Revisiting Factorisation

Much of our discussion in abstract algebra amounts to jacking our understanding of the integers to the maximum limit, and articulating the “movement” within the integers that match the “movement” in possibly other sets (or in this part of the discussion, integral domains).

One property that strikes us must be the unique factorisation feature that the integers enjoy. We call integral domains with $1$ that share this property unique factorisation domains (UFDs).

Henceforth, let $R$ denote an integral domain with $1$ .

Lemma 1. Call $u \in R$ a unit if it has a multiplicative inverse in $R$ . Write $a \sim b$ to mean that there exists a unit $u$ (i.e. an element in $R$ with a multiplicative inverse) such that $a = ub$ . Then

$\mathrm U(R) := \{u \in R : u\ \text{is a unit}\}$

forms a multiplicative group, and $\sim$ forms an equivalence relation on $R$ .

Proof. Straightforward exercise.

Recall that an integral domain is a principal ideal domain (PID) if all of its ideals take the form $\langle r \rangle$ . We claim that PIDs are UFDs. But in order to justify this ambitious claim, we need to develop some nomenclature surrounding UFDs.

Definition 1. We call $R$ a unique factorisation domain (UFD) if for any non-unit $x \in R$ , there exist irreducible elements $p_1,\dots, p_n \in R$ such that

$\displaystyle x = \prod_{i=1}^n p_i.$

Furthermore, for any irreducible elements $q_1,\dots, q_m \in R$ such that

$\displaystyle x = \prod_{j=1}^m q_j,$

we have $m = n$ and after re-labelling, $p_i \sim q_i$ .

Recall that a non-unit $p$ is irreducible if whenever $p = xy$ , at least (and therefore, exactly) one of $x$ or $y$ is a unit.

We’re first going to check that UFDs do carry integer-esque properties with regards to factorisation. Write $a \mid b$ to mean that there exists $q \in R$ such that $b = qa$ .

Lemma 2. Call a nonzero element $p \in R$ prime if $\langle p \rangle \subseteq R$ is a prime ideal. If $R$ is a UFD, then any element $x \neq 0$ is prime if and only if it is irreducible.

Proof. $(\Rightarrow)$ Fix $x \neq 0$ and suppose $x$ is prime. Write $x = uv$ for irreducible elements $u,v$ without loss of generality. Since $x \mid uv$ and $x$ is prime, $x \mid u$ without loss of generality. Write $u = wx$ for some $w \in R$ so that $x = (wx)v = (wv)x$ . Since $R$ is an integral domain, $wv = 1$ . Hence, $v$ is a unit. Similarly, if $x \mid v$ , then $u$ is a unit. Therefore, $x$ is irreducible.

$(\Leftarrow)$ . Suppose $x$ is irreducible. Fix $uv \in \langle p \rangle$ and write $uv = wp$ for some $w \in R$ . Find unique irreducible elements $p_1,\dots p_{m+m}$ such that

$\displaystyle u = \prod_{i=1}^m p_i,\quad v = \prod_{j=1}^n p_{m+j}$

so that

$\displaystyle \prod_{i=1}^{m+n} p_i = u \cdot v = w \cdot p.$

Since $p$ is irreducible, we must have $p_1 \sim p$ without loss of generality. Write $p_i = pz$ for some unit $z$ , so that

$\displaystyle u = \prod_{i=1}^m p_i = p_1 \cdot \prod_{i=2}^m p_i = p \cdot z \cdot \prod_{i=2}^m p_i \in \langle p \rangle.$

Lemma 3. Suppose $R$ is a UFD. Given $a, b \in R$ , there exists some $d \in R$ satisfying the greatest common divisor properties:

$d \mid a$ and $d \mid b$
If $c$ satisfies the property above, then $c \mid d$ .
If $d'$ satisfies the two properties above, then $d \sim d'$ .

