## Exercises from Hartshorne

Work out Hartshorne exercises here!

### 15 Responses to “Exercises from Hartshorne”

1. stevengubkin Says:

1.1.1 (a) Let $Y$ be the plane curve $y=x^2$. Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$.

Proof:

Define $\phi: k[x,y] \rightarrow k[t]$ by $\phi(x) = t, \phi(y) =t^2$. Then $\phi$ is clearly surjective, and $ker(\phi) = (y-x^2)$. Thus $A(Y) = k[x,y]/(y-x^2) \cong k[t]$.

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

Proof:

Assume to the contrary that $A(Z) \cong k[t]$. Then there is a homomorphism $\phi: k[x,y] \rightarrow k[t]$ such that $ker(\phi) = (xy-1)$. But every homomorphism from $k[x,y] \rightarrow k[t]$ is determined by the images of $x$ and $y$. Say $\phi(x) = f(t)$ and $\phi(y) = g(t)$. Then $f(x,y)g(x,y) = 1$, since $xy-1$ is in the kernel. But then $f,g \in k$, and so the kernel of $\phi$ must actually contain both $x-f \text{and} y-g$. But $(x-f, y-g)$ is a maximal ideal (the quotient is easily seen to be a field), so we have $ker(\phi) = (x-f, y-g)$. Clearly $(xy-1) \neq (x-f, y-g)$, so we have a contradiction.

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

2. stevengubkin Says:

1.1.12 Give an example of an irreducible polynomial $f \in \mathbb{R}[x,y]$ whose zero set $Z(f)$ in $\mathbb{A}_\mathbb{R}^2$.

Example: Let $f(x,y) = x^2(x-1)^2 + y^2$. Then $f$ is clearly irreducible, but $Z(f) = \{(0,0),(0,1)\}$, which is composed of two irreducible components.

3. zizhan Says:

1.1.2 The Twisted Cubic Curve. Let $Y \subseteq {\bf A}^3$ 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 $x = t, y=t^2, z= t^3$.

Proof:

Let $k[s]$ denote a polynomial ring in one variable over $k$. Define a ring homomorphism $\varphi: A \rightarrow k[s]$ by $x \rightarrow s, y \rightarrow s^2, z \rightarrow s^3$, $a \rightarrow a, \forall a \in k$, which is clearly surjective. For any $f \in A, \varphi(f) = 0$ iff $f(P) = 0$ for every $P \in Y$. Hence ker$\varphi = I(Y)$ and $\varphi$ induces an isomorphism $A(Y) \cong k[s]$. Since $k[s]$ is an integral domain, so is $A(Y)$. Hence $I(Y)$ is a prime ideal of $A$. Note that $Y = Z(I(Y))$ since for any point $P \not \in Y$, one can easily construct a polynomial of the form $f = a(y-x^2)+b(z-x^3) \in I(Y)$ such that $f(P) \neq 0$ by choosing proper $a$ and $b$ in $k$. Therefore, $Y$ is an affine variety. Furthermore, dim $Y$ = dim $A(Y)$ = dim $k[s] = 1$. We claim that $I(Y)$ is generated by $y-x^2$ and $z-x^3$. It is clear that $y - x^2 \in I(Y)$ and $z - x^3 \in I(Y)$. For any $f \in I(Y)$, by the division algorithm, we have $f = g(y-x^2)+h$ such that $h \in k[x,z]$ and $h = g'(z-x^3)+h'$ such that $h' \in k[x]$. Hence $f = g(y-x^2)+g'(z-x^3)+h'$. Since $f \in I(Y)$, $h'(t) = 0$ for every $t \in k$. Therefore, $h' = 0, f \in (y-x^2, z-x^3)$.

4. Kun Wang Says:

1.1.3 Let $Y$ be the algebraic set in $A^3$ 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.

Proof: Let $P_1=(x,y), P_2=(x,z), P_3=(x^2-y,z-1)$. $P_1, P_2, P_3$ are prime ideals since $k[x,y,z]$ quotient by them are all easily seen to be isomorphic to $k[x]$ which is a domain. Hence $Z(P_1), Z(P_2), Z(P_3)$ are irreducible. By solving the equations $x^2-yz=0, xz-x=0$, we get $Y=Z(P_1)\cup Z(P_2)\cup Z(P_3)$ which is the unique irreducible decomposition of $Y$ into three irreducible components.

