Principal Ideal & PID

2020-2학기, 대학에서 ‘현대대수1’ 수업을 듣고 공부한 바를 정리한 글입니다. 지적은 언제나 환영입니다 :)

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Definition. Principal Ideal

Let $R$ be a commutative ring with unity, and $a \in R$.

the ideal $\left< a \right> = \{ ra \mid r \in R \}$ is called a “principal ideal generated by $a$”

For an ideal $N \trianglelefteq R$, if $N = \left< a \right>$ for some $a \in R$, then $N$ is a Principal Ideal.

Example.

Every ideal of a ring $\mathbb{Z}$ is of the form $n\mathbb{Z}$.

And $n\mathbb{Z}$ is an ideal $\left< n \right>$ generated by $a$.

따라서 $\mathbb{Z}$의 모든 ideal은 principal ideal임!

Example.

$F[x]$의 ideal 중 하나인 $\left< x \right>$를 생각해보자.

$\left< x \right>$은 unity $1$을 포함하지 않는다. 즉 $\left< x \right>$에는 non-constant function들만을 가지고 있다.

또한, $\left< x \right>$는 $x \in F[x]$의 곱에 의해 생성되는 ideal이므로 principal ideal이다.

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Princial Ideal Domain; PID

Definition

An Integral Domain whose every ideal is a Principal Ideal.

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Theorem 27.24

If $F$ is a field, every ideal in $F[x]$ is principal. ($F[x]$ is PID)

proof.

Let $N \trianglelefteq F[x]$.

(Case 1)

If $N = \{ 0 \}$, then $N = \left< 0 \right>$.

Supp. $N \ne 0$, and let $g(x) \ne 0 \in N$ with minimal degree in $N$.

(Case 2) $\deg g(x) = 0$

Then, $g(x) \in F$ is a constant function. 그리고 $N$이 group이므로 $N$은 상수함수들의 집합이다. 따라서 $N = \left< 1 \right>$

따라서 Ideal인 $N$에 대해 $1 \in N$이므로 앞에서 보인 정리에 의해 $N = F[x]$이다.

($N = F[x]$이라는게 말이 되나 근데??)

ㄴ Yes! 잘 생각해보니까 말이 된다. ideal $N$을 애초에 $F[x]$로 잡으면 $N = F[x]$가 된다.

Q. 그런데 $\left< 1 \right> = F[x]$인 걸까?

A. Principal Ideal인 $\left< 1 \right>$을 cyclic group의 notation과 헷갈린 것 같다.

$\left< 1 \right>$는

\[\left< 1 \right> = \{ f(x) \cdot 1 \mid f(x) \in F[x] \}\]

이므로 $\left< 1 \right> = F[x]$가 된다.

(Case 3) $\deg g(x) \ge 1$

let $\forall \; f(x) \in N$.

then, by “Division Algorithm” $f(x) = g(x) q(x) + r(x)$ where $r(x) = 0$ or $\deg r(x) < \deg g(x)$.

$f(x), g(x) \in N$이므로 ideal $N$의 정의에 따라 $f(x) - g(x)q(x) = r(x) \in N$이다.
($q(x) \in F[x]$에 대해 $N \cdot q(x) \subseteq N$이다. 따라서 $g(x) q(x) \in N$이다.)

$g(x)$는 정의상 $N$의 non-zero minimal degree elt이므로 $r(x) \in N$라면, $r(x) = 0$이 되어야 한다.

따라서 $f(x) = g(x) q(x)$이고, 이것은 $N = \left< g(x) \right>$를 의미한다.

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Theorem

Let $F$ be a Field, and $0 \ne p(x) \in F[x]$.

$\left< p(x) \right>$ is maximal ideal

$\iff$ $p(x)$ is irreducible in $F[x]$.

주로 왼쪽 방향의 명제를 더 자주 사용한다!!

proof.

($\implies$)

Supp. $\left< p(x) \right>$ is a Maximal Ideal.

Assume that $p(x)$ is reducible, then

$\implies$ $p(x) = f(x)g(x)$ for some $f(x), g(x) \in F[x]$

$\implies$ $\left< p(x) \right> < \left< f(x) \right> \trianglelefteq F[x]$

Since $\left< p(x) \right>$ is a Maximal Ideal, $\left< f(x) \right> = F[x] = \left< 1 \right>$

Thus, $f(x) = 1$. This is contradict to our assumption of $p(x)$ is reducible.

$\therefore$ $p(x)$ is irreducible. $\blacksquare$

($\impliedby$)

Supp. $p(x)$ is irreducible.

(Goal) $\left< p(x) \right>$ is Maxiaml element.

Let assume there is a bigger ideal $\left< g(x) \right>$.

\[\left< p(x) \right> < \left< g(x) \right> \trianglelefteq F[x]\]

Then, $p(x) \in \left< g(x) \right>$.

This means $p(x) = g(x) \cdot h(x)$ for some $h(x) \in F[x]$.

Sine $p(x)$ is irreducible, $g(x)$ or $h(x)$ is an unit.

(Case 1) If $h(x)$ is an unit, then $\left< p(x) \right> = \left< g(x) \right>$.

