Exercises from Hartshorne

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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}, hencef(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}, hencef(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}.

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