5. Kun Wang Says:

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

Proof: The diagonal set $\bigtriangleup =\{(a,a)\in A^2: a\in A\}=Z(f=x-y)$ is closed in the Zariski topology on $A^2$. For any base open set $U\times V$ of the product topology on $A^2$, $U\cap V\neq \emptyset$ since their complements in $A^1$ is finite while $A^1$ is infinite ($k$ is algebraicly closed). Thus $\exists a_0\in U\cap V$. Hence $(a_0,a_0)\in (U\times V)\cap \bigtriangleup$. Therefore we have proved that any open set in the product topology has nonempty intersection with $\bigtriangleup$. Now if $\bigtriangleup$ is closed in the product topology, then its complement is open in the product topology and they don’t intersect. Contradiction! Thus $\bigtriangleup$ is not closed in the product topology which proves the statement.

6. Kun Wang Says:

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

Proof: Suppose $B\cong k[x_1,\cdots, x_n]/I$, where $\sqrt I=I$. Clearly $B$ is a finitely generated k-algebra. Now suppose there exists nonzero $f+I\in k[x_1,\cdots, x_n]/I$ s.t. $f^r\in I$, then $f\in \sqrt I=I$ which implies $f+I=0$ in $B$. Contradiction! Hence $B$ has no nilpotent.

Now suppose $B$ is a finitely generated k-algebra with no nilpotent elements. Then $B\cong k[x_1,\cdots, x_n]/I$ for some $n$ and ideal $I$ in $k[x_1,\cdots, x_n]$. Now suppose $f^r\in I$, then $(f+I)^r=0$ in $B=k[x_1,\cdots, x_n]/I$ which implies $f+I=0$ in $k[x_1,\cdots, x_n]/I$ by assumption. Hence $f\in I$. Thus $\sqrt I=I$. Therefore $B$ is the coordinate ring of $Z(I)$.

7. Kun Wang Says:

1.1.6 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 $\overline {Y}$ is also irreducible.

Proof. Let $U$ be open and nonempty in the irreducible space $X$. The case $U=X$ is trival. Now if $U\neq X$. Suppose $U$ is not dense, then $\overline{U}$ and $U^c$ are nonempty and proper closed subsets of $X$ which cover $X$. This contradicts to the fact that $X$ is irreducible. Hence $U$ is dense. Now suppose $U=F_1\cup F_2$, where $F_1, F_2$ are closed in $U$. Then there exist $L_1, L_2$ closed in $X$ s.t. $F_1=L_1\cap U, F_2=L_2\cap U$. Note $X=\overline{U}=\overline{F_1\cup F_2}\subseteq{L_1\cup L_2}$. Hence, say, $X=L_1$ since $X$ is irreducible. This implies $F_1=L_1\cap U=U$ and thus $U$ is irreducible.

Now suppose $\overline{Y}=F_1\cup F_2$ with $F_1, F_2$ closed in $\overline{Y}$. Note since $\overline{Y}$ is closed in $X$, $F_1, F_2$ also closed in $X$. However $Y=(Y\cap F_1)\cup (Y\cap F_2)$, we get, say, $Y=Y\cap F_1$ since $Y$ is irreducible. Therefore $Y\subseteq F_1$ which imples $\overline{Y}\subseteq F_1$. Hence $\overline{Y}=F_1$ and $\overline{Y}$ is irreducible.

8. zizhan Says:

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

Proof: Every irreducible component $Y$ of $Z(\mathfrak{a})$ is an affine variety. Furthermore, as $A$ is a Noetherian ring, $\mathfrak{p} = I(Y)$ is a minimal prime ideal belonging to $\mathfrak{a}$ since if $\mathfrak{a} \subseteq \mathfrak{q} \subsetneqq \mathfrak{p}$ and $\mathfrak{q}$ is a prime ideal, then $Z(\mathfrak{a}) \supseteq Z(\mathfrak{q}) \supsetneqq Z(\mathfrak{p}) = Y$, which contradicts the fact that $Y$ is an irreducible component of $Z(\mathfrak{a})$. Since $\mathfrak{a}$ can be generated by $r$ elements, $\text{height } \mathfrak{p} \leq r$ (Atiyah-Macdonald (11.16)). By (1.8A), $\text{height }\mathfrak{p}+\text{dim } A/\mathfrak{p} = \text{dim }A$, and by (1.7), $\text{dim } Y = \text{dim } A/\mathfrak{p}$. Hence $\text{dim } Y = n - \text{height }\mathfrak{p} \geq n-r$.