This contradicts to $\left< p(x) \right> < \left< g(x) \right>$.

(Case 2) Therefore $g(x)$ is a unit, then this means $1 \in \left< g(x) \right>$ and $\left< g(x) \right> = \left< 1 \right> = F[x]$.

This means $\left< p(x) \right>$ is a Maximal Ideal. $\blacksquare$

Example.

$x^2 - 2$ is irreducible in $\mathbb{Q}[x]$.

Therefore, $\left< x^2 - x\right>$ is a Maximal Ideal.

Therefore, $\mathbb{Q}[x] / \left< x^2 - 2 \right>$ is a Field.

\[\mathbb{Q}[x] / \left< x^2 - 2 \right> \cong \mathbb{Q}[\sqrt{2}] = \{ a + b \sqrt{2} \mid a, b \in \mathbb{Q} \}\]

proof.

Define an evaluation homomorphism $\phi_{\sqrt{2}}$

\[\begin{aligned} \phi_{\sqrt{2}}: \mathbb{Q}[x] &\longrightarrow \mathbb{Q}[\sqrt{2}]\\ f(x) &\longmapsto f(\sqrt{2}) \end{aligned}\]

(1) $\phi_{\sqrt{2}}$ is a homomorphism.

생-략

(2) $\phi_{\sqrt{2}}$ is onto

for $a+b\sqrt{2} \in \mathbb{[\sqrt{2}]}$, there exist an inverse image.

for example $a+bx \in \mathbb{Q}[x]$ is one of it.

(Claim) $\ker \phi = \left< x^2 - 2 \right>$.

Since $(\sqrt{2})^2 - 2 = 0$, $\left< x^2 - 2 \right> \subseteq \ker \phi$

For $p(x) \in \ker \phi$, $p(\sqrt{2}) = 0$ and $p(-\sqrt{2}) = 0$.

And also $p(x)$ have rational coefficients, therefore $(x^2 - 2) \mid p(x)$.

This means $p(x) \in \left< x^2 - 2 \right>$.

$\implies$ $\ker \phi \subseteq \left< x^2 - 2 \right>$.

$\therefore \ker \phi = \left< x^2 - 2 \right>$.

By FHT,

\[\begin{aligned} \mathbb{Q}[x] / \ker \phi &\cong \phi[\mathbb{Q}[x]] \\ \mathbb{Q}[x] / \left< x^2 - 2 \right> &\cong \mathbb{Q}[\sqrt{2}] \end{aligned}\]

$\blacksquare$

Example.

$x^2 + 1$ is irreducible in $\mathbb{R}[x]$.

$\mathbb{R}[x] / \left< x^2 + 1 \right> \cong \mathbb{C}$ is a Field.

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Theorem

Let $p(x) \in F[x]$

$p(x)$ irreducible in $F[x]$

$\iff$ $p(x)$ is a prime element of $F[x]$.

proof.

($\implies$)

Let $p(x)$ be a irreducible in $F[x]$.

By Theorem above, $\left< p(x) \right>$ is a Maximal Ideal.

By Theorem of Maximal Ideal, $\left< p(x) \right>$ is a Prime Ideal.

$\implies$ $p(x)$ is a Prime element.

(보충) Thm 27.24에서 우리는 Field에서 유도되는 ideal은 모두 principal임을 보였다. 따라서 $F[x]$는 PID가 된다.

이때, PID에서 만족하는 성질 중 하나는 “Prime Ideals are generated by Prime elements”라는 것이다.¹ 즉, $\left< p(x) \right>$가 Prime Ideal임을 확인한다면, $\left< p(x) \right>$를 생성하는 $p(x)$라는 Prime element가 있음을 알 수 있다.

그래서 사실 $\implies$ 방향 명제를 증명하는 과정에서 “Prime Ideals are generated by Prime elements” 명제를 증명할 필요가 있다.

proof. “Every Principal Prime Ideal is generated by a Prime element”

Let $F$ be a PID, and $I \trianglelefteq F$ be a Prime Ideal.

Since $I \in \textrm{PID} = F$, $I = \left< a \right>$ for some $a \in F$.

Since $I$ is a Prime Ideal,

if $nm \in I$, then $n \in I$ or $m \in I$.

(Check) $a \mid nm$ implies $a \mid n$ or $a \mid m$.

Let $nm \in I$, then $a \mid nm$ ($\because I = \left< a \right>$)

• if $n \in I$, then $a \mid n$.
• if $m \in I$, then $a \mid m$.

Therefore, $a$ is a Prime element, and Every Principal Prime Ideal is generated by a Prime element. $\blacksquare$

($\impliedby$)

In General, a Prime element is Irreducible. Theorem

위의 명제가 시사하는 바는 매우 엄청나다!!

$p(x)$가 irreducible이라고 가정한 것이 $p(x)$가 prime이라는 걸 유도했기 때문이다!!

이것은 PID에선 irreducibility와 primality가 일치함을 시사한다!!!

Corollary

In PID, irreducible $\equiv$ prime

증명은 위의 명제의 $\implies$ 방향에 의해 보장된다.

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