posted 6 Jan 2022

Unique Factorization in a Principal Ideal Domain

Notes Number Theory Unique Factorization in a Principal Ideal Domain

§ Principal Ideal Domain

For elements $a_1, a_2, \dots, a_n$ of the ring $R,$ we define

(a_1, a_2, \dots, a_n) = Ra_1 + Ra_2 + \dots + Ra_n = \left\{\sum_{i=1}^n r_ia_i\,|\,r_i \in R\right\}.

Then, $(a_1, a_2, \dots, a_n)$ is an ideal.

If an ideal $I$ is equal to $(a_1, a_2, \dots, a_n)$ for some elements $a_i \in I,$ we say that $I$ is finitely generated. If $I = (a)$ for some $a \in I,$ we say that $I$ is a principal ideal.

Principal Ideal Domain

If every ideal of $R$ is principal, then we say that $R$ is a principal ideal domain (PID).

Every Euclidean domain is a PID, but not every PID is an Euclidean domain, i.e. PID $\supset$ Euclidean domain. However finding counterexamples to PID $=$ Euclidean domain is somewhat hard. In practice, it is useful to show that a given ring is an Euclidean domain, which gives us the added bonus of showing that it is a PID.

§ Terminology

Division

If $a, b \in R, b \neq 0,$ then we say that $b$ divides $a$ if $\exists c \in R, a = bc$ and write $b | a.$

In terms of ideals, $a | b \iff (b) \subseteq (a).$

Associate

$a, b \in R$ are associates if $a = bu$ for some unit $u.$

A unit is defined as $u \in R$ such that $(u) = R.$ And in terms of ideals, $a$ and $b$ are associates iff $(a) = (b).$

Irreducibility

An element $p \in R$ is irreductible if $a | p$ implies that $a$ is either a unit or an associate of $p.$

Prime

A nonunit $p \in R \setminus\{0\}$ is said to be prime if $p | ab$ implies that $p | a$ or $p | b.$

In terms of ideals, $p$ prime iff $ab \in (p) \implies a \in (p) \lor b \in (p).$

§ Greatest Common Divisor

Greatest Common Divisor

$d \in R$ is said to be a (gcd) of two elements $a, b \in R$ if

$d | a$ and $d | b,$
$d' | a \wedge d' | b \implies d' | d.$

If both $d$ and $d'$ are gcds of $a$ and $b,$ then $d$ is associate to $d'.$

Let $R$ be a PID and $a, b \in R.$ Then $a$ and $b$ have a gcd and $(a, b) = (d).$

Form the ideal $(a, b).$

Since $R$ is a PID there exists $d$ such that $(a, b) = (d).$ Since $(a) \subseteq (d)$ and $(b) \subseteq (d)$ we have that $d | a$ and $d | b.$

If $d' | a$ and $d' | b,$ then $(a) \subseteq (d')$ and $(b) \subseteq (d').$ Then, $(d) = (a, b) \subseteq (d')$ and $d' | d.$ Therefore $(a, b) = (d).$

§ Coprimality

Coprime

$a$ and $b$ are coprime iff their common divisors are units.

Corollary 1

If $R$ is a PID and $a, b \in R$ are coprime, then $(a, b) = R.$

Corollary 2

If $R$ is a PID, and $p \in R$ is irreductible, then $p$ prime.

Sps. that $p | ab$ and that $p \nmid a.$ Since $p \nmid a$ it follows that they only share units as common divisors. By corollary 1 $(a, p) = R,$ so $(ab, pb) = (b).$ Since $ab \in (p)$ and $pb \in (p)$ we have $(b) \subseteq (p).$ Therefore $p | b$ and in a PID, prime elements $\iff$ irreductible elements.

In a PID we will use prime and irreductible interchangeably.

§ Order

Let $(a_1) \subseteq (a_2) \subseteq (a_3) \subseteq \dots$ be an ascending chain of ideals. Then there exists $k \in \mathbb{N}$ such that $(a_k) = (a_{k+l})$ for $l = 0, 1, 2, \dots.$ That is, the chain breaks in finitely many steps.

