Appendix A. Ring Theory（环论）
The following appendices present some of the background material used in this book. In this appendix, we present the aspects of ring theory that we need in this book. We go into detail about unique factorization domains and polynomials in multiple variables, mostly for Chapter V. The irreducibility tests are used in the text mainly for dealing with polynomials over \(\mathbb{Q}\) and over the rational function field \(k(x)\) over a field \(k\).
Throughout this book, we make the assumption that all rings have a multiplicative identity. We start off with a review of the characteristic of a ring. Let \(R\) be a ring. The characteristic of \(R\), denoted \(\mathrm{char}(R)\), is the order of \(1\) in the additive group \((R,+)\), provided that this order is finite; if it is infinite, we set \(\mathrm{char}(R) = 0\). There is a map \(\varphi : \mathbb{Z} \to R\) given by \(\varphi(n) = n \cdot 1\), the sum of \(1\) with itself \(n\) times. It is clear that this map is a ring homomorphism. The kernel of \(\varphi\) is generated by a positive integer \(m\), and \(m\) is precisely \(\mathrm{char}(R)\). Thus, \(\mathbb{Z}/m\mathbb{Z}\) is isomorphic to a subring of \(R\). Moreover, this subring is easily seen to be the unique minimal subring of \(R\); recall that we assume that our rings have an identity. This ring is called the prime subring of \(R\).
We remind the reader that all references to Theorems, Lemmas, etc., made in each appendix refer to that appendix unless it is explicitly stated that they come from a section of the main text.
1 Prime and Maximal Ideals
Let \(R\) be a commutative ring. A prime ideal of \(R\) is an ideal \(P \neq R\), such that if \(a, b \in R\) with \(ab \in P\), then either \(a \in P\) or \(b \in P\). For example, if \(p\) is a prime number, then the ideal \(p\mathbb{Z}\) is a prime ideal of \(\mathbb{Z}\). A maximal ideal of \(R\) is an ideal \(M \neq R\), such that if \(I\) is any ideal of \(R\) with \(M \subseteq I \subseteq R\), then either \(I = M\) or \(I = R\); that is, \(M\) is maximal if \(M\) is not contained in any proper ideal other than itself. Again, if \(p\) is a prime number, then \(p\mathbb{Z}\) is a maximal ideal of \(\mathbb{Z}\). This can be seen from the fact that the gcd of two integers can be written as a linear combination of the integers. If \(p\mathbb{Z} \subseteq I\) and \(I \neq p\mathbb{Z}\), let \(a \in I\) with \(a \notin p\mathbb{Z}\). Then \(p\) does not divide \(a\), so \(\gcd(a,p) = 1\). Therefore, \(1 = ax + py\) for some \(x,y \in \mathbb{Z}\). This means that \(1 \in I\), since \(a, p \in I\); hence, \(I = \mathbb{Z}\). This proves that \(p\mathbb{Z}\) is indeed a maximal ideal of \(\mathbb{Z}\).
Prime and maximal ideals can be characterized in terms of quotient rings. This characterization is often a very useful way to deal with these ideals.
Proposition 1.1 Let \(R\) be a commutative ring with \(1\).
- If \(P\) is a proper ideal of \(R\), then \(P\) is a prime ideal of \(R\) if and only if \(R/P\) is an integral domain.
- If \(M\) is a proper ideal of \(R\), then \(M\) is a maximal ideal of \(R\) if and only if \(R/M\) is a field.
Proof. Let \(P\) be a prime ideal. To show that \(R/P\) is an integral domain, suppose that \(\alpha, \beta \in R/P\) with \(\alpha\beta = 0\). Then \(\alpha = a + P\) and \(\beta = b + P\) for some \(a, b \in R\). The condition \(\alpha\beta = 0\) in \(R/P\) means \((a+P)(b+P) = 0+P\), so \(ab \in P\). Since \(P\) is a prime ideal, either \(a \in P\) or \(b \in P\), so \(a+P = 0+P\) or \(b+P = 0+P\). Thus, \(R/P\) is an integral domain. The converse follows from the same arguments; if \(R/P\) is an integral domain and \(ab \in P\), then \((a+P)(b+P) = 0\) in \(R/P\), so \(a+P = 0+P\) or \(b+P = 0+P\); thus, \(a \in P\) or \(b \in P\).
