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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 \(R\). We say that \(a\) divides \(b\), and write \(a \mid b\), if there exists an element \(c \in R\) such that \(b = ac\). 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 \(a = ub\).

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 \(p = ab\), either \(a\) or \(b\) is a unit. Furthermore, \(p\) is prime if whenever \(p \mid ab\) either \(p \mid a\) or \(p \mid b\).

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 \(x^2\), \(y^2\), and \(xy\). Each of these elements is irreducible in \(R\); however, \(xy\) is not prime, since \(xy\) divides \(x^2 y^2\) but does not divide either \(x^2\) or \(y^2\).

The Fundamental Theorem of Arithmetic states that every positive integer \(n \gt 1\) can be factored into a product of prime numbers \(p_1 \cdots p_k\), 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 \(D\).

  2. Let \(a = p_1 \cdots p_r = q_1 \cdots q_s\), 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 \(j = 1, \ldots, r\).

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 \(\nu( z) = |z|^2 = a^2 + 3 b^2\). It is clear that \(\nu(z) \geq 0\) with equality when \(z = 0\). Also, from our knowledge of complex numbers we know that \(\nu(z w) = \nu(z) \nu(w)\). It is easy to show that if \(\nu(z) = 1\), then \(z\) is a unit, and that the only units of \({\mathbb Z}[ \sqrt{3}\, i ]\) are 1 and \(-1\).

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).\end{equation*} We must show that each of these factors is an irreducible element in \({\mathbb Z}[ \sqrt{3}\, i ]\). If 2 is not irreducible, then \(2 = z w\) for elements \(z, w\) in \({\mathbb Z}[ \sqrt{3}\, i ]\) where \(\nu( z) = \nu(w) = 2\). 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 \(1 + \sqrt{3}\, i\), 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 \(\langle a \rangle = \{ ra : r \in R \}\). An integral domain in which every ideal is principal is called a principal ideal domain, or PID.

Proof

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.

Proof
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 \(I = \{ 5 f(x) + x g(x) : f(x), g(x) \in {\mathbb Z}[x] \}\). We can easily show that \(I\) is an ideal of \({\mathbb Z}[x]\). Suppose that \(I = \langle p(x) \rangle\). Since \(5 \in I\), \(5 = f(x) p(x)\). In this case \(p(x) = p\) must be a constant. Since \(x \in I\), \(x = p g(x)\); consequently, \(p = \pm 1\). However, it follows from this fact that \(\langle p(x) \rangle = {\mathbb Z}[x]\). But this would mean that 3 is in \(I\). Therefore, we can write \(3 = 5 f(x) + x g(x)\) for some \(f(x)\) and \(g(x)\) in \({\mathbb Z}[x]\). Examining the constant term of this polynomial, we see that \(3 = 5 f(x)\), which is impossible.

SubsectionEuclidean Domains

We have repeatedly used the division algorithm when proving results about either \({\mathbb Z}\) or \(F[x]\), 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 for each \(a \in D\) there is a nonnegative integer \(\nu(a)\) satisfying the following conditions.

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

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

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} \}.\end{equation*} We usually measure the size of a complex number \(a + bi\) by its absolute value, \(|a + bi| = \sqrt{ a^2 + b^2}\); 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 \({\mathbb Z}[i]\). Let \(z, w \in {\mathbb Z}[i]\). Then \(\nu( zw) = |zw|^2 = |z|^2 |w|^2 = \nu(z) \nu(w)\). Since \(\nu(z) \geq 1\) for every nonzero \(z \in {\mathbb Z}[i]\), \(\nu( z) \leq \nu(z) \nu(w)\).

Next, we must show that for any \(z= a+bi\) and \(w = c+di\) in \({\mathbb Z}[i]\) with \(w \neq 0\), there exist elements \(q\) and \(r\) in \({\mathbb Z}[i]\) such that \(z = qw + r\) with either \(r=0\) or \(\nu(r) \lt \nu(w)\). We can view \(z\) and \(w\) as elements in \({\mathbb Q}(i) = \{ p + qi : p, q \in {\mathbb Q} \}\), the field of fractions of \({\mathbb Z}[i]\). 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 \({\mathbb Q}(i)\). 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 \(|n_i / (a^2 + b^2)| \leq 1/2\). For example, we write \begin{align*} \frac{9}{8} & = 1 + \frac{1}{8}\\ \frac{15}{8} & = 2 - \frac{1}{8}. \end{align*} Thus, \(s\) and \(t\) are the “fractional parts” of \(z w^{-1} = (m_1 + m_2 i) + (s + ti)\). We also know that \(s^2 + t^2 \leq 1/4 + 1/4 = 1/2\). Multiplying by \(w\), we have \begin{equation*}z = z w^{-1} w = w (m_1 + m_2 i) + w (s + ti) = q w + r,\end{equation*} where \(q = m_1 + m_2 i\) and \(r = w (s + ti)\). Since \(z\) and \(qw\) are in \({\mathbb Z}[i]\), \(r\) must be in \({\mathbb Z}[i]\). Finally, we need to show that either \(r = 0\) or \(\nu(r) \lt \nu(w)\). However, \begin{equation*}\nu(r) = \nu(w) \nu(s + ti) \leq \frac{1}{2} \nu(w) \lt \nu(w).\end{equation*}

Proof

SubsectionFactorization in \(D\lbrack x \rbrack\)

One of the most important polynomial rings is \({\mathbb Z}[x]\). 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 \(D[x]\). Then the content of \(p(x)\) is the greatest common divisor of \(a_0, \ldots, a_n\). We say that \(p(x)\) is primitive if \(\gcd(a_0, \ldots, a_n ) = 1\).

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.

Proof

The following corollaries are direct consequences of Lemma 18.26.

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 \(\sqrt{-1}\).

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.