Unique Factorization Domain 📂Abstract Algebra

Unique Factorization Domain

Definitions ¹

A domain $D$, that is neither $0$ nor a field, in which every element (except for 0 and units) has a unique factorization into irreducible elements is said to be a Unique Factorization Domain.
In a Unique Factorization Domain $D$, for $a_{1} , \cdots , a_{n}$ if $d \mid a_{i}$ and every divisor of $a_{i}$ divides $d$, then $d$ is called the Greatest Common Divisor of $a_{1} , \cdots , a_{n}$, denoted by $\gcd$.
Let a polynomial in a Unique Factorization Domain $D$ be denoted by $f(x) := a_{0} + a_{1} x + \cdots + a_{n} x^{n}$. If $\gcd ( a_{0} , a_{1} , \cdots , a_{n} ) = 1$, then $f(x) \in D [ x ]$ is said to be Primitive.

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

Theorems ²

[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 the 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 often shortened to UFD because it’s quite lengthy.

UFD

The existence of a finite factorization of elements means that a given element can be represented as a product of a finite number of irreducible elements. The utility of UFDs lies in the ability to decompose larger objects into more manageable pieces. Even if we cannot specifically identify these components in all cases, the mere existence of such factorization is greatly helpful. It makes the domain conform to our “common sense” of computation.

There are many examples of UFDs. For instance, as mentioned in Theorem [2], the ring of integers $\mathbb{Z}$ is one. However, consider a simple extension field $\mathbb{Z} ( \sqrt{ - 5 } )$ obtained by adding $\sqrt{-5}$ to the ring of integers. Here, while $21 \in \mathbb{Z} ( \sqrt{ - 5 } )$ has a prime factorization $21 = 3 \cdot 7$, $21 = ( 1 + 2 \sqrt{-5}) ( 1 - 2 \sqrt{-5}) $ is also possible making the factorization not unique, and we can easily verify that $\mathbb{Z} ( \sqrt{ - 5 } )$ is not a Unique Factorization Domain.

Primitive Function?

Being primitive in this context has nothing to do with the notion of a primitive function in calculus, which refers to not being able to factor out coefficients uniformly across the polynomial, unlike being able to encapsulate $(3 x^2 + 6 x + 3) \in \mathbb{Z} [ x ]$ as $3 ( x^2 + 2x + 1)$.

Fundamental Theorem of Arithmetic

Unlike the statement in number theory, the fact that the ring of integers $\mathbb{Z}$ is a UFD is summarized here. Obviously, a myriad of concepts are mobilized for this declaration, but higher-level number theory often expresses such ideas in algebraic terminologies, making the study of algebra indispensable. Even if one does not major in algebra, lacking knowledge in it is like being in the dark.

Gauss’s Lemma

Gauss’s Lemma is more interesting than one might think. For example, considering $(5x + 1) , (2x^2 + 3x + 1) \in \mathbb{Z} [ x ]$, its product is $( 10 x^3 + 17 x^2 + 8 x + 1 )$, which can’t obviously be factored by some greatest common divisor $a \in \mathbb{Z}$. One might expect to stumble upon at least one counter-example eventually, but thanks to Gauss’s Lemma, such futile efforts are unnecessary.

Proof

[1]

Part 1. Existence

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

Part 2. Uniqueness

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

Since irreducible elements in a PID are prime, for some $1 \le j \le s$, $p_{1} \mid q_{j}$ must hold. $$ p_{1} p_{2} \cdots p_{r} = p_{1} u_{1} q_{2} \cdots q_{s} $$ Cancelling $p_{1}$ from both sides gives $$ p_{2} \cdots p_{r} = u_{1} q_{2} \cdots q_{s} $$ Repeatedly applying the same logic up to $i=r$ yields $$ 1 = u_{1} \cdots u_{r} q_{r+1} \cdots q_{s} $$ Since $q_{r+1} \cdots q_{s}$ is irreducible, it must divide $r=s$.

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[2]

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

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[3]

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

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

Since $f(x)$ is primitive, $\gcd ( a_{0} , \cdots , a_{n} ) = 1$, and $p$ can’t divide all of $a_{0} , \cdots , a_{n}$. Let the first coefficient that $p$ fails to divide among those be $a_{r}$.
Similarly, since $g(x)$ is primitive, $\gcd ( b_{0} , \cdots , b_{m} ) = 1$, and $p$ can’t divide all of $b_{0} , \cdots , b_{m}$. Let the first coefficient that $p$ fails to divide among those be $b_{s}$.

Then the coefficient of the $( r + s)$ 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 definition,

for $a_{r}$: $$ p \mid ( a_{0} b_{r+s} + \cdots + a_{r-1} b_{s+1} ) $$
for $b_{s}$: $$ p \mid ( a_{r+1} b_{s-1} + \cdots + a_{r+s} b_{0} ) $$

However, since $p \nmid a_{r} b_{s}$, the given $p$ can’t divide $f(x) g(x)$. The same applies to all irreducible elements, making $f(x) g(x)$ primitive.

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[4]

Let the degree of $f(x) \in D[x]$ be $n$.

Then, $f(x)$ can be factored into $$ f (x) = g_{1} (x) \cdots g_{r} (x)) $$ And for each $i = 1 , \cdots , r$, the factors can be expressed as the product of a primitive function $h_{i} (x) \in D[x]$ and $c_{i} \in D$, as in $$ g_{i} (x) = c_{i} h_{i} (x) $$ This $c_{i}$ is called the Content of $g_{i} (x)$, leading to $$ f(x) = c_{1} h_{1} (x) \cdots c_{r} h_{r} (x) $$ Since the content is unique for given $g_{i} (x)$ and $h_{i} (x)$, the factorization of $f(x)$ is unique disregarding the order and scalar multiplication of factors.

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[5]

Factorization of Polynomials:
[2]: If $F$ is a field, all non-constant polynomials $f(x) \in F [ x ]$ can be factored into irreducible elements, and this factorization is unique.

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

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