Chapter 1 – Fulton


Discussion of exercises from Algebraic Curves – Chapter 1 – Fulton

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8 Responses to “Chapter 1 – Fulton”

  1. Jose Cervantes Says:

    Exercise 1.6 Show that any algebraically closed field is infinite.

    Proof: Let {k} be an algebraically closed field. If {k} is finite, then we can write {k=\{ a_1,a_2, \cdots, a_r\}}. We can consider:

    \displaystyle  p(X)=(X-a_1)\cdots(X-a_r)+1.

    Since {k} is algebraic closed, there exits a root of {p(X)} in {k}, but {p(a)=1\neq 0} for all {a\in k}. It is a contradiction. Hence {k} is infinite. \Box

  2. Jose Cervantes Says:

    Exercise 1.7 Let {k} be a field, {F\in k[ X_1,\cdots, X_n ], a_1,\cdots, a_n\in k}.

    (a) Show that {F=\sum \lambda_{(i)}(X_1-a_1)^{i_1}\cdots(X_n-a_n)^{i_n}, \lambda_{(i)}\in k}.

    Proof: Use induction over {n}. When {n=1} then {F\in k[ X_1]}, by the division algorithm in {k[ X_1]} we have that {F=\sum \lambda_{(i)}(X_1-a_1)^{i}}, where {\lambda_{(i)}\in k}.

    Assume that it is true for {n=r}. Let {F\in k[ X_1,\cdots, X_{r+1} ]= k[ X_1,\cdots, X_{r} ][X_{r+1}]} then by the division algorithm in {k[ X_1,\cdots, X_{r} ][X_{r+1}]} we obtain:

    \displaystyle  F=\sum G_{(i)}(X_{r+1}-a_{r+1})^{i},\quad\text{where } G_{(i)}\in k[ X_1,\cdots, X_{r} ].

    Using the induction hypothesis in every {G_{(i)}}, we get that:

    \displaystyle F=\sum \lambda_{(i)}(X_1-a_1)^{i_1}\cdots(X_{r+1}-a_{r+1})^{i_{r+1}},\quad\text{where } \lambda_{(i)}\in k.

    Hence the assertion is true. \Box

    (b) If {F(a_1,\cdots, a_n)=0}, show that {F=\sum_{i=1}^n G_{(i)}(X_{i}-a_{i})} for some (not unique) {G_{(i)}\in k[ X_1,\cdots, X_n ]}

    Proof: By part (a), {F(a_1,\cdots, a_n)=0} implies that {\lambda_{(i)}=0} when {i_1=\cdots=i_n=0}. Consequently, if { \lambda_{(i)}(X_1-a_1)^{i_1}\cdots(X_n-a_n)^{i_n}\ne 0} then there exists {r} such that {i_r\ne 0}. Then we can factor out {(X_r-a_r)} from { \lambda_{(i)}(X_1-a_1)^{i_1}\cdots(X_n-a_n)^{i_n}} and regroup the summation. Hence:

    \displaystyle F=\sum_{i=1}^n G_{(i)}(X_{i}-a_{i}),\quad\text{where } G_{(i)}\in k[ X_1,\cdots, X_{n} ].


  3. Jose Cervantes Says:

    Exercise 1.8 Show that the algebraic subsets of {\mathbb{A}^1(k)} are just finite subsets, together with {\mathbb{A}^1(k)} itself.

    Proof: Suppose {V(S)} is an algebraic subset of {\mathbb{A}^1(k)}. If {V(S)\ne \mathbb{A}^1(k)} then there exists a non-zero polynomial {F_0\in S \subset k[X_1]}. {F_0} has at most a finite number of roots, and thus {V(F_0)} and {V(S)=\cap_{F\in S}V(F)} are finite. Hence the algebraic subsets of {\mathbb{A}^1(k)} are just finite subsets, together with {\mathbb{A}^1(k)} itself. \Box

  4. Jose Cervantes Says:

    Exercise 1.10 Give an example of a countable collection of algebraic sets whose union is not algebric

