## Section18.2Factorization in Integral Domains

The building blocks of the integers are the prime numbers. If $$F$$ is a field, then irreducible polynomials in $$F[x]$$ play a role that is very similar to that of the prime numbers in the ring of integers. Given an arbitrary integral domain, we are led to the following series of definitions.

Let $$R$$ be a commutative ring with identity, and let $$a$$ and $$b$$ be elements in $$R\text{.}$$ We say that $$a$$ divides $$b\text{,}$$ and write $$a \mid b\text{,}$$ if there exists an element $$c \in R$$ such that $$b = ac\text{.}$$ A unit in $$R$$ is an element that has a multiplicative inverse. Two elements $$a$$ and $$b$$ in $$R$$ are said to be associates if there exists a unit $$u$$ in $$R$$ such that $$a = ub\text{.}$$

Let $$D$$ be an integral domain. A nonzero element $$p \in D$$ that is not a unit is said to be irreducible provided that whenever $$p = ab\text{,}$$ either $$a$$ or $$b$$ is a unit. Furthermore, $$p$$ is prime if whenever $$p \mid ab$$ either $$p \mid a$$ or $$p \mid b\text{.}$$

### Example18.8.

It is important to notice that prime and irreducible elements do not always coincide. Let $$R$$ be the subring (with identity) of $${\mathbb Q}[x, y]$$ generated by $$x^2\text{,}$$ $$y^2\text{,}$$ and $$xy\text{.}$$ Each of these elements is irreducible in $$R\text{;}$$ however, $$xy$$ is not prime, since $$xy$$ divides $$x^2 y^2$$ but does not divide either $$x^2$$ or $$y^2\text{.}$$

The Fundamental Theorem of Arithmetic states that every positive integer $$n \gt 1$$ can be factored into a product of prime numbers $$p_1 \cdots p_k\text{,}$$ where the $$p_i$$'s are not necessarily distinct. We also know that such factorizations are unique up to the order of the $$p_i$$'s. We can easily extend this result to the integers. The question arises of whether or not such factorizations are possible in other rings. Generalizing this definition, we say an integral domain $$D$$ is a unique factorization domain, or UFD, if $$D$$ satisfies the following criteria.

1. Let $$a \in D$$ such that $$a \neq 0$$ and $$a$$ is not a unit. Then $$a$$ can be written as the product of irreducible elements in $$D\text{.}$$

2. Let $$a = p_1 \cdots p_r = q_1 \cdots q_s\text{,}$$ where the $$p_i$$'s and the $$q_i$$'s are irreducible. Then $$r=s$$ and there is a $$\pi \in S_r$$ such that $$p_i$$ and $$q_{\pi(j)}$$ are associates for $$j = 1, \ldots, r\text{.}$$

### Example18.9.

The integers are a unique factorization domain by the Fundamental Theorem of Arithmetic.

### Example18.10.

Not every integral domain is a unique factorization domain. The subring $${\mathbb Z}[ \sqrt{3}\, i ] = \{ a + b \sqrt{3}\, i\}$$ of the complex numbers is an integral domain (Exercise 16.7.12, Chapter 16). Let $$z = a + b \sqrt{3}\, i$$ and define $$\nu : {\mathbb Z}[ \sqrt{3}\, i ] \rightarrow {\mathbb N} \cup \{ 0 \}$$ by $$\nu( z) = |z|^2 = a^2 + 3 b^2\text{.}$$ It is clear that $$\nu(z) \geq 0$$ with equality when $$z = 0\text{.}$$ Also, from our knowledge of complex numbers we know that $$\nu(z w) = \nu(z) \nu(w)\text{.}$$ It is easy to show that if $$\nu(z) = 1\text{,}$$ then $$z$$ is a unit, and that the only units of $${\mathbb Z}[ \sqrt{3}\, i ]$$ are $$1$$ and $$-1\text{.}$$

