Unique Factorization Domain 📂Abstract Algebra

Unique Factorization Domain

Definitions¹

If for every element of an integral domain $D$ that is neither $0$ nor a unit there uniquely exists a finite factorization, then $D$ is called a unique factorization domain, or UFD.
For $a_{1} , \cdots , a_{n}$ in a unique factorization domain $D$, if $d \mid a_{i}$ and every divisor of $a_{i}$ divides $d$, then $d$ is called the greatest common divisor of $a_{1} , \cdots , a_{n}$ and is written as $\gcd$.
Let a polynomial in a unique factorization domain $D$ be $f(x) := a_{0} + a_{1} x + \cdots + a_{n} x^{n}$. If $\gcd ( a_{0} , a_{1} , \cdots , a_{n} ) = 1$, then $f(x) \in D [ x ]$ is said to be primitive.

The unity is the identity $1$ with respect to multiplication, and a unit is an element that has a multiplicative inverse.

Theorem²

[1]: Every PID is a UFD.
[2] Fundamental Theorem of Arithmetic: $\mathbb{Z}$ is a UFD.
[3] Gauss’s Lemma: If $D$ is a UFD, then a product of primitive polynomials in $D [ x ]$ is also primitive.
[4]: If $D$ is a UFD, then $D [ x ]$ is also a UFD.
[5]: If $F$ is a field, then $F[ x_{1} , \cdots , x_{n} ]$ is a UFD.

Explanation

The term “unique factorization domain” is so long that the abbreviation UFD is usually used instead.

UFD

That an element has a finite factorization means that the given element can be expressed as a product of finitely many irreducible elements. The reason UFDs are useful is that they allow us to think of some large object by breaking it apart. Given the nature of algebra, even if we cannot say exactly what those pieces are, the mere existence of such a factorization is already a great help. Thanks to this, a UFD becomes a domain where our “common sense” computations actually work.

There are plenty of examples of unique factorization domains. As mentioned in Theorem [2], the ring of integers $\mathbb{Z}$ is one right away. Now consider the simple extension $\mathbb{Z} ( \sqrt{ - 5 } )$ obtained by adjoining $\sqrt{-5}$ to the ring of integers. Here $21 \in \mathbb{Z} ( \sqrt{ - 5 } )$ has the prime factorization $21 = 3 \cdot 7$, while $21 = ( 1 + 2 \sqrt{-5}) ( 1 - 2 \sqrt{-5}) $ is also possible, so the factorization is not unique, and one can easily check that $\mathbb{Z} ( \sqrt{ - 5 } )$ is not a unique factorization domain.

Primitive Functions?

A function being primitive has nothing whatsoever to do with the antiderivative in calculus; it refers to a function whose coefficients cannot be factored out, unlike $(3 x^2 + 6 x + 3) \in \mathbb{Z} [ x ]$, which can be factored entirely by $3$ as in $3 ( x^2 + 2x + 1)$.

Fundamental Theorem of Arithmetic

Unlike the statement in number theory, it is summarized as saying that the ring of integers $\mathbb{Z}$ is a UFD. Of course, an enormous number of concepts were mobilized to get here, so in the end it says the same thing, but in higher-level number theory many things are expressed in the language of algebra like this, which makes studying algebra essential. Even if you do not major in algebra, not knowing algebra makes you a fool.

Gauss’s Lemma

Gauss’s Lemma is a more surprising theorem than one might think. For example, considering $(5x + 1) , (2x^2 + 3x + 1) \in \mathbb{Z} [ x ]$, their product is $( 10 x^3 + 17 x^2 + 8 x + 1 )$, from which indeed no greatest common divisor $a \in \mathbb{Z}$ can be factored out. It might seem that if you kept trying, you would stumble upon at least one counterexample, but thanks to Gauss’s Lemma there is no need for such futile effort.

Proof

[1]

Part 1. Existence

If $D$ is a PID, then $d \in D$ can be expressed as a finite product $a = p_{1} \cdots p_{r}$ of irreducible elements $p_{1} , \cdots , p_{r}$.

