable further for giving us an interesting application of Krull's Hauptidealsatz (Hartshorne's Theorem 1.11A).

First let us fix some notation: since $H$ is a hypersurface write $I(H)=(f)$, and let the irreducible components of $Y\cap H$ be denoted $G_i$.

Now, we will work in the affine coordinate ring $A(Y)=A/I(Y)$ of the variety $Y$. In particular, we claim that the canonical image of $f$ (the generator of $I(H)$) in this ring is neither a zero-divisor nor a unit. For the first statement, since $Y\not\subseteq H$ taking ideals gives $(f)\not\subseteq I(Y)$, so $f\notin I(Y)$. Thus, under the canonical map, $f$ is not sent to the additive identity of $A(Y)$. Now, we use the assertion that $Y\cap H\neq\emptyset$ to prove that $f$ is not a unit. If the image $\bar f$ was a unit in $A(Y)$ and we had a prime ideal $\mathfrak q$ containing $\bar f$, then $\mathfrak q=(1)$. Thus $\mathfrak q$ is not minimal - in particular, no minimal prime ideals in $A(Y)$ contain $\bar f$. We will show momentarily that this implies $Y\cap H=\emptyset$.

Our next goal is to study the algebraic representation of the $G_i$ under the correspondence. Clearly $I(G_i)$ is a prime ideal in $A$, but under the quotient map $A\to A(Y)$ the image $\mathfrak p_i$ is a prime ideal. Note that it contains $\bar f$, since $f\in I(G_i)$ and inclusion is preserved under quotient maps. It is minimal over $(\bar f)$ by the irreducibility of $G_i$, and in particular we see $\bar f$ is a unit. By the Hauptidealsatz, $\operatorname{height}\mathfrak p_i=1$ and by Theorem 1.8Ab) we see $$\dim A(Y)/\mathfrak p_i=\dim A(Y)-\operatorname{height}\mathfrak p_i=r-1.$$ Since $A(Y)$ is the coordinate ring of $G_i$ we see $\dim G_i=r-1$ as desired.

Chapter 1, Exercise 1.7: Basic properties of noetherian spaces

(a) Show that the following conditions are equivalent for a topological space $X$:
    (i) $X$ is noetherian; (ii) every nonempty family of closed subsets has a minimal element; (iii) $X$ satisfies the ascending chain condition for open subsets; (iv) every nonempty family of open subsets has a maximal element.

(b) A noetherian topological space is quasi-compact, i.e., every open cover has a finite subcover.

(c) Any subset of a noetherian topological space is noetherian in its induced topology.

(d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology.

(a) We will verify that (i) is equivalent to (ii), that (iii) is equivalent to (iv), and that (i) is equivalent to (iii).

For the first equivalence, suppose $X$ is noetherian and let $\{X_\alpha\}$ be a family of closed sets that is nonempty. Pick a member $X_0\in\{X_\alpha\}$: if $X_0$ is minimal we are done. Otherwise, there is another set $X_1\in\{X_\alpha\}$ such that $X_1\subset X_0$. This process is iterated inductively. If no $X_i$ is minimal, then we construct a chain $\dots\subset X_2\subset X_1\subset X_0$ which does not stabilize, a contradiction. Conversely, if $X$ satisfies (ii) and $X_1\supset X_2\supset\dots$ is any descending chain, then the set $\{X_i\}$ has a minimal element and consequently the chain stabilizes.

The proof of the second equivalence is exactly the same, switching around the $\subset$ signs.

For the third equivalence, if $X$ is noetherian and $X_0\subset X_1\subset\dots$ is an ascending open chain in $X$, then $X_0^c\supset X_1^c\supset\dots$ is a descending closed chain which must stabilize, and since the complements stabilize so do the original sets. The other direction is identical, and the exercise is thus completed.

(b) Let $\{U_\alpha\}$ be any open cover for $X$ and consider the set $\Sigma$ of finite unions of the $U_\alpha$. Since $X$ is noetherian, $\Sigma$ has a maximal element $U=U_{\alpha_1}\cup\dots\cup U_{\alpha_r}$ by (a). For each $\alpha$, $U_\alpha\subseteq U$, and thus the elements of $U$ form a finite subcover.

(c) Equivalently by (a), we verify the ascending chain condition for open sets. Let $Y\subset X$ be arbitrary, and let $Y\cap U_0\subset U_1\subset\dots$ be an ascending chain of open sets in $Y$ where each $U_i$ is open in $X$. Then $U_0\subset U_1\subset\dots$ is an ascending chain in $X$ which stabilizes by assumption. Thus the intersection of this chain with $Y$ also stabilizes.