We claim that $$4$$ has two distinct factorizations into irreducible elements:

\begin{equation*} 4 = 2 \cdot 2 = (1 - \sqrt{3}\, i) (1 + \sqrt{3}\, i)\text{.} \end{equation*}

We must show that each of these factors is an irreducible element in $${\mathbb Z}[ \sqrt{3}\, i ]\text{.}$$ If $$2$$ is not irreducible, then $$2 = z w$$ for elements $$z, w$$ in $${\mathbb Z}[ \sqrt{3}\, i ]$$ where $$\nu( z) = \nu(w) = 2\text{.}$$ However, there does not exist an element in $$z$$ in $${\mathbb Z}[\sqrt{3}\, i ]$$ such that $$\nu(z) = 2$$ because the equation $$a^2 + 3 b^2 = 2$$ has no integer solutions. Therefore, $$2$$ must be irreducible. A similar argument shows that both $$1 - \sqrt{3}\, i$$ and $$1 + \sqrt{3}\, i$$ are irreducible. Since $$2$$ is not a unit multiple of either $$1 - \sqrt{3}\, i$$ or $$1 + \sqrt{3}\, i\text{,}$$ $$4$$ has at least two distinct factorizations into irreducible elements.

### SubsectionPrincipal Ideal Domains

Let $$R$$ be a commutative ring with identity. Recall that a principal ideal generated by $$a \in R$$ is an ideal of the form $$\langle a \rangle = \{ ra : r \in R \}\text{.}$$ An integral domain in which every ideal is principal is called a principal ideal domain, or PID.

(1) Suppose that $$a \mid b\text{.}$$ Then $$b = ax$$ for some $$x \in D\text{.}$$ Hence, for every $$r$$ in $$D\text{,}$$ $$br =(ax)r = a(xr)$$ and $$\langle b \rangle \subset \langle a \rangle\text{.}$$ Conversely, suppose that $$\langle b \rangle \subset \langle a \rangle\text{.}$$ Then $$b \in \langle a \rangle\text{.}$$ Consequently, $$b =a x$$ for some $$x \in D\text{.}$$ Thus, $$a \mid b\text{.}$$

(2) Since $$a$$ and $$b$$ are associates, there exists a unit $$u$$ such that $$a = u b\text{.}$$ Therefore, $$b \mid a$$ and $$\langle a \rangle \subset \langle b \rangle\text{.}$$ Similarly, $$\langle b \rangle \subset \langle a \rangle\text{.}$$ It follows that $$\langle a \rangle = \langle b \rangle\text{.}$$ Conversely, suppose that $$\langle a \rangle = \langle b \rangle\text{.}$$ By part (1), $$a \mid b$$ and $$b \mid a\text{.}$$ Then $$a = bx$$ and $$b = ay$$ for some $$x, y \in D\text{.}$$ Therefore, $$a = bx = ayx\text{.}$$ Since $$D$$ is an integral domain, $$x y = 1\text{;}$$ that is, $$x$$ and $$y$$ are units and $$a$$ and $$b$$ are associates.

(3) An element $$a \in D$$ is a unit if and only if $$a$$ is an associate of $$1\text{.}$$ However, $$a$$ is an associate of $$1$$ if and only if $$\langle a \rangle = \langle 1 \rangle = D\text{.}$$

Suppose that $$\langle p \rangle$$ is a maximal ideal. If some element $$a$$ in $$D$$ divides $$p\text{,}$$ then $$\langle p \rangle \subset \langle a \rangle\text{.}$$ Since $$\langle p \rangle$$ is maximal, either $$D = \langle a \rangle$$ or $$\langle p \rangle = \langle a \rangle\text{.}$$ Consequently, either $$a$$ and $$p$$ are associates or $$a$$ is a unit. Therefore, $$p$$ is irreducible.

