# Unique Factorization

Through this article and the next few ones, we will explore unique factorization in rings. The inspiration, of course, comes from ℤ. Here is an application of unique factorization. Warning: not all steps may make sense to the reader at this point of time.

Sample Problem.

Find all integer solutions to $x^2 + 2 = y^3$.

Solution (Sketch)

Factor the equation to obtain $y^3 = (x + \alpha)(x - \alpha)$ where $\alpha = \sqrt{-2}$. Use the fact that $\mathbb Z[\alpha]$ satisfies unique factorization; deduce that $x + \alpha$ and $x-\alpha$ must be coprime (since x, y are odd). Hence $x + \alpha = a^3$ and $x - \alpha = b^3$ for some $a,b \in \mathbb Z[\alpha]$ with $ab = y$. This gives $2\sqrt{-2} = a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Now we factor $2\sqrt{-2}$ in all possible ways and solve a pair of simultaneous equations in a and b. E.g.

$a-b = 2\sqrt{-2}, \ a^2 + ab + b^2 = 1\ \implies \ (a, b) = (1 + \sqrt{-2}, 1 - \sqrt{-2})$

which gives $(x,y) = (-5, 3).$

Before starting, here are some preliminary definitions.

Throughout this article, A denotes an integral domain.

Definition.

Recall that A is a unit if it is invertible under product.

• Elements $x,y\in A$ are associated (written as $x\sim y$) if there is a unit $u\in A$ such that $y = ux$.
• For $x,y\in A$ we say x divides y (written as $x|y$) if there exists $z\in A$ such that $y = xz$.
• Let $x\in A - \{0\}$, $x$ not a unit. We say it is reducible if we can find $y,z\in A$ which are non-unit such that $x=yz$. Otherwise, $x$ is irreducible.

All the above conditions can be expressed in terms of ideals.

• x and y are associated if and only if (x) = (y) as ideals of A. Thus, the relation is an equivalence relation.
• x divides y if and only if $(y) \subseteq (x)$ as ideals. Thus $x|y$ and $y|x$ if and only if $x\sim y$.
• x is irreducible if and only if when (x) = (y)(z) as a product of ideals, either (x) = (y) or (x) = (z). In particular, if x is irreducible, so are all its associates.

The descriptions in terms of principal ideals are much cleaner.

## Examples

1. Suppose $A = \mathbb Z[i] = \{a + b\sqrt{-1} : a,b\in \mathbb Z\}$. The only units are {-1, +1, –i, +i}. So each non-zero element has 4 associates, e.g.

$(1+2i, -2+i, -1-2i, 2-i)$ are associates, $(1-2i, -2-i, -1+2i, 2+i)$ are associates.

Note that 1 + 2i and 1 – 2i are not associates. On the other hand, 1 + i and 1 – i are associates. (Verify this!)

2. Suppose $A = \mathbb Z[\sqrt 2] = \{a + b\sqrt 2 : a,b \in \in \mathbb Z\}$. The unit group is infinite since it contains $\pm (1 + \sqrt 2)^n$ for all $n\in \mathbb Z$. Thus every non-zero element has infinitely many associates.

# Factoring Into Irreducibles

Let $x\in A$ be non-zero, non-unit. If x is reducible, we can factor it as xyz where yz are non-zero and non-unit. Again, if either of them is reducible, we factor them further. If this process terminates, we obtain x as a factor of irreducibles.

Unfortunately, there are cases where it does not.

Example

Let $A = \mathbb R[x^{1/n} : n = 1, 2, \ldots]$. Now $x\in A$ can be factorized indefinitely:

$x=x^{1/2} \cdot x^{1/2} = (x^{1/4} \cdot x^{1/4}) \cdot (x^{1/4}\cdot x^{1/4}) = \ldots$

Here is a condition which ensures that factorization terminates.

Proposition 1.

Partially order the principal ideals of A by reverse inclusion:

$(x) \le (y)\iff (y)\subseteq (x)$.

If this poset is noetherian, then every non-unit $x\in A-\{0\}$ can be factored as a product of irreducibles.

Note

By proposition 1 here, the above condition is equivalent to either of the following.

• Any non-empty collection of principal ideals of A has a maximal element.
• In any chain of principal ideals $(x_1) \subseteq (x_2) \subseteq \ldots$ of A, there exists n such that $(x_n) = (x_{n+1}) = \ldots$.

One also says A satisfies ascending chain condition (a.c.c.) on the set of principal ideals.

Proof

Let S be the collection of all non-unit $x\in A-\{0\}$ which cannot be factored as a product of irreducibles. If S is non-empty, the set of principal ideals (x) for $x\in S$ has a maximal element (z) with respect to ⊆.