(d) Note that a noetherian space endowed with the discrete topology must be a finite set, since if $X$ was an infinite space with the discrete topology then we could trivially find non-stabilizing ascending chains of open subsets since every subset is open in $X$. Thus we need only show that a noetherian Hausdorff space has the discrete topology.

It is enough to show that each $Y\subset X$ is closed, or that $Y^c$ is open. For $y\in Y$, by Hausdorffness we can construct an open set $B_y$ such that the set of all $B_y$ as $y\in Y$ is an open cover for $Y$, and such that $B_y$ is disjoint from a neighborhood $C_x$ of any $x\in Y^c$. By (b) and (c), pick a finite subcover $\{B_{y_i}\}$ of $Y$. The $B_{y_i}$ correspond to subsets $C_{x_i}$ of $Y^c$. The set $C=\bigcup C_{x_i}$ is disjoint from $Y$, and contains all points of $Y^c$ by construction. Thus $Y^c=C$ is open.

Chapter 1, Exercise 1.6: Subsets and closures of irreducible spaces

$\renewcommand{\bar}{\overline}$ Any nonempty open subset of an irreducible topological space is dense and irreducible. If $Y$ is a subset of a topological space $X$, which is irreducible in its induced topology, then the closure $\bar Y$ is also irreducible.

This is just a topology problem, and even then is not a particularly difficult one. For the first statement, we will show separately that an open $Y\subseteq X$ is dense and that it is irreducible. For density, note that $\bar Y$ and $Y^c$ are both closed and $\bar Y\cup Y^c=X$, so by irreducibility $X$ equals one of the factors: $\bar Y=X$ or $Y^c=X$. The latter implies $Y=\emptyset$ which we assume is false, thus $\bar Y=X$ and $Y$ is dense. Now, suppose we could write $Y=Y_1\cup Y_2$ where each $Y_i$ is closed. Then $X=Y_1\cup Y_2\cup Y^c$, all of which are closed, so again since $Y$ is nonempty we have $X=Y_1$ or $X=Y_2$. In either case $Y=X$ which is irreducible, so $Y=Y_1$ or $Y=Y_2$.

Now, let $Y\subset X$ be irreducible in the subspace topology and let $\bar Y = Y_1\cup Y_2$, both of which are closed. A simple calculation shows $$(Y_1\cap Y)\cup(Y_2\cap Y)=Y,$$ and both are closed in the subspace topology on $Y$. Thus $Y=Y\cap Y_1$ or $Y=Y\cap Y_2$, meaning that, without loss of generality, $Y\subset Y_1$. Then $$Y_1=Y_1\cap\bar Y=\bar{Y_1\cap Y}=\bar Y$$ as desired.

Chapter 1, Exercise 1.5: Affine coordinate rings and finitely-generated reduced $k$-algebras

Show that a $k$-algebra $B$ is isomorphic to the affine coordinate ring of some algebraic set in $\mathbb A^n$, for some $n$, if and only if $B$ is a finitely generated $k$-algebra with no nilpotent elements.

There are two directions to prove here. First, if $A=k[x_1,\dots,x_n]$ and $B=A/I(Y)$ is the affine coordinate ring of any algebraic set $Y$, we must find a canonical realization of $B$ as a $k$-algebra. To do this, we consider the composition $$k\hookrightarrow A\twoheadrightarrow A/I(Y)$$ where the first map takes elements of $k$ to their associated constant polynomials and the second map is the canonical quotient map. The first of these maps is finitely generated since it is just the polynomial algebra, and the second map is finitely generated since $A$ is noetherian (Hilbert's basis theorem). Thus the composition is also finitely generated. Further, since $I(Y)$ is a radical ideal the quotient $A/I(Y)$ is reduced.

Conversely, if $k\to B$ is any finitely generated $k$-algebra then we can realize $B$ as the quotient of some polynomial ring $k[x_1,\dots,x_n]$ by an ideal $\mathfrak a$. Since $A/\mathfrak a=B$ is finitely generated by assumption, the ideal $\mathfrak a$ is also finitely generated. It is a radical ideal since $B$ is a reduced ring. Set $Y=Z(\mathfrak a)$, then $$A/I(Y) = A/\sqrt{\mathfrak a}=A/\mathfrak a=B$$ as desired.