    Proof: Let {k=\mathbb{C}} and {V_n=V(X-n)=\{n\}} for all integers {n}. {V_n} is an algebraic set of {A^1(k)} and {|V_n| < \infty}. Thus {\{V_n\}_{n\in\mathbb{Z}}} is a countable collection of algebraic sets, and {\cup_{_{\mathbb{Z}}}V_n=\mathbb{Z}}. We have that {|\cup_{_{\mathbb{Z}}}V_n|=\infty} and {\cup_{_{\mathbb{Z}}}V_n\ne A^1(k)}. By problem {1.8}, {\cup_{_{\mathbb{Z}}}V_n} is not algebraic. \Box

  5. Rudy Says:

    Fulton 1.14 Let {F} be a nonconstant polynomial in {k[X_1, \dots,X_n]}, {k} an algebraically closed field.

    Claim 1: {\mathbb{A}^{n}(k) \setminus V(F)} is infinite if {n \ge 1}. Proof: It’s clear that as {k} is an algebraically closed field {k} is infinite. This implies, of course, that {\mathbb{A}^n(k)} is infinite. Now suppose {\mathbb{A}^n(k) \setminus V(F)} is finite. Then, there exists {G \in k[X_1, \dots, X_n]} such that {V(G) = \mathbb{A}^n(k) \setminus V(F)}. This implies {V(FG) = \mathbb{A}^n(k)}, but this only occurs when {FG = 0}. As {G} is clearly nonzero, we must have {F = 0}. But {F} is nonconstant by hypothesis. We conclude {\mathbb{A}^n(k) \setminus V(F)} must be infinite. \Box Claim 2: If {n \geq 2}, then {V(F)} is infinite. Proof: For every {(n-1)}-tuple {(a_1, \dots, a_{n-1}) \in k^{n-1}} we have {F(X_1, a_1, \dots, a_{n-1}) \in k[X_1]}. Because {k} is algebraically closed, there exists {a \in k} such that {F(a, a_1, \dots, a_{n-1}) = 0}. Thus we have shown for every {(n-1)}-tuple {(a_1, \dots, a_{n-1})} there exists {a \in k} such that {(a, a_1, \dots, a_{n-1})} is a root of {F}. Thus {V(F)} is infinite. \Box

  6. Jose Cervantes Says:

    Exercise 1.16 Let {V, W} be algebraic sets in {\mathbb{A}^n(k)}. Show that {V=W} if an only if {I(V)=I(W)}.

    Proof: For two algebraic sets {V} and {W}, we have {V\subseteq W} (resp. {W\subseteq V}) if and only if {I(V)\supseteq I(W)} (resp. {I(W)\supseteq I(V)}), because {V(I(V))=V} and {V(I(W))=W} . Hence {V=W} implies {I(V)=I(W)} \Box

  7. Rudy Says:

    Fulton 1.20 Claim 1: For any ideal {I \in k[X_1, \dots, X_n]}, {V(I) = }Rad{(I)}. Proof: {I \subset }Rad{(I) \implies V(}Rad{(I)) \subset V(I)}, trivially.

    Conversely, say {\textbf{x} \in V(I)} Then for {F \in }Rad{(I)} we have {F^m(\textbf{x}) = 0} for precisely the {m} such that {F^m \in I}. We need to show that {F(\textbf{x}) = 0}. It’s very easy to see {I(\textbf{x})} is a prime ideal. For {a,b \in k[X_1, \dots, X_n]} are such that {ab \in I(\textbf{x}) \iff a(\textbf{x})b(\textbf{x}) = 0 \in k \iff a(\textbf{x}) = 0} or {b(\textbf{x}) = 0 \iff a} or {b \in I(\textbf{x})}. Thus {F^m \in I(\textbf{x}) \implies F \in I(\textbf{x})}. We conclue {V(I) \subset V(}Rad{(I))}. \Box

    Claim 2: Rad{(I) \subset I(V(I))}. Proof: Suppose {F^m \in I}. Then for any {\textbf{x} \in V(I)} we have {F(\textbf{x})^m = F^m(\textbf{x}) = 0 \in k \implies F(\textbf{x}) = 0 \implies F \in I(V(I))}. We conclude Rad{(I) \subset I(V(I))}. \Box

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