9. zizhan Says:

1.1.8 Let $Y$ be an affine variety of dimension $r$ in ${\bf A}^n$. Let $H$ be a hypersurface in ${\bf A}^n$, and assume $Y \nsubseteq H$. Then every irreducible component of $Y \cap H$ has dimension $r-1$.

Proof: Since $H$ is a hypersurface in ${\bf A}^n$, $H = Z(f)$ for some nonconstant irreducible polynomial $f$. Let $\mathfrak{p} = I(Y)$. We note that $I(Y \cap H) = \sqrt{I(Y)+I(H)} = \sqrt{\mathfrak{p}+(f)}$. It follows that every irreducible component of $Y \cap H$ is defined by a minimal prime ideal $\mathfrak{q}$ of $A$ that contains both $\mathfrak{p}$ and $f$. Let $\overline{f}$ and $\overline{\mathfrak{q}}$ denote the image of $f$ and $\mathfrak{q}$ in $A/\mathfrak{p}$ under the natural map from $A$ to $A/\mathfrak{p}$. Then $\overline{\mathfrak{q}}$ is a minimal prime ideal of $A/\mathfrak{p}$ containing $\overline{f}$, and $A/\mathfrak{q} \cong (A/\mathfrak{p})/\overline{\mathfrak{q}}$. Hence by (1.8A), $\text{dim } A/\mathfrak{q} = \text{dim } A/\mathfrak{p} - \text{height }\overline{\mathfrak{q}}$. Since $Y \nsubseteq H$, $\overline{f}$ is nonzero in $A/\mathfrak{p}$. Hence by (1.11A), $\text{height }\overline{\mathfrak{q}} = 1$. Therefore, $\text{dim } A/\mathfrak{q} = r-1$.

10. zizhan Says:

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

Proof: Consider any closed irreducible subsets $F_0,...,F_l$ of $Y$ such that $\emptyset \neq F_0 \subsetneq F_1 \subsetneq \cdots \subsetneq F_l \subseteq Y$. We claim that every $F_i, 0 \leq i \leq l$ is also irreducible in $X$. Suppose $F_i = C \cup D$ for some nonempty closed subsets $C$ and $D$ of $X$, then since $F_i \subseteq Y$, $C$ and $D$ are also closed subsets of $Y$, which contradicts the fact that $F_i$ is irreducible in $Y$. It follows that every $\overline{F}_i$ is a closed irreducible subset of $X$ and $\emptyset \neq \overline{F}_0 \subsetneq \overline{F}_1 \subsetneq \cdots \subsetneq \overline{F}_l \subseteq Y$. Hence $\text{dim } Y \leq \text{dim } X$.

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

Proof: By (a), $\text{dim }X \geq \text{sup dim } \{U_i\}$. Now consider any closed irreducible subsets $F_0,...,F_l$ of $X$ such that $\emptyset \neq F_0 \subsetneq F_1 \subsetneq \cdots \subsetneq F_l \subseteq X$. Since $\{U_i\}$ is an open cover of $X$, there is $i$ such that $F_0 \cap U_i \neq \emptyset$. We claim that $F_i = \overline{F_i \cap U_i}$ for every $i$. Since $F_i$ is closed and $F_i \supseteq F_i \cap U_i, F_i \supseteq \overline{F_i \cap U_i}$. Suppose $F_i \supsetneq \overline{F_i \cap U_i}$. Then $F_i = \overline{F_i \cap U_i} \cup (F_i - (F_i \cap U_i))$ where $F_i - (F_i \cap U_i) = F_i \cap (X - U_i)$ is a nonempty closed subset, which contradicts the fact $F_i$ is irreducible. Therefore, $F_i = \overline{F_i \cap U_i}$. Hence $F_i \cap U_i$ is irreducible, and we have $\emptyset \neq F_0 \cap U_i \subsetneq F_1 \cap U_i \subsetneq \cdots \subsetneq F_l \cap U_i \subseteq U_i$. It follows that $\text{dim } X \leq \text{sup dim } \{U_i\}$. Hence $\text{dim }X=\text{sup dim } \{U_i\}$.