Chapter 1, Exercise 1.4: Product of topologies on $\mathbb A^1\times\mathbb A^1$

If we identify $\mathbb A^2$ with $\mathbb A^1\times\mathbb A^1$ in the natural way, show that the Zariski topology on $\mathbb A^2$ is not the product topology of the Zariski topologies on the two copies of $\mathbb A^1$.

This is clearly much more a problem in topology than in algebra, and in particular we can reduce it to a fairly well-known statement in general topology. We note that the diagonal $Z(y-x)$ in $\mathbb A^2$ is an algebraic set and thus closed. Recall that a topological space is Hausdorff if and only if its diagonal $\Delta =\{(x,x):x\in X\}$ is closed in the product $\mathbb A^1\times\mathbb A^1$. Thus, if $\mathbb A^1$ is not a Hausdorff space, then we have found a distinction.

To do this, we will use the fact that the Zariski topology on $\mathbb A^1$ is just the finite complement topology. More generally, we will show that the finite complement topology on any infinite set is not Hausdorff. Consider any such infinite space $X$ endowed with the finite complement topology and suppose $x,y\in X$ are separated by neighborhoods $x\in U,y\in V$. Then $(U\cap V)^c = X$ since $U\cap V=\emptyset$. By De Morgan, $U^c\cup V^c = X$, but $U^c,V^c$ are both finite and $X$ is infinite so this is a contradiction. Hence this topology is not Hausdorff as desired.

Chapter 1, Exercise 1.3: A Multi-Component Algebraic Set

Let $Y$ be the algebraic set in $\mathbb A^3$ defined by the two polynomials $x^2-yz$ and $xz-x$. Show that $Y$ is a union of three irreducible components. Describe them and find their prime ideals.

This is a relatively relaxing calculation problem to solve before entering the more abstract part of the first problem set. We are looking for three irreducible components of $Y$. To do this, we first consider two cases: $x=0$ and $x\neq 0$.

For the case $x=0$, the second equation becomes trivial and the first equation becomes $yz=0$. The algebraic set $Z(yz)$ is obviously just $Z(y)\cup Z(z)$, and further imposing the condition $x=0$ gives two irreducible components $Y_1=Z(x,y)$ and $Y_2=Z(x,z)$. These correspond geometrically to the $z$ and $y$ axes respectively, and have prime ideals $I(Y_1)=(x,y)$ and $I(Y_2)=(x,z)$.

Next, we have the case $x\neq 0$. In this case we can divide by $x$ in the second equation to obtain $z-1=0$ or $z=1$. Plugging this into the first equation, we find $y=x^2$. There appears to be a problem here however - how are we supposed to turn the condition $x\neq 0$ into a polynomial equation? The good news is that we don't have to: the point in which $x=0$ is $(0,0,1)$ which is already in $Y_1$. Thus we have the final irreducible component $Y_3=Z(y-x^2,z-1)$ with prime ideal $I(Y_3)=(y-x^2,z-1)$. This corresponds to a parabola lifted up one unit above the $xy$ plane along the $z$ axis. By construction, $Y=Y_1\cup Y_2\cup Y_3$ as desired.

Chapter 1, Exercise 1.2: The Twisted Cubic Curve

Let $Y\subseteq\mathbb A^3$ be the set $Y=\{(t,t^2,t^3):t\in k\}$. Show that $Y$ is an affine variety of dimension $1$. Find generators for the ideal $I(Y)$. Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$. We say that $Y$ is given by the parametric representation $x=t,y=t^2,z=t^3$.

At first, this problem looks like it should be extremely trivial: we are given a certain set, and we are asked to show that it is an affine variety of dimension $1$. The trick, however, appears when we realize this set is given in a parametric form, and it is in fact up to us to find the polynomials that specify or cut out this twisted cubic. Intuitively, we might expect that there are two, since the twisted cubic is a curve in $\mathbb A^3$ and thus we will have to "cut out" two dimensions. To find these polynomials, we can exploit the fact that $x=t$ along the curve and thus that $y=t^2=x^2$ and so on: in particular, we can see that $Y=Z(y-x^2,z-x^3)$. Applying $I$ to both sides, we find as expected that $I(Y)=(y-x^2,z-x^3)$. This solves one part of the problem.

Next we will show $A(Y)\cong k[x]$. As in the previous exercise, define a surjective map $k[x,y,z]\to k[x]$ taking $f(x,y,z)\mapsto f(x,x^2,x^3)$. Clearly by the Nullstellensatz this has kernel $I(Y)$, so we induce an isomorphism $A(Y)=A/I(Y)\cong k[x]$.

