Gaussβs Lemma
2020-2νκΈ°, λνμμ βνλλμ1β μμ μ λ£κ³ 곡λΆν λ°λ₯Ό μ 리ν κΈμ λλ€. μ§μ μ μΈμ λ νμμ λλ€ :)
μ κΉ λ§₯λ½μ κ³λ€μ΄μλ©΄, βGaussβs Lemmaβλ μλμ λͺ μ λ₯Ό μ¦λͺ νλ κ³Όμ μμ λ§λλ μ€κ°λ¨κ³μ΄λ€.
μλμ μ΄μ λΉμ·ν λλμ μ 리λ€μ μ’ λͺ¨μλ΄€λ€.
- $R$ is a ring $\implies$ $R[x]$ is also a ring.
- $D$ is an integral domain $\implies$ $D[x]$ is also an integral domain.
- $F$ is a field $\iff$ $F[x]$ is a PID. (Thm 27.24)
- μ΄λ λͺ¨λ Fieldλ PIDμ΄λ―λ‘ μ λͺ μ μμ βfieldβλ₯Ό βPIDβλ‘ λ°κΏλ μ±λ¦½νλ€.
- $D$ is an integral domain with an irreducible elt $\implies$ $D[x]$ is not a PID.
- $F$ is a field $\implies$ $F[x_1, \dots, x_n]$ is not a PID for $n \ge 2$.
μ°λ¦¬λ μ§κΈκΉμ§ μμ°μμμμ GCDλ₯Ό μ νλ€. νμ§λ§ μμμ μ΄ν΄λ³Έ βμ°μ μ κΈ°λ³Έμ 리βμ λ°λ₯΄λ©΄ μμ°μ μ§ν©λ κ²°κ΅μ UFDμ ν μμ λΆκ³Όνλ€. μμ°μ μ§ν©μμ μ μν GCDλ₯Ό UFDλ‘ νμ₯νμ¬ λ€μ μ μν΄λ³΄μ.
Definition. GCD in UFD
Let $D$ be a UFD, and $a_1$, $a_2$, β¦, $a_n$ be non-zero elts of $D$.
$d$ is a GCD of all of $a_i$,
if $d \mid a_i$ for $i=1, β¦, n$ and all $dβ \in D$ that divides all the $a_i$ also divides $d$.
μμ°μ μ§ν©μμλ GCDκ° μ μΌνκ² μ‘΄μ¬νμ§λ§, μμ°μ μ§ν©λ₯Ό ν¬κ΄νλ κ°λ μΈ UFDμμ GCDλ λμ΄μ μ μΌνμ§ μλ€.
UFDμμ λ κ°μ GCD $d$, $dβ$κ° μ‘΄μ¬νλ€κ³ κ°μ ν΄λ³΄μ. κ·Έλ¬λ©΄, GCDμ μ μμ μν΄ $d \mid dβ$, $dβ \mid d$κ° μ±λ¦½νλ€.
μ¦, μλ‘ λ€λ₯Έ λ GCD $d$, $dβ$κ° associate νλ€.
Simple Example.
μλ‘ λ€λ₯Έ λ GCDμ λν μλ μκ°λ³΄λ€ κ°λ¨ν μ°Ύμ μ μλ€. λ μ μ $6$μ $-8$μ λν GCDλ $2$μ $-2$μ΄λ€.
μ¦, μ μ $\mathbb{Z}$λ UFDμ΄λ©΄μ μλ‘ λ€λ₯Έ λ GCDλ₯Ό μ°Ύμ μ μλ€. λ°λ©΄μ μμ°μ $\mathbb{N}$λ UFDμ΄μ§λ§, GCDκ° μ μΌνκ² κ²°μ λλ€.
Primitive polynomial
Definition. content & primitive part
The content of (polynomial with integer coeffi-.) = $\gcd$ of it coeffi-.
The primitive part of polynomial = quotient of polynomial by its content.
λ°λΌμ (polynomial) = (content) x (primitive part)
Definition. Primitive Polynomial
A polynomial is primitive, if its content equals 1.
Gaussβs Lemma
Lemma 45.23
If $D$ is a UFD,
then for every non-constant $f(x) \in D[x]$, we have $f(x) = (c)g(x)$, where $c \in D$ and primitive $g(x) \in D[x]$.
(μ¦, UFDμμ λͺ¨λ non-constant $f(x)$λ primitiveμ μμκ³±μ΄λΌλ λ§μ΄λ€.)
proof.
Let $f(x) \in D[x]$ be a non-constant polynomial; $f(x) = a_0 + a_1 x + \cdots a_n x^n$
Let $c$ be a gcd of all $a_i$.
Then for each $i$, we have $a_i = c \cdot q_i$ for some $q_i \in D$.
By the distributive law, we have $f(x) = (c) g(x)$.
By definition of gcd $c$, the left polynomial $g(x)$ is a primitive polynomial. $\blacksquare$
Lemma 45.25 Gaussβs Lemma
If $D$ is a UFD, then a product of two primitive polynomials in $D[x]$ is again primitive.
proof.
Let $f(x) = a_0 + a_1 x + \cdots a_n x^n$ and $g(x) = b_0 + b_1 x + \cdots b_m x^m$ be primitive in $D[x]$,
and let $h(x) = f(x)g(x)$.
Let $p$ be an irreducible in $D$.
μ΄λ―Έ $f(x)$, $g(x)$κ° primitiveμ΄λ―λ‘ $p$κ° $a_i$ μ λΆλ₯Ό, λ $b_j$ μ λΆλ₯Ό λλμ§λ λͺ» νλ€.
Let $a_r$ be the first coefficient of $f(x)$ not divisible by $p$;
that is $p \mid a_i$ for $i < r$, but $p \not\mid a_r$.
Similarly, let $b_s$ be the first coefficient of $g(x)$ not divisible by $p$.
The coefficient of $x^{r+s}$ in $h(x) = f(x)g(x)$ is
\[c_{r+s} = (a_0 b_{r+s} + \cdots + a_{r-1} b_{s+1}) + a_r b_s + (a_{r+1} b_{s-1} \cdots a_{r+s} b_0)\]$p \mid a_i$ for $i < r$μ΄λ―λ‘ $p \mid (a_0 b_{r+s} + \cdots + a_{r-1} b_{s+1})$
λ§μ°¬κ°μ§λ‘ $p \mid b_j$ for $j < s$μ΄λ―λ‘ $p \mid (a_{r+1} b_{s-1} \cdots a_{r+s} b_0)$
νμ§λ§, $p$κ° $a_r$, $b_s$λ₯Ό λλμ§ λͺ» νλ―λ‘ $p \not\mid a_r b_s$μ΄λ€.
μ’ ν©νλ©΄, μ΄λ€ irreducible $p \in D$μΌμ§λΌλ $f(x)g(x)$μ κ³μλ₯Ό λλμ§ λͺ» νλ μ§μ μ΄ μκΈ° λλ¬Έμ $f(x)g(x)$μ κ³μλ μ΄λ€ irreducible $p$λΌλ common divisorλ‘ κ°μ§ μ μλ€.
λ°λΌμ $f(x)g(X)$λ primitiveλ€. $\blacksquare$
μ΄μ μ΄ Gaussβs Lemmaλ₯Ό νμ©ν΄ λ³Έλμ λͺ©μ μΈ
λ₯Ό μ¦λͺ ν΄λ³΄μ!
λ€μ ν¬μ€νΈ: Poylnomial over UFD