[ ALGE I ] Introduction to the Plane Algebraic Curve
Well i'm going to retell what i've studied from the algebraic geometry books so far.. one of my favorite book is the Basic Algebraic Geometry: Varieties in Projective Space By Shafarevich.
Let $\mathbb{F}$ be an algebraic closed field. An algebraic curve $X$ is the set of points $(x.y)$ defined by the equation \[f(x,y)=0\]
where $f(x,y) \in \mathbb{F}[x,y]$.
When $f(x,y)$ is an irreducible polynomial, we say $X$ is an algebraic irreducible curve.
The degree of $X$ is equal to $\text{deg}f$. The condition that $\mathbb{F}$ must be an algebraic closed field is necessary for the uniqueness of the degree of $X$.
For example the curve $(0,0)$ in the field of real numbers is defined by infinitely many equations that is $x^{2n}+y^{2n}=0$ for any positive integers $n$. For even $n$ the curve is irreducible, while whenever $n$ is odd, the curve is reducible.This confusion doesn't happen when $\mathbb{F}$ is an algebraic closed field. As we can prove it by the following very useful lemma:
Lemma: Let $\mathbb{F}$ be a field, given an irreducible polynomial $f \in \mathbb{F}[x,y]$. Then for any polynomial $g \in \mathbb{F}[x,y]$ which is not divisible by $f$, the system of equations $f(x,y)=g(x,y)=0$ has only finitely many solutions.
Proof: We view $f$ and $g$ as the element of $u[y]$, where $u$ is the field of rational function of $x$ with coefficient in $\mathbb{F}$. By Gauss Lemma, it is easy to see that $f$ remains irreducible in $u[x]$ and also $f$ and $g$ are still relatively prime in $u[y]$. Thus there exists $\lambda, \mu \in u[y]$ such that
\[f \lambda + g \mu =1\]
Multiplying by the common denominator $a(y) \in \mathbb{F}[y]$ we will get
\[f(x,y) \lambda(x,y) + g(x,y) \mu(x,y) = a(y)\]
If $(\alpha, \beta)$ is the solution of the system of equations. Then $a(\beta)=0$, and there are only finitely many possibility of such $\beta$. Thus $\alpha$ is the root of equation $f(x, \beta)=0$, and there are also a finitely many value of them. The result follows.
Applying the lemma, let $X$ be an irreducible algebraic curve in an algebraic closed field. If there are two such equations which define $X$ say $f(x,y)=0$ and $g(x,y)=0$, since $f(x,y)=0$ has infinitely many solution in the algebraic closed field then the systems of equations $f(x,y)=g(x,y)=0$ also has infinitely many solutions. Hence the lemma says that $f(x,y)=c g(x,y)$. Thus the two equations are essentially the same.
The same is true for arbitrary curve $X$. Suppose that its equation is $h(x,y)=0$. Since $\mathbb{F}[x,y]$ is a Unique Factorization Domain, therefore
\[h=f_1^{k_1} f_2^{k_2} \cdots f_n^{k_n}\]
for irreducible polynomial $f_i \in \mathbb{F}[x,y]$. The points such that $h(x,y)=0$ are also such points so that $f_i(x,y)=0$ for some $i$. Let $X_i$ be a curve defined by $f_i (x,y)=0$, then we have \[X= \bigcup_{i=1}^n X_i \]
where $X_i$ is an irreducible curves.
This decomposition is called the irreducible components decomposition. Since the equation of each $X_i$ is irreducible and defined by the equation which only may differ by constant multiple, so any other equation of $X$ is also differ from $h$ by the constant multiple.
Let $\mathbb{F}$ be an algebraic closed field. An algebraic curve $X$ is the set of points $(x.y)$ defined by the equation \[f(x,y)=0\]
where $f(x,y) \in \mathbb{F}[x,y]$.
When $f(x,y)$ is an irreducible polynomial, we say $X$ is an algebraic irreducible curve.
The degree of $X$ is equal to $\text{deg}f$. The condition that $\mathbb{F}$ must be an algebraic closed field is necessary for the uniqueness of the degree of $X$.
For example the curve $(0,0)$ in the field of real numbers is defined by infinitely many equations that is $x^{2n}+y^{2n}=0$ for any positive integers $n$. For even $n$ the curve is irreducible, while whenever $n$ is odd, the curve is reducible.This confusion doesn't happen when $\mathbb{F}$ is an algebraic closed field. As we can prove it by the following very useful lemma:
Lemma: Let $\mathbb{F}$ be a field, given an irreducible polynomial $f \in \mathbb{F}[x,y]$. Then for any polynomial $g \in \mathbb{F}[x,y]$ which is not divisible by $f$, the system of equations $f(x,y)=g(x,y)=0$ has only finitely many solutions.
Proof: We view $f$ and $g$ as the element of $u[y]$, where $u$ is the field of rational function of $x$ with coefficient in $\mathbb{F}$. By Gauss Lemma, it is easy to see that $f$ remains irreducible in $u[x]$ and also $f$ and $g$ are still relatively prime in $u[y]$. Thus there exists $\lambda, \mu \in u[y]$ such that
\[f \lambda + g \mu =1\]
Multiplying by the common denominator $a(y) \in \mathbb{F}[y]$ we will get
\[f(x,y) \lambda(x,y) + g(x,y) \mu(x,y) = a(y)\]
If $(\alpha, \beta)$ is the solution of the system of equations. Then $a(\beta)=0$, and there are only finitely many possibility of such $\beta$. Thus $\alpha$ is the root of equation $f(x, \beta)=0$, and there are also a finitely many value of them. The result follows.
Applying the lemma, let $X$ be an irreducible algebraic curve in an algebraic closed field. If there are two such equations which define $X$ say $f(x,y)=0$ and $g(x,y)=0$, since $f(x,y)=0$ has infinitely many solution in the algebraic closed field then the systems of equations $f(x,y)=g(x,y)=0$ also has infinitely many solutions. Hence the lemma says that $f(x,y)=c g(x,y)$. Thus the two equations are essentially the same.
The same is true for arbitrary curve $X$. Suppose that its equation is $h(x,y)=0$. Since $\mathbb{F}[x,y]$ is a Unique Factorization Domain, therefore
\[h=f_1^{k_1} f_2^{k_2} \cdots f_n^{k_n}\]
for irreducible polynomial $f_i \in \mathbb{F}[x,y]$. The points such that $h(x,y)=0$ are also such points so that $f_i(x,y)=0$ for some $i$. Let $X_i$ be a curve defined by $f_i (x,y)=0$, then we have \[X= \bigcup_{i=1}^n X_i \]
where $X_i$ is an irreducible curves.
This decomposition is called the irreducible components decomposition. Since the equation of each $X_i$ is irreducible and defined by the equation which only may differ by constant multiple, so any other equation of $X$ is also differ from $h$ by the constant multiple.
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