Finally, the remaining parts of the problem follow immediately: $Y$ is an affine variety since $k[x]$ is an integral domain, and thus $I(Y)$ is prime. Further, we have $$\dim Y = \dim A(Y) = \dim k[x]=1$$ and we are done.

Chapter 1, Exercise 1.1: Classification of Affine Conics

(a) Let $Y$ be the plane curve $y=x^2$ (i.e. $Y$ is the zero set of the polynomial $f=y-x^2$). Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$.

(b) Let $Z$ be the plane curve $xy=1$. Show that $A(Z)$ is not isomorphic to a polynomial ring in one variable over $k$.

*(c) Let $f$ be any irreducible quadratic polynomial in $k[x,y]$, and let $W$ be the conic defined by $f$. Show that $A(W)$ is isomorphic to $A(Y)$ or $A(Z)$. Which one is it when?

(a) First note $I(Y)=(y-x^2)$ by the Nullstellensatz, then $A(Y)=k[x,y]/(y-x^2)$. Consider the surjective map $k[x,y]\to k[x]$ taking $f(x,y)\mapsto f(x,x^2)$. This has kernel $(y-x^2)$, so it induces an isomorphism $A(Y)=k[x,y]/(y-x^2)\cong k[x]$.

(b) We will show that $A(Z)\cong k[x,\frac 1 x]$. Note that this is not isomorphic to $k[x]$, for example because the polynomial $f=x$ is a nonconstant unit in $k[x,\frac 1 x]$. Once again, $I(Z)=(xy-1)$ and $A(Z)=k[x,y]/(xy-1)$. Consider the surjective map $k[x,y]\to k[x,\frac 1 x]$ taking $f(x,y)\mapsto f(x,\frac 1 x)$. This has kernel $(xy-1)$, so we induce the required isomorphism once again.

(c) This is Hartshorne's first starred exercise and as such the most elegant solution I am told requires some projective geometry. However, I have come up with an elementary solution following the classification of affine conics. Such a classification for algebraically closed fields of characteristic zero is a rather sluggish piece of algebraic manipulation, so the reader is referred to the beginning of chapter 5 of Gibson's Elementary Geometry of Algebraic Curves for the calculations. The result is simple enough, however: conics over such a field can be classified as affinely isomorphic to one of

• $y^2-x=0$ (parabolas);
• $y^2-x^2-1=0$ (conics);
• $y^2-x^2=0$ (line pairs);
• $y^2-1=0$ (repeated lines);
• $y^2=0$ (repeated line).

Irreducibility is preserved under these manipulations, and thus we can discard the three later cases in which the polynomial factors. The two remaining cases are parabolas and conics. The first is exactly $Y$ and thus of course has the same coordinate ring. The second equation can be written $$(x+y)(x-y)=1,$$ and under the algebra-preserving automorphism of $k[x,y]$ taking $x\mapsto x+y$ and $y\mapsto x-y$ this becomes $xy=1$ and thus corresponds to case b).

Introduction

This page will cover, hopefully completely, my journey through the vast field of algebraic geometry via Hartshorne's text. Most of the posts here will consist of solutions to the hundreds of exercises proposed in the book, but occasionally if an insight, question, or interesting problem strikes my eye I may write an expository post discussing the concept. By the time I am finished, I hope to have a full set of solutions in this repository for others to use in the future. I feel no need to write a longer introductory post, so I will simply cover my basic notions here before moving on.

Post Structure. Each exercise-based post will contain exactly one exercise. If the exercise is divided into multiple parts a), b), c), et cetera, all parts will appear in one post. The title will consist of the chapter number, the exercise number, and a brief 3-10 word summary of the idea of the problem (sometimes provided by Hartshorne, mostly not). For example, the post corresponding to the book's third exercise would be Chapter 1, Exercise 1.3: A Multi-Component Algebraic Set. Within the post will first be a statement of the problem, and then the solution divided into lemmas and the final proof as is necessary.

Expository posts will be labeled as such. For example, a post on properties of graded rings would be titled Expository: Properties of Graded Rings or similar. Such posts will consist of nothing more than the discussion involved.

Starred Exercises. For the most part, I will attempt to complete exercises in order, doing each one without stopping and using only techniques discussed in the book up to the given point. However, in the special case of exercises that have been starred as difficult I may skip them, creating a post with no solution and come back to them later, perhaps with more tools at my disposal.

Double-Starred Exercises. Double-starred exercises correspond to open problems in the world of Hartshorne. I will create a post for each such exercise and either write a brief commentary on the problem or simply mention that it is a double-starred and thus open problem.

Thanks for stopping by!