Conversely, let $$p$$ be irreducible. If $$\langle a \rangle$$ is an ideal in $$D$$ such that $$\langle p \rangle \subset \langle a \rangle \subset D\text{,}$$ then $$a \mid p\text{.}$$ Since $$p$$ is irreducible, either $$a$$ must be a unit or $$a$$ and $$p$$ are associates. Therefore, either $$D = \langle a \rangle$$ or $$\langle p \rangle = \langle a \rangle\text{.}$$ Thus, $$\langle p \rangle$$ is a maximal ideal.

Let $$p$$ be irreducible and suppose that $$p \mid ab\text{.}$$ Then $$\langle ab \rangle \subset \langle p \rangle\text{.}$$ By Corollary 16.40, since $$\langle p \rangle$$ is a maximal ideal, $$\langle p \rangle$$ must also be a prime ideal. Thus, either $$a \in \langle p \rangle$$ or $$b \in \langle p \rangle\text{.}$$ Hence, either $$p \mid a$$ or $$p \mid b\text{.}$$

We claim that $$I= \bigcup_{i = 1}^\infty I_i$$ is an ideal of $$D\text{.}$$ Certainly $$I$$ is not empty, since $$I_1 \subset I$$ and $$0 \in I\text{.}$$ If $$a, b \in I\text{,}$$ then $$a \in I_i$$ and $$b \in I_j$$ for some $$i$$ and $$j$$ in $${\mathbb N}\text{.}$$ Without loss of generality we can assume that $$i \leq j\text{.}$$ Hence, $$a$$ and $$b$$ are both in $$I_j$$ and so $$a - b$$ is also in $$I_j\text{.}$$ Now let $$r \in D$$ and $$a \in I\text{.}$$ Again, we note that $$a \in I_i$$ for some positive integer $$i\text{.}$$ Since $$I_i$$ is an ideal, $$ra \in I_i$$ and hence must be in $$I\text{.}$$ Therefore, we have shown that $$I$$ is an ideal in $$D\text{.}$$

Since $$D$$ is a principal ideal domain, there exists an element $$\overline{a} \in D$$ that generates $$I\text{.}$$ Since $$\overline{a}$$ is in $$I_N$$ for some $$N \in {\mathbb N}\text{,}$$ we know that $$I_N = I = \langle \overline{a} \rangle\text{.}$$ Consequently, $$I_n = I_N$$ for $$n \geq N\text{.}$$

Any commutative ring satisfying the condition in Lemma 18.14 is said to satisfy the ascending chain condition, or ACC. Such rings are called Noetherian rings, after Emmy Noether.

Existence of a factorization. Let $$D$$ be a PID and $$a$$ be a nonzero element in $$D$$ that is not a unit. If $$a$$ is irreducible, then we are done. If not, then there exists a factorization $$a = a_1 b_1\text{,}$$ where neither $$a_1$$ nor $$b_1$$ is a unit. Hence, $$\langle a \rangle \subset \langle a_1 \rangle\text{.}$$ By Lemma 18.11, we know that $$\langle a \rangle \neq \langle a_1 \rangle\text{;}$$ otherwise, $$a$$ and $$a_1$$ would be associates and $$b_1$$ would be a unit, which would contradict our assumption. Now suppose that $$a_1 = a_2 b_2\text{,}$$ where neither $$a_2$$ nor $$b_2$$ is a unit. By the same argument as before, $$\langle a_1 \rangle \subset \langle a_2 \rangle\text{.}$$ We can continue with this construction to obtain an ascending chain of ideals

\begin{equation*} \langle a \rangle \subset \langle a_1 \rangle \subset \langle a_2 \rangle \subset \cdots\text{.} \end{equation*}

By Lemma 18.14, there exists a positive integer $$N$$ such that $$\langle a_n \rangle = \langle a_N \rangle$$ for all $$n \geq N\text{.}$$ Consequently, $$a_N$$ must be irreducible. We have now shown that $$a$$ is the product of two elements, one of which must be irreducible.