Part 2. Uniqueness

Suppose that $a = q_{1} \cdots q_{s}$ is also possible for other irreducible elements $q_{1} , \cdots , q_{s}$.

Since an irreducible element of a PID is a prime element, we must have $p_{1} \mid q_{j}$ for some $1 \le j \le s$. $$ p_{1} p_{2} \cdots p_{r} = p_{1} u_{1} q_{2} \cdots q_{s} $$ Cancelling $p_{1}$ from both sides of the above gives $$ p_{2} \cdots p_{r} = u_{1} q_{2} \cdots q_{s} $$ Repeating the same argument up to $i=r$ yields $$ 1 = u_{1} \cdots u_{r} q_{r+1} \cdots q_{s} $$ Since $q_{r+1} \cdots q_{s}$ are irreducible elements, it must be that $r=s$.

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[2]

Every ideal of $\mathbb{Z}$ is of the form $\left< n \right> = n \mathbb{Z}$, so it is a PID, and by Theorem [1] it is a UFD.

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[3]

$$ \begin{align*} f(x) &:= a_{0} + a_{1} x + \cdots + a_{n} x^n \\ g(x) &:= b_{0} + b_{1} x + \cdots + b_{m} x^m \end{align*} $$ Write the primitive polynomials $f(x) , g(x) \in D[x]$ as above.

Let $p \in D$ be an irreducible element.

Since $f(x)$ is primitive, $\gcd ( a_{0} , \cdots , a_{n} ) = 1$, and $p$ cannot divide all of $a_{0} , \cdots , a_{n}$. Accordingly, for $i = 0, 1 , \cdots , n$, let $a_{r}$ be the first coefficient such that $p$ fails to divide $a_{i}$.
Since $g(x)$ is primitive, $\gcd ( b_{0} , \cdots , b_{m} ) = 1$, and $p$ cannot divide all of $b_{0} , \cdots , b_{m}$. Accordingly, for $j = 0, 1 , \cdots , m$, let $b_{s}$ be the first coefficient such that $p$ fails to divide $b_{j}$.

Then the coefficient of the $( r + s)$-th degree term of $f(x)g(x)$ is $$ c_{r+s} = ( a_{0} b_{r+s} + \cdots + a_{r-1} b_{s+1} ) + a_{r} b_{s} + ( a_{r+1} b_{s-1} + \cdots + a_{r+s} b_{0} ) $$ and

by the definition of $a_{r}$, $$ p \mid ( a_{0} b_{r+s} + \cdots + a_{r-1} b_{s+1} ) $$
by the definition of $b_{s}$, $$ p \mid ( a_{r+1} b_{s-1} + \cdots + a_{r+s} b_{0} ) $$

However, since $p \nmid a_{r} b_{s}$, the given $p$ cannot divide $f(x) g(x)$. The same holds for every irreducible element, so $f(x) g(x)$ is primitive.

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[4]

Suppose the degree of $f(x) \in D[x]$ is $n$.

Then $f(x)$ factors into $r \le n$ factors and can be expressed as $$ f (x) = g_{1} (x) \cdots g_{r} (x)) $$ Moreover, for $i = 1 , \cdots , r$, each factor can be expressed as the product $$ g_{i} (x) = c_{i} h_{i} (x) $$ of a primitive function $h_{i} (x) \in D[x]$ and $c_{i} \in D$. Such a $c_{i}$ is called the content of $g_{i} (x)$, and through this we obtain $$ f(x) = c_{1} h_{1} (x) \cdots c_{r} h_{r} (x) $$ Since the content is unique for given $g_{i} (x)$ and $h_{i} (x)$, $f(x)$ factors uniquely up to the order of factors and multiplication by constants.

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[5]

Factorization of Polynomials:
[2]: If $F$ is a field, then every non-constant $f(x) \in F [ x ]$ factors into a product of irreducible elements, and this factorization is unique.

By the definition of a UFD, $F [x_{1} ]$ is a UFD, and by Theorem [4], $F [ x_{1} , x_{2} ]$ is also a UFD. Repeating this for finitely many $x_{3} , \cdots , x_{n}$ yields the conclusion that $F[ x_{1} , \cdots , x_{n} ]$ is a UFD.

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