For the second statement, suppose that \(M\) is a maximal ideal of \(R\). We need to show that each nonzero element of \(R/M\) is invertible. Take \(a+M \in R/M\) with \(a+M \neq 0+M\). Then \(a \notin M\), so the ideal \(M + aR\) is properly larger than \(M\). By maximality, this forces \(M + aR = R\), so \(1 = m + ar\) for some \(m \in M\) and \(r \in R\). Then \((a+M)(r+M) = 1+M\), since \(m \in M\). Therefore, \(a+M\) is invertible, so \(R/M\) is a field. Conversely, suppose that \(R/M\) is a field. Let \(I\) be an ideal of \(R\) with \(M \subset I \subseteq R\). We need to show that \(I = R\). Let \(a \in I - M\). Then \(a+M \neq 0+M\); hence, \(a+M\) is invertible. Thus, there is a \(b \in R\) with \((a+M)(b+M) = 1+M\), so \(ab - 1 \in M\). Since \(a \in I\) and \(M \subseteq I\), this forces \(1 \in I\), so \(I = R\). Therefore, \(M\) is a maximal ideal of \(R\). □
From this proposition, we see that any maximal ideal is prime, but the converse may not be true. Our main use of these concepts will be for the study of polynomials. It follows from the results of Section 3 below that if \(F\) is a field and \(R = F[x]\) is the ring of polynomials over \(F\), then any prime ideal of \(R\) is maximal and is generated by an irreducible polynomial.
Example 1.2 By calculations similar to those before the proposition, one can show that an ideal \(a\mathbb{Z}\) of \(\mathbb{Z}\) is a prime ideal if and only if \(a\) is a prime number. Moreover, an ideal \(I\) of \(\mathbb{Z}\) is maximal if and only if \(I\) is prime.
Let \(R = \mathbb{Z}[x]\), the ring of polynomials in \(x\) over \(\mathbb{Z}\). The ideal \(xR\) is prime, since \(R/xR \cong \mathbb{Z}\) is an integral domain. Moreover, \(xR\) is not maximal, since \(R/xR\) is not a field. Equivalently, \(xR\) is not maximal, since \(xR\) is properly contained in the proper ideal \(xR + 2R\) generated by \(x\) and \(2\).
The proposition above also gives us some information about the characteristic of a ring. If \(R\) is an integral domain, then the map \(\varphi : \mathbb{Z} \to R\) that sends \(n\) to \(n \cdot 1\) is a ring homomorphism, and \(\mathrm{im}(\varphi)\) is a subring of \(R\). Thus, \(\mathbb{Z}/\ker(\varphi)\) is an integral domain, so \(\ker(\varphi)\) is a prime ideal. But \(\ker(\varphi)\) is generated by \(\mathrm{char}(R)\), so \(\mathrm{char}(R)\) is either \(0\) or a prime number.
2 Unique Factorization Domains
The main ring theoretic properties about polynomials we require in Galois theory are that the ring \(F[x]\) of polynomials in a variable \(x\) over a field \(F\) be a principal ideal domain (PID) and be a unique factorization domain (UFD). While these facts can be proved relatively easily, we go into some detail about UFDs primarily to deal with polynomials in more than one variable, a case we need in Chapter V.
Let \(R\) be an integral domain. If \(a, b \in R\), we say that \(a\) divides \(b\) if \(b = ac\) for some \(c \in R\). A nonunit \(a \in R\) is said to be irreducible if whenever \(a = bc\), then either \(b\) or \(c\) is a unit. A nonunit \(a \in R\) is said to be prime if whenever \(a\) divides \(bc\), then \(a\) divides \(b\) or \(a\) divides \(c\). Equivalently, \(a\) is prime if the principal ideal \(aR\) is a prime ideal. If \(a\) is prime, then we show that \(a\) is irreducible. If \(a = bc\), then \(a\) divides \(bc\); hence, \(a\) divides \(b\) or \(c\). If \(a\) divides \(b\), then \(b = ad\) for some \(d\). Consequently, \(1 = dc\), so \(c\) is a unit. On the other hand, if \(a\) divides \(c\), then the same argument shows that \(b\) is a unit. However, irreducible elements need not be prime. Perhaps the easiest example is in the ring \(\mathbb{Z}[\sqrt{-5}]\). In this ring, \(6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\). With some calculation, we can see that \(2\) is irreducible and that \(2\) does not divide either of \(1 \pm \sqrt{-5}\). Since \(2\) divides their product, \(2\) is not prime in \(\mathbb{Z}[\sqrt{-5}]\).
Definition 2.1 An integral domain \(R\) is a unique factorization domain (UFD) if every nonzero nonunit of \(R\) can be factored uniquely into a product of irreducible elements.
Some words about this definition are in order. What does it mean to factor an element uniquely? In \(\mathbb{Z}\), the integer \(6\) factors as \(6 = 2 \cdot 3\) and as \(6 = (-2) \cdot (-3)\). The four elements \(\pm 2\) and \(\pm 3\) are prime according to our definition. This means we have to be more precise in our meaning. If \(a, b \in R\) such that \(a\) divides \(b\) and \(b\) divides \(a\), we say that \(a\) and \(b\) are associates. Equivalently, \(a\) and \(b\) are associates if \(aR = bR\) (see Problem 4). Therefore, two associates differ by multiplication by a unit. In studying divisibility, units are trivial; hence, we would like not to have to worry about them. Therefore, we say that an element \(a\) factors uniquely into a product of irreducible elements if \(a\) is a product of irreducibles, and if \(a = \pi_1^{e_1} \cdots \pi_n^{e_n} = \theta_1^{f_1} \cdots \theta_m^{f_m}\) with each \(\pi_i\) and \(\theta_j\) irreducible, then \(n = m\), and after reordering, \(e_i = f_i\) and \(\pi_i R = \theta_i R\) for each \(i\). Therefore, unique factorization means unique up to multiplication by units.