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

Example: Let $X = \{a,b\}$ and the open subsets of $X$ are $\emptyset, X$ and $\{a\}$, which clearly defines a topology on $X$. Let $U = \{a\}$. Then $U$ is a dense open subset since $\overline{U} = X$. It is easy to see that $\text{dim} U = 0$ and $\text{dim } X = 1$.

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

Proof: Suppose $Y \neq X$. Then by the same argument as in (a), any chain of distinct irreducible closed subsets of $Y$ is also chain of distinct irreducible closed subsets of $X$. Since $X$ is irreducible and $Y \neq X$, one can add $X$ to the end of the chain. It follows that $\text{dim }X \geq \text{dim }Y+1$. Since $X$ is finite-dimensional, we have $\text{dim }Y \neq \text{dim }X$, a contradiction.

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

Example: Let $X = [0,1]$. Define the closed subsets of $X$ to be $\emptyset, X$ and $\{[1/n,1] | n \in \mathbb{N}^+\}$, which clearly gives a topology on $X$. Furthermore, each closed subset is clearly irreducible. Since any descending chain of closed subsets is lower bounded by $[1/2,1]$, $X$ is noetherian. On the other hand, for any integer $n>1$, there is a chain $[1/2,1] \subset [1/3,1] \subset \cdots$ $\subset [1/n,1] \subset [0,1]$ of distinct irreducible closed subsets. Hence $X$ is infinite dimensional.

11. Kun Wang Says:

1.1.7 (a) Show that the following conditions are equivalent for a topological space $X$:
(i) $X$ is notherian (ii) Every nonempty family of closed subsets has a minimal element (iii) X satisfies the ascending chian condition for open subsets (iv) Every nonempty family of open subsets has a maximal element;
(b) A notherian topological space is quasi-compact
(c) Any subset of a notherian topological space is notherian in its induced topology
(d) A notherian space which is also Hausdorff must be a finite set with the discrete topology

Proof: (a) (i)$\Rightarrow$(ii) Suppose not, then for such a family, we can construct a non-ending descending chain of closed subsets starts any element of this family; (ii)$\Rightarrow$(iii) Suppose not, choose a non-ending ascending chain of open sets, then the complement of this chain has no minimal element; (iii)$\Rightarrow$(iv) Suppose not, then for such a family, we can construct a non-ending ascending chain of open subsets starts any element of this family (iv)$\Rightarrow$(i) Then every family of nonempty closed subsets has a minimal element, thus every descending chain of closed subsets stablizes since the set they form has a minimal element

(b) Suppose $X$ is a notherian space which is not quasi-compact, then there exists an open cover $\{U_\alpha\}$ of $X$ has no finite subcover. Choose $U_1$ from the cover, it doesn’t cover $X$. Hence we can choose $U_2$ from the cover s.t. $U_1\subsetneqq U_1\cup U_2$. Still $U_1\cup U_2$ doesn’t cover $X$, we can choose $U_3$ from the cover with $U_1\subsetneqq U_1\cup U_2\subsetneqq U_1\cup U_2\cup U_3$ and $U_1\cup U_2\cup U_3$ doesn’t cover $X$. Proceeding in this way, we can construct a non-stationary ascending chain of open sets which contradicts to the fact that $X$ is notherian.

(c) Let $Y$ be a subset of $X$. Let $F_0\supset F_1\supset F_2\supset\cdots$ be a descending chain of closed subets of $Y$. Now $\overline{F_0}\supset \overline{F_1}\supset \overline{F_2}\supset\cdots$ is a descending chain of closed subsets in $X$, hence stationary. Now note $F_i=\overline{F_i}\cap Y$, Therefore $F_0\supset F_1\supset F_2\supset\cdots$ is stationary. Hence $Y$ is notherian

(d) Let $X$ be a notherian space which is Hausdorff. Suppose $A\subset X$ has more than two points. $A$ is Hausdorff since $X$ is. Choose $a\neq b$ in $A$. Then there exist non-intersect open subsets $U, V$ of $A$ s.t. $a\in U, b\in V$. Then $A=(A-U)\cup (A-V)$ and thus $A$ is reducible. We have proved that those irreducible subsets of $X$ are one-point subsets. Since $X$ is notherian, $X$ is a finite union of irreducible components, which in this case, is a finite union of closed points. Hence $X$ is a finite set with discrete topology.

