## Chapter 1 – Fulton

Discussion of exercises from Algebraic Curves – Chapter 1 – Fulton

Here is a link to the book:

http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

### 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:

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

$\Box$

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${(I) \subset I(V(I))}$. $\Box$