Now suppose that $$a = c_1 p_1\text{,}$$ where $$p_1$$ is irreducible. If $$c_1$$ is not a unit, we can repeat the preceding argument to conclude that $$\langle a \rangle \subset \langle c_1 \rangle\text{.}$$ Either $$c_1$$ is irreducible or $$c_1 = c_2 p_2\text{,}$$ where $$p_2$$ is irreducible and $$c_2$$ is not a unit. Continuing in this manner, we obtain another chain of ideals

\begin{equation*} \langle a \rangle \subset \langle c_1 \rangle \subset \langle c_2 \rangle \subset \cdots\text{.} \end{equation*}

This chain must satisfy the ascending chain condition; therefore,

\begin{equation*} a = p_1 p_2 \cdots p_r \end{equation*}

for irreducible elements $$p_1, \ldots, p_r\text{.}$$

Uniqueness of the factorization. To show uniqueness, let

\begin{equation*} a = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s\text{,} \end{equation*}

where each $$p_i$$ and each $$q_i$$ is irreducible. Without loss of generality, we can assume that $$r \lt s\text{.}$$ Since $$p_1$$ divides $$q_1 q_2 \cdots q_s\text{,}$$ by Corollary 18.13 it must divide some $$q_i\text{.}$$ By rearranging the $$q_i$$'s, we can assume that $$p_1 \mid q_1\text{;}$$ hence, $$q_1 = u_1 p_1$$ for some unit $$u_1$$ in $$D\text{.}$$ Therefore,

\begin{equation*} a = p_1 p_2 \cdots p_r = u_1 p_1 q_2 \cdots q_s \end{equation*}

or

\begin{equation*} p_2 \cdots p_r = u_1 q_2 \cdots q_s\text{.} \end{equation*}

Continuing in this manner, we can arrange the $$q_i$$'s such that $$p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r\text{,}$$ to obtain

\begin{equation*} u_1 u_2 \cdots u_r q_{r + 1} \cdots q_s = 1\text{.} \end{equation*}

In this case $$q_{r + 1} \cdots q_s$$ is a unit, which contradicts the fact that $$q_{r + 1}, \ldots, q_s$$ are irreducibles. Therefore, $$r = s$$ and the factorization of $$a$$ is unique.

#### Example18.17.

Every PID is a UFD, but it is not the case that every UFD is a PID. In Corollary 18.31, we will prove that $${\mathbb Z}[x]$$ is a UFD. However, $${\mathbb Z}[x]$$ is not a PID. Let $$I = \{ 5 f(x) + x g(x) : f(x), g(x) \in {\mathbb Z}[x] \}\text{.}$$ We can easily show that $$I$$ is an ideal of $${\mathbb Z}[x]\text{.}$$ Suppose that $$I = \langle p(x) \rangle\text{.}$$ Since $$5 \in I\text{,}$$ $$5 = f(x) p(x)\text{.}$$ In this case $$p(x) = p$$ must be a constant. Since $$x \in I\text{,}$$ $$x = p g(x)\text{;}$$ consequently, $$p = \pm 1\text{.}$$ However, it follows from this fact that $$\langle p(x) \rangle = {\mathbb Z}[x]\text{.}$$ But this would mean that $$3$$ is in $$I\text{.}$$ Therefore, we can write $$3 = 5 f(x) + x g(x)$$ for some $$f(x)$$ and $$g(x)$$ in $${\mathbb Z}[x]\text{.}$$ Examining the constant term of this polynomial, we see that $$3 = 5 f(x)\text{,}$$ which is impossible.

### SubsectionEuclidean Domains

We have repeatedly used the division algorithm when proving results about either $${\mathbb Z}$$ or $$F[x]\text{,}$$ where $$F$$ is a field. We should now ask when a division algorithm is available for an integral domain.

Let $$D$$ be an integral domain such that there is a function $$\nu : D \setminus \{0\} \to \mathbb N$$ satisfying the following conditions.