Since z cannot be factored as a product of irreducibles, it is not irreducible. Write zxy where xy are non-unit and non-zero. Then $(z) \subsetneq (x)$ and $(z) \subsetneq (y)$ so by maximality of (z) we have $x,y\not\in S$. Hence x and y can be expressed as a product of finitely many irreducibles; since zxy so does z, a contradiction. ♦

But how does one check the above abstract condition for a ring A? There are two answers to this.

• Most rings we are interested in actually satisfy a.c.c. on all ideals. Called noetherian rings, we will see later that they are everywhere.
• More immediately, we will give a criterion for unique factorizability which also implies a.c.c. on principal ideals.

# UFDs

Finally, here is our main object of interest.

Definition.

An integral domain A is called a unique factorization domain (UFD) if every proper principal ideal $(z) \subsetneq A$ can be factored as a product of irreducible principal ideals.

$(z) = (x_1) (x_2) \ldots (x_k)$

uniquely up to permutation.

Here is a non-example. Suppose $A = \mathbb Z[\sqrt{-5}]$. In this ring, we have

$6 = 2 \times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$

It is easy to see that no two of these elements are associates. Thus to see that unique factorization fails, it remains to show all of them are irreducible. For that, take the norm function $N : A \to \mathbb Z$ where $N(a+b\sqrt{-5}) = a^2 + 5b^2.$ An easy calculation shows:

• $N(xy) = N(x) N(y)$,
• $N(x) \ge 0$ for all x,
• $N(x) = 1 \iff x = \pm 1$.

Now the above elements have norms

$N(2) = 4, \ N(3) = 9, \ N(1+\sqrt{-5}) = N(1 - \sqrt{-5}) = 6.$

If any of them had a proper factor, such a factor would have norm 2 or 3. But it is easy to see no element of norm 2 or 3 exists.

Exercise A

Prove that $\mathbb Z[\sqrt{10}]$ is not a UFD.

# Prime Elements

We end this section by stating a necessary and sufficient condition for unique factorization to hold.

Definition.

Let $x\in A - \{0\}$. We say x is prime if $(x)$ is a prime ideal of A.

Thus x is prime if and only if:

• whenever yz is a multiple of x, either y or z is a multiple of x.

Lemma 1.

A prime element $x\in A - \{0\}$ is irreducible.

Proof

Suppose x = yz, where yz are non-zero. Then yz is a multiple of x so either y or z is a multiple of x. Assume the former, so y is a multiple of x, but since xyz we have x is a multiple of y also. Thus x and y are associates and z is a unit. ♦

Finally, the main result we want is as follows.

Theorem 1.

Let A be a domain which satisfies a.c.c. on principal ideals.

Then A is a unique factorization domain if and only if all irreducible elements are prime.

Proof

(⇒) Let A be a UFD and $x\in A$ be irreducible. To show x is prime, suppose $yz \in (x)$ and $y\not\in (x)$; we need to show $z\in (x)$. Write $yz = ax$ for some $a\in A$. Write y, z and a as a product of irreducibles. By unique factorization, since x does not occur in the factoring of y, it must occur in the factoring of z. Thus $z\in (x)$.

(⇐) Suppose all irreducibles are prime. For any $a\in A$, suppose we have factorizations $a = x_1 x_2 \ldots x_k = y_1 y_2 \ldots y_l,$ where $x_i, y_j \in A$ are all irreducible, and hence prime. Write this in terms of principal ideals:

$(x_1) (x_2) \ldots (x_k) = (y_1) (y_2) \ldots (y_l).$

Now since $y_1 \ldots y_l \in (x_1)$, one of the terms, say $y_1$ must be in $(x_1)$. Hence $y_1$ is a multiple of $x_1$; by irreducibility we have $(x_1) = (y_1)$. So $x_1$ and $y_1$ are associates and upon division we have

$u x_2 \ldots x_k = y_2 \ldots y_l\ (u = \text{unit}) \implies (x_2)\ldots (x_k) = (y_2)\ldots (y_l).$

Repeat until we run out of terms on one side so we get (1). The other side must also be empty so we have established unique factorization. ♦

In the next article, we will find many examples of UFDs based on this criterion.

Example

Let us show that $A = \mathbb Z[\sqrt {-5}]$ is not a UFD by finding an irreducible element which is not prime: 2. By the norm argument above, 2 is irreducible but it is not prime because

$A/(2) = \mathbb Z[X]/(X^2 + 5, 2) = \mathbb F_2[X]/(X^2 + 5) = \mathbb F_2[X]/(X^2 + 1)$

is not an integral domain.

Exercise B

Find all prime ideals of $A = \mathbb Z[\sqrt{-5}]$ which contain 2 or 3.

[Hint (highlight to read): prime ideals of A containing 2 correspond to prime ideals of A/(2).]

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