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 $y^2\text{.}$

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 16.6.12, 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:

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.

Lemma18.11

Let $D$ be an integral domain and let $a, b \in D\text{.}$ Then

1. $a \mid b$ if and only if $\langle b \rangle \subset \langle a \rangle\text{.}$

2. $a$ and $b$ are associates if and only if $\langle b \rangle = \langle a \rangle\text{.}$

3. $a$ is a unit in $D$ if and only if $\langle a \rangle = D\text{.}$

Proof

(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. However, $a$ is an associate of 1 if and only if $\langle a \rangle = \langle 1 \rangle = D\text{.}$

Theorem18.12

Let $D$ be a PID and $\langle p \rangle$ be a nonzero ideal in $D\text{.}$ Then $\langle p \rangle$ is a maximal ideal if and only if $p$ is irreducible.

Proof

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.

Corollary18.13

Let $D$ be a PID. If $p$ is irreducible, then $p$ is prime.

Proof

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 $p \mid b\text{.}$

Lemma18.14

Let $D$ be a PID. Let $I_1, I_2, \ldots$ be a set of ideals such that $I_1 \subset I_2 \subset \cdots\text{.}$ Then there exists an integer $N$ such that $I_n = I_N$ for all $n \geq N\text{.}$

Proof

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 $n \geq N\text{.}$

Theorem18.15

Every PID is a UFD.

Proof

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

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

This chain must satisfy the ascending chain condition; therefore,

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

Uniqueness of the factorization. To show uniqueness, let

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,

or

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

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.

Corollary18.16

Let $F$ be a field. Then $F[x]$ is a UFD.

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.

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

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

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

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

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,

Theorem18.21

Every Euclidean domain is a principal ideal domain.

Proof

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 $I = \langle b \rangle\text{.}$

Corollary18.22

Every Euclidean domain is a unique factorization domain.

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; however, the polynomial $q(x) = 4 x^2 - 6 x + 8$ is not primitive since the content of $q(x)$ is 2.

Theorem18.24Gauss's Lemma

Let $D$ be a UFD and let $f(x)$ and $g(x)$ be primitive polynomials in $D[x]\text{.}$ Then $f(x) g(x)$ is primitive.

Proof

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

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.

Lemma18.25

Let $D$ be a UFD, and let $p(x)$ and $q(x)$ be in $D[x]\text{.}$ Then the content of $p(x) q(x)$ is equal to the product of the contents of $p(x)$ and $q(x)\text{.}$

Proof

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 $cd\text{.}$

Lemma18.26

Let $D$ be a UFD and $F$ its field of fractions. Suppose that $p(x) \in D[x]$ and $p(x) = f(x) g(x)\text{,}$ where $f(x)$ and $g(x)$ are in $F[x]\text{.}$ Then $p(x) = f_1(x) g_1(x)\text{,}$ where $f_1(x)$ and $g_1(x)$ are in $D[x]\text{.}$ Furthermore, $\deg f(x) = \deg f_1(x)$ and $\deg g(x) = \deg g_1(x)\text{.}$

Proof

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 $a_1, b_2 \in D$ 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 $\deg g(x) = \deg g_1(x)\text{.}$

Corollary18.27

Let $D$ be a UFD and $F$ its field of fractions. A primitive polynomial $p(x)$ in $D[x]$ is irreducible in $F[x]$ if and only if it is irreducible in $D[x]\text{.}$

Corollary18.28

Let $D$ be a UFD and $F$ its field of fractions. If $p(x)$ is a monic polynomial in $D[x]$ with $p(x) = f(x) g(x)$ in $F[x]\text{,}$ then $p(x) = f_1(x) g_1(x)\text{,}$ where $f_1(x)$ and $g_1(x)$ are in $D[x]\text{.}$ Furthermore, $\deg f(x) = \deg f_1(x)$ and $\deg g(x) = \deg g_1(x)\text{.}$

Theorem18.29

If $D$ is a UFD, then $D[x]$ is a UFD.

Proof

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

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

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.

Corollary18.30

Let $F$ be a field. Then $F[x]$ is a UFD.

Corollary18.31

The ring of polynomials over the integers, ${\mathbb Z}[x]\text{,}$ is a UFD.

Corollary18.32

Let $D$ be a UFD. Then $D[x_1, \ldots, x_n]$ is a UFD.

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 (${\mathbb Z}[x]$).