1. If $$a$$ and $$b$$ are nonzero elements in $$D\text{,}$$ then $$\nu(a) \leq \nu(ab)\text{.}$$

2. Let $$a, b \in D$$ and suppose that $$b \neq 0\text{.}$$ Then there exist elements $$q, r \in D$$ such that $$a = bq + r$$ and either $$r = 0$$ or $$\nu(r) \lt \nu(b)\text{.}$$

Then $$D$$ is called a Euclidean domain and $$\nu$$ is called a Euclidean valuation.

#### Example18.18.

Absolute value on $${\mathbb Z}$$ is a Euclidean valuation.

#### Example18.19.

Let $$F$$ be a field. Then the degree of a polynomial in $$F[x]$$ is a Euclidean valuation.

#### Example18.20.

Recall that the Gaussian integers in Example 16.12 of Chapter 16 are defined by

\begin{equation*} {\mathbb Z}[i] = \{ a + b i : a, b \in {\mathbb Z} \}\text{.} \end{equation*}

We usually measure the size of a complex number $$a + bi$$ by its absolute value,. $$|a + bi| = \sqrt{ a^2 + b^2}\text{;}$$ however, $$\sqrt{a^2 + b^2}$$ may not be an integer. For our valuation we will let $$\nu(a + bi) = a^2 + b^2$$ to ensure that we have an integer.

We claim that $$\nu( a+ bi) = a^2 + b^2$$ is a Euclidean valuation on $${\mathbb Z}[i]\text{.}$$ Let $$z, w \in {\mathbb Z}[i]\text{.}$$ Then $$\nu( zw) = |zw|^2 = |z|^2 |w|^2 = \nu(z) \nu(w)\text{.}$$ Since $$\nu(z) \geq 1$$ for every nonzero $$z \in {\mathbb Z}[i]\text{,}$$ $$\nu( z) \leq \nu(z) \nu(w)\text{.}$$

Next, we must show that for any $$z= a+bi$$ and $$w = c+di$$ in $${\mathbb Z}[i]$$ with $$w \neq 0\text{,}$$ there exist elements $$q$$ and $$r$$ in $${\mathbb Z}[i]$$ such that $$z = qw + r$$ with either $$r=0$$ or $$\nu(r) \lt \nu(w)\text{.}$$ We can view $$z$$ and $$w$$ as elements in $${\mathbb Q}(i) = \{ p + qi : p, q \in {\mathbb Q} \}\text{,}$$ the field of fractions of $${\mathbb Z}[i]\text{.}$$ Observe that

\begin{align*} z w^{-1} & = (a +b i) \frac{c -d i}{c^2 + d^2}\\ & = \frac{ac + b d}{c^2 + d^2} + \frac{b c -ad}{c^2 + d^2}i\\ & = \left( m_1 + \frac{n_1}{c^2 + d^2} \right) + \left( m_2 + \frac{n_2}{c^2 + d^2} \right) i\\ & = (m_1 + m_2 i) + \left( \frac{n_1}{c^2 + d^2} + \frac{n_2}{c^2 + d^2}i \right)\\ & = (m_1 + m_2 i) + (s + ti) \end{align*}

in $${\mathbb Q}(i)\text{.}$$ In the last steps we are writing the real and imaginary parts as an integer plus a proper fraction. That is, we take the closest integer $$m_i$$ such that the fractional part satisfies $$|n_i / (a^2 + b^2)| \leq 1/2\text{.}$$ For example, we write

\begin{align*} \frac{9}{8} & = 1 + \frac{1}{8}\\ \frac{15}{8} & = 2 - \frac{1}{8}\text{.} \end{align*}