Let $I = \bigcup_{i=1}^{\infty} (a_i),$ obviously $I$ is an ideal so $I = (a)$ for some $a \in R.$ But $a \in \bigcup_{i=1}^{\infty} (a_i)$ implies that $a \in (a_k)$ for some $k.$ This shows that $I = (a) \subseteq (a_k).$ It then follows that $I = (a_k) = (a_{k + l}) = \dots$

Proposition 5

Every nonzero nonunit of $R$ is a product of irreductibles.

Let $a \in R \setminus \{0\},$ where $a$ is a nonunit, then we want to prove the following cases.

$a$ is divisible by irreductibles

In the case that $a$ itself is irreductible, we are done. Otherwise, $a = a_1b_1$ where $a_1, b_1$ are nonunits. If $a_1$ irreductible, done. Otherwise $a_1 = a_2b_2$ where $a_2, b_2$ are nonunits. If $a_2$ irreductible, done. Otherwise, continue as before.

Notice that $(a) \subset (a_1) \subset (a_2) \subset \dots$ and by lemma 5 we have that this is a finite chain, so there must exist some $k$ such that $a_k$ is irreductible.
$a$ is the product of irreductibles

If $a$ is irreductible, done. Otherwise let $p_1$ be some irreductible such that $p_1 | a.$ Then $a = p_1c_1.$ If $c_1$ is a unit, done. Otherwise, continue as before. Again, notice that $(a) \subset (c_1) \subset (c_2) \subset \dots,$ which is a finite chain by lemma 5. There must exist some $k$ for which $a = p_1p_2\dots p_k c_k,$ where $c_k$ a unit and $p_kc_k$ is irreductible.

We use the following lemma to construct the $\text{ord}$ function akin to those for $\mathbb{Z}$ and $k[ x ].$

Let $p$ be prime and $a \neq 0.$ Then there exists $n \in \mathbb{Z}$ such that $p^{n} | a$ but $p^{n + 1} \nmid a.$

Sps. that there does not exist an $n$ satisfying the lemma, then for each integer $m > 0$ there would be an element $b_m$ such that $a = p^mb_m.$ Then $pb_{m+1} = b_m$ so that $(b_1) \subset (b_2) \subset (b_3) \subset$ which would be an infinite chain which contradicts lemma 5.

Then the integer $n$ defined in lemma 6 is uniquely determined by $p$ and $a,$ and we set $n = \text{ord}_p a.$

§ Unique Factorization Proof

If $a, b \in R \setminus \{0\},$ then $\text{ord}_p (ab) = \text{ord}_p a + \text{ord}_p b.$

Let $\alpha = \text{ord}_p a$ and $\beta = \text{ord}_p b.$ Then $a = p^{\alpha}c$ and $b = p^{\beta}d$ with $p \nmid c$ and $p \nmid d.$ Thus $ab = p^{\alpha + \beta}cd.$ Since $p$ prime, $p \nmid cd.$ Therefore $\text{ord}_p ab = \alpha + \beta = \text{ord}_p a + \text{ord}_p b$

Let $S \subset R$ be a set of primes with the properties

$\forall p \in R \exists s \in R,$ such that $p$ and $s$ are associates.
$\nexists a, b \in S$ such that $a$ and $b$ are associates.

To generate the set $S$ we can choose a representative prime from each class of associate primes. In $\mathbb{Z}$ we can construct the set $S$ by choosing all positive primes. In $k[ x ]$ we chose $S$ to be the set of monic irreductible polynomials.

There does not exist a general way to construct $S$ for an arbitrary PID.

Let $R$ be a PID and $S$ a set of primes with the aforementioned properties.

Then if $a \in R \setminus \{0\},$ then we can write

a = u \prod_p p^{e(p)}, \tag{\dag}

where $u$ is a unit and the product is over all $p \in S.$ The unit $u$ and the exponents $e(p)$ are uniquely determined by $a.$ In fact, $e(p) = \text{ord}_p q.$

The existence of such a decomposition follows from proposition 5.

To show the uniqueness, let $q \in S$ and apply $\text{ord}_q$ to both sides of eq. ( $\dag$ ). Using lemma 7, we get

\text{ord}_q a = \text{ord}_q u + \sum_p e(p)\,\text{ord}_q p.

By definition of $\text{ord}_q$ we get that $\text{ord}_q u = 0$ and that $\text{ord}_q p = 0$ if $q \neq p$ and $1$ otherwise. So $\text{ord}_q a = e(q).$ Since exponents $e(q)$ are uniquely determined, so is the unit $u.$