12. zizhan Says:

1.3.1 (a) Show that any conic in ${latex \bf A^2}$ is isomorphic either to ${\bf A^1}$ or ${\bf A^1}-\{0\}$. (cf Ex. 1.1).

Proof: By Ex.1.1.1(c), for any conic $W$ in ${\bf A^2}$, $A(W)$ is isomorphic to either $A(Y)$ or $A(Z)$ defined in 1.1.1(a) and (b). Since $A(Y)$ is isomorphic to $k[t]$, a polynomial ring in one variable over $k$, and $A({\bf A^1}) = k[t]/\{0\}$, $Y \cong \bf A^1$ by Corollary (3.7). We claim that $Z \cong {\bf A^1}-\{0\}$. Define $\varphi: Z \rightarrow {\bf A^1}-\{0\}$ by $(x,1/x) \rightarrow x$, which is clearly bijective. The close subsets of ${\bf A^1}-\{0\}$ are any finite subsets without including 0. Every close subset of $Z$ is the intersection of a finite set of polynomials in $k[x,y]$ (since $k[x,y]$ is noetherian) with the curve $xy = 1$ and hence contains only finite points. It follows that the close subsets of $Z$ and ${\bf A^1}-\{0\}$ has a natural correspondence under $\varphi$. Hence $\varphi$ is bicontinuous. For any open set $V \subset Z$ and for every regular function $f: V \rightarrow k$, the function $f \circ \varphi^{-1}: \varphi(V) \rightarrow k$ is clearly a regular function since a rational function in terms of $x$ and $1/x$ is also a rational function in terms of $x$. Hence $\varphi^{-1}$ is a morphism. It is easy to see that $\varphi$ is also a morphism. Therefore, $\varphi$ is an isomorphism of varieties.

(b) Show that ${\bf A^1}$ is not isomorphic to any proper open subset of itself. (This result is generalized by (Ex.6.7) below.)

Proof: Let $X$ denote a proper open subset of ${\bf A^1}$. Then $X = {\bf A^1} - \{a_1,...,a_n\}$ for some $a_1,...,a_n \in k$. Let $f(x) = (x-a_1)(x-a_2)...(x-a_n)$. Then $1/f$ is regular on $X$. Suppose $X \cong {\bf A^1}$. Then $k[x] \cong \mathcal{O}(X)$ as $k$-algebras by Proposition (3.5). However, only elements in $k$ are invertible in $k[x]$, while $f \not \in k$ is invertible in $\mathcal{O}(X)$, a contradiction!

13. Tao Says:

1.2.1
Prove the “homogeneous Nullstellensatz”, which says that 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 $f^{q} \in a$ for some $q > 0$.
Proof:
Let $a\subseteq S$ be a homogeneous ideal and $f\in S$ a homogeneous polynomial with
deg f > 0, such that $f(P) = 0$ for all
$P\in Z(a)$ in $\mathbf{P}^{n}$. Then f vanishes for all representatives
for points in Z(a) in $\mathbf{A}^{n}$, so $f^{q} \in a$ for some $q > 0$ when applying Theorem
1.3A fot Z(a) as being in $\mathbf{A}^{n}$. Then $f^{q} \in a$ for some $q > 0$ as homogeneous polynomial and homogeneous
ideal in $\mathbf{P}^{n}$.

14. Tao Says:

1.2.2.
For a homogeneous ideal $a\subseteq S$, show that the following conditions are equivalent:
(i) $Z(a) = \emptyset$;
(ii)$\sqrt{a}=$ either $S$ or the ideal $S_{+}=\bigoplus_{d>0}S_{d}$;
(iii) $a\supseteq S_{d}$ for some $d > 0$.

(i)$\Rightarrow$(ii): If $f\in S$ and deg $f >0$, $f(P) = 0$ for all
$P\in Z(a)$ since $Z(a) = \emptyset$, hence by 1.2.1, $\exists q>0$, $\ni f^{q} \in a$. Hence $S_{+} \subseteq \sqrt{a}$.
If $\sqrt{a}$ contains any element of $S_{0}$, then it contains the whole $S_{0}$ since $\sqrt{a}$ is an ideal, so $\sqrt{a}=$ either $S$ or the ideal $S_{+}=\bigoplus_{d>0}S_{d}$.