Thus, $$s$$ and $$t$$ are the “fractional parts” of $$z w^{-1} = (m_1 + m_2 i) + (s + ti)\text{.}$$ We also know that $$s^2 + t^2 \leq 1/4 + 1/4 = 1/2\text{.}$$ Multiplying by $$w\text{,}$$ we have

\begin{equation*} z = z w^{-1} w = w (m_1 + m_2 i) + w (s + ti) = q w + r\text{,} \end{equation*}

where $$q = m_1 + m_2 i$$ and $$r = w (s + ti)\text{.}$$ Since $$z$$ and $$qw$$ are in $${\mathbb Z}[i]\text{,}$$ $$r$$ must be in $${\mathbb Z}[i]\text{.}$$ Finally, we need to show that either $$r = 0$$ or $$\nu(r) \lt \nu(w)\text{.}$$ However,

\begin{equation*} \nu(r) = \nu(w) \nu(s + ti) \leq \frac{1}{2} \nu(w) \lt \nu(w)\text{.} \end{equation*}

Let $$D$$ be a Euclidean domain and let $$\nu$$ be a Euclidean valuation on $$D\text{.}$$ Suppose $$I$$ is a nontrivial ideal in $$D$$ and choose a nonzero element $$b \in I$$ such that $$\nu(b)$$ is minimal for all $$a \in I\text{.}$$ Since $$D$$ is a Euclidean domain, there exist elements $$q$$ and $$r$$ in $$D$$ such that $$a = bq + r$$ and either $$r = 0$$ or $$\nu(r) \lt \nu(b)\text{.}$$ But $$r = a - bq$$ is in $$I$$ since $$I$$ is an ideal; therefore, $$r = 0$$ by the minimality of $$b\text{.}$$ It follows that $$a = bq$$ and $$I = \langle b \rangle\text{.}$$

### SubsectionFactorization in $$D\lbrack x \rbrack$$

One of the most important polynomial rings is $${\mathbb Z}[x]\text{.}$$ One of the first questions that come to mind about $${\mathbb Z}[x]$$ is whether or not it is a UFD. We will prove a more general statement here. Our first task is to obtain a more general version of Gauss's Lemma (Theorem 17.14).

Let $$D$$ be a unique factorization domain and suppose that

\begin{equation*} p(x) = a_n x^n + \cdots + a_1 x + a_0 \end{equation*}

in $$D[x]\text{.}$$ Then the content of $$p(x)$$ is the greatest common divisor of $$a_0, \ldots, a_n\text{.}$$ We say that $$p(x)$$ is primitive if $$\gcd(a_0, \ldots, a_n ) = 1\text{.}$$

#### Example18.23.

In $${\mathbb Z}[x]$$ the polynomial $$p(x)= 5 x^4 - 3 x^3 + x -4$$ is a primitive polynomial since the greatest common divisor of the coefficients is $$1\text{;}$$ however, the polynomial $$q(x) = 4 x^2 - 6 x + 8$$ is not primitive since the content of $$q(x)$$ is $$2\text{.}$$

Let $$f(x) = \sum_{i=0}^{m} a_i x^i$$ and $$g(x) = \sum_{i=0}^{n} b_i x^i\text{.}$$ Suppose that $$p$$ is a prime dividing the coefficients of $$f(x) g(x)\text{.}$$ Let $$r$$ be the smallest integer such that $$p \notdivide a_r$$ and $$s$$ be the smallest integer such that $$p \notdivide b_s\text{.}$$ The coefficient of $$x^{r+s}$$ in $$f(x) g(x)$$ is

\begin{equation*} c_{r + s} = a_0 b_{r + s} + a_1 b_{r + s - 1} + \cdots + a_{r + s - 1} b_1 + a_{r + s} b_0\text{.} \end{equation*}

Since $$p$$ divides $$a_0, \ldots, a_{r-1}$$ and $$b_0, \ldots, b_{s-1}\text{,}$$ $$p$$ divides every term of $$c_{r+s}$$ except for the term $$a_r b_s\text{.}$$ However, since $$p \mid c_{r+s}\text{,}$$ either $$p$$ divides $$a_r$$ or $$p$$ divides $$b_s\text{.}$$ But this is impossible.

