# Difference between revisions of "Unique factorization domain"

A unique factorization domain is an integral domain in which an analog of the fundamental theorem of arithmetic holds. More precisely an integral domain $R$ is a unique factorization domain if for any nonzero element $r\in R$ which is not a unit:

• $r$ can be written in the form $r=p_1p_2\cdots p_n$ where $p_1,p_2,\ldots,p_n$ are (not necessarily distinct) irreducible elements in $R$.
• This representation is unique up to units and reordering, that is if $r = p_1p_2\cdots p_n = q_1q_2\cdots q_m$ where $p_1,p_2,\ldots,p_n$ and $q_1,q_2\ldots,q_m$ are all irreducibles then $m=n$ and there is some permutation $\sigma$ of $\{1,2\ldots,n\}$ such that for each $k$ there is a unit $u_k$ such that $p_k = u_kq_{\sigma(k)}$.

One of the most significant results about unique factorization domains is that any principal ideal domain (and hence any euclidean domain) is a unique factorization domain. This automatically implies that many well-known rings are unique factorization domains including:

• The ring of integers, $\mathbb Z$ (so in this sense, this result is a generalization of the fundamental theorem of arithmetic)
• The Gaussian integers, $\mathbb Z[i]$
• The polynomial ring $F[x]$ over any field $F$.

## Proof

First we note that in any principal ideal domain, $R$, the irreducible elements are precisely the prime elements. One implication (that any prime element is irreducible) is known to be true for any integral domain. For the other direction, let $m\in R$ be irreducible. Then as $R$ is a principal ideal domain, $(m)$ must be a maximal ideal. But now in a commutative ring with unity maximal ideals are prime. Thus $(m)$ is prime, and hence $m$ is prime.

Now let $R$ be any principal domain. First we shall show that any nonzero non-unit in $R$ can be factored into irreducibles. Assume that this is not the case. Then let $S\subseteq R$ be the set of all non-units in $R$ which cannot be written as a product of irreducibles. Consider any $r\in S$. Clearly $r$ itself cannot be irreducible, so we may write $r=ab$ for some nonzero $a,b\in R$, neither of which are units. Now if neither $a$ nor $b$ is in $S$, then both $a$ and $b$ can be written as a product of irreducibles, and hence so can $r$. This is a contradiction, so at least one of $a$ and $b$ must be in $S$. WLOG let this be $a$. Now $a|r$, so $(r)\subseteq (a)$ and also $(r)\ne (a)$ as $b$ is not a unit. Thus $(r)\subset (a)$ and we have proved the following proposition:

• If $r\in S$ then there is some element $r'\in S$ such that $(r)\subset (r')$.

So now we can construct a sequence $r_1,r_2,\ldots$ of elements of $S$ such that $$(r_1)\subset (r_2)\subset (r_3) \subset \cdots$$ (simply let $r_n = (r_{n-1})'$). But now as $R$ is a principal ideal domain, and hence Noetherian, this contradicts the ascending chain condition and is therefore impossible. Thus every element of $R$ can indeed be written as the product of irreducibles.

It now remains to show that such representations are unique. For any nonzero $r\in R$, let $f(r)$ be the smallest integer $n$ such that $r$ can be written as the product of $n$ irreducibles (this is guaranteed to exist by the previous work). We proceed by strong induction on $f(r)$.

If $f(r) = 0$ then $r$ is a unit. So now assume that $r$ has some other factorization $r = p_1p_2\cdots p_m$ (clearly $m>0$ or this factorization would not be different from the factorization $r=r$). Let $P = p_2\cdots p_m$ (or $P=1$ if $m=1$). Then we have $r = p_1P\Rightarrow 1 = p_1(Pr^{-1})$, which implies that $p_1$ is a unit, a contradiction. So the satement is true for $f(r)=0$.

Now assume that we have unique factorization for any $r$ with $f(r). Consider some $s\in R$ with $f(s)=n$. Assume that $$s = p_1p_2\cdots p_n=q_1q_2\cdots q_m,$$ for irreducibles $p_1,p_2,\ldots,p_n$ and $q_1,q_2\ldots,q_m$ (note that by the definition of $f(s)$, $n\le m$). Now as $p_n$ is irreducible by the above note it must also be prime. Hence as $p_n|q_1q_2\cdots q_m$ we must have $p_n|q_k$ for some $k$. Renumbering the $q_i$'s if necessary, we may assume WLOG that $k=m$. So now $p_n|q_m$, and so $q_m=p_nc$ for some $c\in R$. But $q_m$ is irreducible and $p_n$ is not a unit, so $c$ must be a unit. Plugging this back into our expressions for $s$, we get: $$s = p_1p_2\cdots p_{n-1}p_n=q_1q_2\cdots q_{m-1}q_m = q_1q_2\cdots q_{m-1}(p_nc) = (q_1c)q_2\cdots q_{m-1}p_n.$$ Now as $R$ is an integral domain we get $p_1p_2\cdots p_{n-1}=(cq_1)q_2\cdots q_{m-1}$. Letting $s' = p_1p_2\cdots p_{n-1}$ we get $f(s')\le n-1$, so by the inductive hypothesis (since $cq_1$ is clearly irreducible) there is a unique factorization for $s'$. Thus $m=n$ and there is a permutation $\sigma$ of $\{1,2,\ldots,n-1\}$ and units $u_1,u_2,\ldots,u_{n-1}\in R$ such that: \begin{align*} q_1 &= (u_1c^{-1}) p_{\sigma(1)}\\ q_2 &= u_2p_{\sigma(2)}\\ &\cdots\\ q_{n-1} &= u_{n-1}p_{\sigma(n-1)}\\ \end{align*} Combining this with the fact that $q_n = cp_n$ proves that the representation of $s$ is unique and finishes the induction.

Therefore factorization into irreducibles in $R$ is unique, and therefore $R$ is a unique factorization domain.