Proof. Find irreducible elements $p_1,\dots, p_n, q_1,\dots, q_n$ such that

$\displaystyle a = \prod_{i=1}^n p_i^{\alpha_i}, \quad b = \prod_{i=1}^n q_i^{\beta_i}.$

where $\alpha_i \geq 0, \beta_i \geq 0$ , and $p_i^0 := 1, q_i^0:= 1$ . Suppose that $p_i \sim q_i$ for each $i$ . Define

$\displaystyle d := \prod_{i=1}^n p_i^{\min\{ \alpha_i, \beta_i \}}.$

It is obvious that $d \mid a$ . For $d \mid b$ , find units $u_i$ such that $q_i = p_i u_i$ . Note that $u_i^{\beta_i}$ will always be a limit, so that

$\displaystyle d \mid \prod_{i=1}^n u_i^{\beta_i} \cdot \prod_{i=1}^n p_i^{\beta_i} = \prod_{i=1}^n q_i^{\beta_i} = b.$

Now suppose $c \mid a$ and $c \mid b$ . Write

$\displaystyle c = \prod_{i=1}^n p_i^{\gamma_i},\quad \gamma_i \geq 0.$

Then $\gamma_i \leq \min\{\alpha_i,\beta_i\}$ , so that $c \mid d$ .

Finally, given $d'$ satisfying the first two properties, we must have $d \mid d'$ and $d' \mid d$ . Find elements $u,v \in R$ such that $d = ud'$ and $d' = vd$ . Substituting, $d = (uv)d$ . Since $R$ is an integral domain with $d \neq 0$ , we must have $uv = 1$ , so that $u,v$ are units, and hence, $d \sim d'$ .

Lemma 4. Suppose $R$ is a PID. Fix a collection $\{I_n\}$ of non-decreasing ideals, i.e. $I_n \subseteq I_{n+1}$ for any $n \in \mathbb N$ . Then there exists some $N \in \mathbb N$ such that for $k > N$ , $I_k = I_N$ . In this case, we say that $\{I_n\}$ satisfies the ascending chain condition.

Proof. Write $I_n = \langle x_n \rangle$ for each $n$ . Define $I := \bigcup_{n \in \mathbb N} I_n$ . It is not hard to check that $I \subseteq R$ is an ideal. Since $R$ is a PID, there exists $x \in R$ such that $I = \langle x \rangle$ . Since $x \in I = \bigcup_{n \in \mathbb N} I_n$ , we have $x \in I_N$ for some $N \in \mathbb N$ , so that $I = \langle x \rangle = I_N$ . Then for any $k > N$ , $I_N \subseteq I_k \subseteq I = I_N$ .

Theorem 1. If $R$ is a PID, then $R$ is a UFD.

Proof. Suppose for a contradiction instead that $R$ is not a UFD. Then there exists $x_0 \in R$ that cannot be factorised into irreducible elements. In particular, $x_0$ is not irreducible. Therefore, we can find non-units $x_1, y_1 \in R$ such that $x_0 = x_1 y_1 \in \langle x_1\rangle$ .

Inductively, since $x_k$ is not irreducible, there exist non-units $x_{k+1}, y_{k+1} \in R$ such that $x_k = x_{k+1} y_{k+1} \in \langle x_{k+1}\rangle$ . Since $\langle x_n \rangle \subseteq \langle x_{n+1} \rangle$ for each $n$ , we have constructed a collection $\{ \langle x_n \rangle \}$ of non-decreasing ideals, by Lemma 4, there exists $N \in \mathbb N$ such that $\langle x_k \rangle = \langle x_N \rangle$ for $k > N$ .

In particular, $x_{N+1} \sim x_N$ , which means there exists a unit $u \in R$ such that $x_{N+1} = ux_N$ . On the other hand, by construction, $x_N = x_{N+1} y_{N+1}$ , where $x_{N+1}, y_{N+1}$ are not units. Hence,

$x_{N+1} = (u y_{N+1}) \cdot x_{N+1}.$

Since $R$ is an integral domain, we must have $u y_{N+1} = 1$ , implying that $y_{N+1}$ is a unit, a contradiction.

Therefore, $R$ must be a UFD.

Corollary 1. Let $R$ be an integral domain. If $R$ is a Euclidean domain, then it is a PID, and hence, a UFD.

Why bother with factorisation? Because exceedingly early in our mathematical journey, before even entering university, we factorised polynomials again and again and again. In particular, the real-valued turn out to be a Euclidean domain, and hence becomes a PID and a UFD. These constructions become exceedingly useful when discussing field extensions as preludes to Galois theory.

And so, we turn our attention to the polynomials once again, from the ring-theoretic perspective.

—Joel Kindiak, 30 Jan 26, 1112H

KindiakMath

Revisiting Factorisation

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Revisiting Factorisation

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