Let $$p(x) = c p_1(x)$$ and $$q(x) = d q_1(x)\text{,}$$ where $$c$$ and $$d$$ are the contents of $$p(x)$$ and $$q(x)\text{,}$$ respectively. Then $$p_1(x)$$ and $$q_1(x)$$ are primitive. We can now write $$p(x) q(x) = c d p_1(x) q_1(x)\text{.}$$ Since $$p_1(x) q_1(x)$$ is primitive, the content of $$p(x) q(x)$$ must be $$cd\text{.}$$

Let $$a$$ and $$b$$ be nonzero elements of $$D$$ such that $$a f(x), b g(x)$$ are in $$D[x]\text{.}$$ We can find $$a_1, b_1 \in D$$ such that $$a f(x) = a_1 f_1(x)$$ and $$b g(x) = b_1 g_1(x)\text{,}$$ where $$f_1(x)$$ and $$g_1(x)$$ are primitive polynomials in $$D[x]\text{.}$$ Therefore, $$a b p(x) = (a_1 f_1(x))( b_1 g_1(x))\text{.}$$ Since $$f_1(x)$$ and $$g_1(x)$$ are primitive polynomials, it must be the case that $$ab \mid a_1 b_1$$ by Gauss's Lemma. Thus there exists a $$c \in D$$ such that $$p(x) = c f_1(x) g_1(x)\text{.}$$ Clearly, $$\deg f(x) = \deg f_1(x)$$ and $$\deg g(x) = \deg g_1(x)\text{.}$$

The following corollaries are direct consequences of Lemma 18.26.

Let $$p(x)$$ be a nonzero polynomial in $$D[x]\text{.}$$ If $$p(x)$$ is a constant polynomial, then it must have a unique factorization since $$D$$ is a UFD. Now suppose that $$p(x)$$ is a polynomial of positive degree in $$D[x]\text{.}$$ Let $$F$$ be the field of fractions of $$D\text{,}$$ and let $$p(x) = f_1(x) f_2(x) \cdots f_n(x)$$ by a factorization of $$p(x)\text{,}$$ where each $$f_i(x)$$ is irreducible. Choose $$a_i \in D$$ such that $$a_i f_i(x)$$ is in $$D[x]\text{.}$$ There exist $$b_1, \ldots, b_n \in D$$ such that $$a_i f_i(x) = b_i g_i(x)\text{,}$$ where $$g_i(x)$$ is a primitive polynomial in $$D[x]\text{.}$$ By Corollary 18.27, each $$g_i(x)$$ is irreducible in $$D[x]\text{.}$$ Consequently, we can write

\begin{equation*} a_1 \cdots a_n p(x) = b_1 \cdots b_n g_1(x) \cdots g_n(x)\text{.} \end{equation*}

Let $$b = b_1 \cdots b_n\text{.}$$ Since $$g_1(x) \cdots g_n(x)$$ is primitive, $$a_1 \cdots a_n$$ divides $$b\text{.}$$ Therefore, $$p(x) = a g_1(x) \cdots g_n(x)\text{,}$$ where $$a \in D\text{.}$$ Since $$D$$ is a UFD, we can factor $$a$$ as $$u c_1 \cdots c_k\text{,}$$ where $$u$$ is a unit and each of the $$c_i$$'s is irreducible in $$D\text{.}$$