(ii)$\Rightarrow$(iii): Noticed that our $S = k[x_{0},..., x_{n}]$.Then for each $x_{i}$, there is an integer $q_{i}$ such that $x_{i}^{q_{i}} \in a$. So for
N =max {$q_{i}$}, we have $x_{i}^{N} \in a$. Therefore for $d>(n + 1)N + 1$ , every monomial of degree d contains a multiple of $x_{i}^{N}$
for some i, so $S_{d} \subseteq a$.

(iii)$\Rightarrow$(i): If $a\supseteq S_{d}$ for some $d > 0$, then $a$ contains $x_{i}^{d}$ for $i = 0,...,n$. This implies that $Z(a)$
consists of those points which are 0 for all $x_{i}$, and hence is empty.

15. Tao Says:

1.2.3

(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 $\mathbf{P}^n$, then $I(Y_{1})\supseteq I(Y_{2})$.
(c) For any two subsets $Y_{1}$,$Y_{2}$ of $\mathbf{P}^n$, $I(Y_{1} \cup Y_{2}) = I(Y_{1}) \cap I(Y_{2})$.

(d) If $a \subseteq S$ is a homogeneous ideal with $Z(a) \ne \emptyset$, then $I(Z(a)) = \sqrt{a}$.
(e) For any subset $Y \subseteq \mathbf{P}^n$, $Z(I(Y )) = \bar{Y}$ .

Pf:
(a) If $P \in Z(T_{2})$, then $f(P) = 0$, $\forall f \in T_{2}$. However $T_{1}\subseteq T_{2}$, hence$f(P) = 0$,$\forall f \in T_{1}$. Therefore $P \in Z(T_{1})$.
(b) If $f \in I(Y_{2})$, then $f(P) = 0$, $\forall P \in Y_{2}$. However $Y_{1}\subseteq Y_{2}$, hence$f(P) = 0$,$\forall P \in Y_{1}$. Therefore $f \in I(Y_{1})$.
(c) If $f \in I(Y_{1}\cup Y_{2})$, then $f(P) = 0$,$\forall P \in Y_{1} \cup Y_{2}$, hence $f(P) = 0$,$\forall P \in Y_{1}$ and $f(P) = 0$,$\forall P \in Y_{2}$. Therefore $f \in I(Y_{1})\cap I(Y_{2})$.
If $f \in I(Y_{1})\cap I(Y_{2})$, then $f(P) = 0$,$\forall P \in Y_{1}$ and $\forall P \in Y_{2}$, hence $f(P) = 0$,$\forall P \in Y_{1}\cup Y_{2}$. Therefore $f \in I(Y_{1}\cup Y_{2})$.
(d) Since $Z(a) \ne \emptyset$, $a$ can not contain any deg 0 polynomial.
Then $\forall f \in I(Z(a))$, with positive degree, we have $f(P)=0, \forall P \in Z(a) \subseteq \mathbf{p}^{n}$, by 1.2.1 “homogeneous Nullstellensatz”, $f \in \sqrt{a}$.
Therefore $I(Z(a)) \subseteq \sqrt{a}$.
Conversely, if $f^q \in a$ for some $q$, then $\forall P \in Z(a)$, $f^q (P)=0$, hence $f(P)=0$. Therefore $I(Z(a)) \supseteq \sqrt{a}$.
(e) Use corollary 2.3, $Y$ is covered by the open sets $Y_{i}=Y \cap U_{i}$, $i=0,...,n$, each $Y_{i}$ is homeomorphic to an affine variety. Then for each $Y_{i}$ we have $Z(I(Y_i))=\bar{Y_{i}}$.
Still we have $Z(I(Y))=Z(I(\bigcup_{1}^{n}Y_{i}))=Z(\bigcap_{1}^{n}I(Y_{i})=\bigcup_{1}^{n}Z(I(Y_{i}))$.
So $Z(I(Y))=\bigcup_{1}^{n}Z(I(Y_{i}))=\bigcup_{1}^{n}\bar{Y_{i}}$=closure of $\bigcup_{1}^{n}Y_{i}$=$\bar{Y}$.