We will now show the uniqueness of this factorization. Let

\begin{equation*} p(x) = a_1 \cdots a_m f_1(x) \cdots f_n(x) = b_1 \cdots b_r g_1(x) \cdots g_s(x) \end{equation*}

be two factorizations of $$p(x)\text{,}$$ where all of the factors are irreducible in $$D[x]\text{.}$$ By Corollary 18.27, each of the $$f_i$$'s and $$g_i$$'s is irreducible in $$F[x]\text{.}$$ The $$a_i$$'s and the $$b_i$$'s are units in $$F\text{.}$$ Since $$F[x]$$ is a PID, it is a UFD; therefore, $$n=s\text{.}$$ Now rearrange the $$g_i(x)$$'s so that $$f_i(x)$$ and $$g_i(x)$$ are associates for $$i = 1, \ldots, n\text{.}$$ Then there exist $$c_1, \ldots, c_n$$ and $$d_1, \ldots, d_n$$ in $$D$$ such that $$(c_i / d_i) f_i(x) = g_i(x)$$ or $$c_i f_i(x) = d_i g_i(x)\text{.}$$ The polynomials $$f_i(x)$$ and $$g_i(x)$$ are primitive; hence, $$c_i$$ and $$d_i$$ are associates in $$D\text{.}$$ Thus, $$a_1 \cdots a_m = u b_1 \cdots b_r$$ in $$D\text{,}$$ where $$u$$ is a unit in $$D\text{.}$$ Since $$D$$ is a unique factorization domain, $$m = s\text{.}$$ Finally, we can reorder the $$b_i$$'s so that $$a_i$$ and $$b_i$$ are associates for each $$i\text{.}$$ This completes the uniqueness part of the proof.

The theorem that we have just proven has several obvious but important corollaries.

#### Remark18.33.

It is important to notice that every Euclidean domain is a PID and every PID is a UFD. However, as demonstrated by our examples, the converse of each of these statements fails. There are principal ideal domains that are not Euclidean domains, and there are unique factorization domains that are not principal ideal domains ($${\mathbb Z}[x]$$).

### SubsectionHistorical Note

Karl Friedrich Gauss, born in Brunswick, Germany on April 30, 1777, is considered to be one of the greatest mathematicians who ever lived. Gauss was truly a child prodigy. At the age of three he was able to detect errors in the books of his father's business. Gauss entered college at the age of 15. Before the age of 20, Gauss was able to construct a regular $$17$$-sided polygon with a ruler and compass. This was the first new construction of a regular $$n$$-sided polygon since the time of the ancient Greeks. Gauss succeeded in showing that if $$N= 2^{2^n} + 1$$ was prime, then it was possible to construct a regular $$N$$-sided polygon.

Gauss obtained his Ph.D. in 1799 under the direction of Pfaff at the University of Helmstedt. In his dissertation he gave the first complete proof of the Fundamental Theorem of Algebra, which states that every polynomial with real coefficients can be factored into linear factors over the complex numbers. The acceptance of complex numbers was brought about by Gauss, who was the first person to use the notation of $$i$$ for $$\sqrt{-1}\text{.}$$

Gauss then turned his attention toward number theory; in 1801, he published his famous book on number theory, Disquisitiones Arithmeticae. Throughout his life Gauss was intrigued with this branch of mathematics. He once wrote, “Mathematics is the queen of the sciences, and the theory of numbers is the queen of mathematics.”

In 1807, Gauss was appointed director of the Observatory at the University of Göttingen, a position he held until his death. This position required him to study applications of mathematics to the sciences. He succeeded in making contributions to fields such as astronomy, mechanics, optics, geodesy, and magnetism. Along with Wilhelm Weber, he coinvented the first practical electric telegraph some years before a better version was invented by Samuel F. B. Morse.

Gauss was clearly the most prominent mathematician in the world in the early nineteenth century. His status naturally made his discoveries subject to intense scrutiny. Gauss's cold and distant personality many times led him to ignore the work of his contemporaries, making him many enemies. He did not enjoy teaching very much, and young mathematicians who sought him out for encouragement were often rebuffed. Nevertheless, he had many outstanding students, including Eisenstein, Riemann, Kummer, Dirichlet, and Dedekind. Gauss also offered a great deal of encouragement to Sophie Germain (1776–1831), who overcame the many obstacles facing women in her day to become a very prominent mathematician. Gauss died at the age of 78 in Göttingen on February 23, 1855.