While irreducible elements may not be prime, they are in a UFD. This fact will be used frequently when dealing with polynomial rings.
Lemma 2.2 Let \(R\) be a UFD. If \(\pi \in R\) is irreducible, then \(\pi\) is prime.
Proof. Suppose that \(\pi\) divides \(ab\). Then \(ab = \pi c\) for some \(c\). If \(c = \theta_1^{f_1} \cdots \theta_m^{f_m}\) is the factorization of \(c\) into irreducibles, then \(\pi c = \pi \theta_1^{f_1} \cdots \theta_m^{f_m}\) is the factorization of \(\pi c = ab\) into irreducibles. However, if we look at the factorization of \(a\) and \(b\), by uniqueness \(\pi\) must occur in one of these factorizations. Therefore, \(\pi\) divides \(a\) or \(\pi\) divides \(b\), so \(\pi\) is prime. □
There are some equivalent definitions of a UFD. Some of these are addressed in the problems at the end of this appendix. One characterization, due to Kaplansky, is presented now, and we will use it to show that the ring of polynomials over a field is a UFD. This is a prime ideal theoretic characterization of a UFD, and is quite useful in proving facts about UFDs. Another characterization is that a ring is a UFD if and only if each nonunit can be factored into a product of primes. We use this characterization in the proof of the following theorem, although we leave its proof to the reader (Problem 9).
Theorem 2.3 (Kaplansky) Let \(R\) be an integral domain. Then \(R\) is a UFD if and only if each nonzero prime ideal of \(R\) contains a nonzero principal prime ideal.
Proof. Suppose that \(R\) is a UFD, and let \(P\) be a nonzero prime ideal of \(R\). If \(a \in P\) with \(a \neq 0\), let \(a = \pi_1 \cdots \pi_n\) be a prime factorization of \(a\). Since \(P\) is a prime ideal and \(a \in P\), one of the \(\pi_i\) must be in \(P\). Therefore, \(P\) contains the principal prime ideal \((\pi_i)\).
Conversely, suppose that every nonzero prime ideal of \(R\) contains a nonzero principal prime ideal. Let
If \(S = R - \{0\}\), then \(R\) is a UFD by Problem 9. If not, there is an \(a \in R - S\) with \(a \neq 0\). Let \(I\) be an ideal of \(R\) containing \(a\) that is maximal under inclusion among the ideals disjoint from \(S\). Such an ideal exists by an easy application of Zorn's lemma. We claim that \(I\) is a prime ideal. Assuming this for the moment, by hypothesis \(I\) contains a prime element \(\pi\). However, \(\pi \in S\), since \(\pi\) is prime. But \(I \cap S = \varnothing\), a contradiction. Therefore, \(S = R - \{0\}\), and so \(R\) is a UFD.
We are then left with showing that \(I\) is prime. First, we note that \(S\) is closed under multiplication. If \(I\) is not prime, then there are \(b, c \in R - I\) with \(bc \in I\). Then \(I + bR\) and \(I + cR\) are larger than \(I\), so, by maximality, both intersect \(S\). Say \(x \in S \cap (I+bR)\) and \(y \in S \cap (I+cR)\). If \(x = u_1 + br_1\) and \(y = u_2 + cr_2\) with \(u_i \in I\) and \(r_i \in R\), then \(xy = u_1(u_2+cr_2) + bcr_1r_2 \in I\), since \(bc \in I\). But \(xy \in S\), since \(S\) is closed under multiplication. This forces \(S \cap I \neq \varnothing\), a contradiction. Therefore, \(I\) is prime. □
We finish this section with a short discussion of greatest common divisors.
Definition 2.4 Let \(R\) be a UFD. If \(a, b \in R\) are nonzero elements, then a greatest common divisor of \(a\) and \(b\) is an element \(d\) such that
- \(d\) divides \(a\) and \(d\) divides \(b\);
- if \(e\) divides \(a\) and \(e\) divides \(b\), then \(e\) divides \(d\).
The gcd of two elements is not unique if it exists. However, by the second condition in the definition, it follows that any two gcds of \(a\) and \(b\) are associates. We often abuse language and call an element \(d\) the gcd of \(a\) and \(b\), and write \(d = \gcd(a,b)\).
The definition of gcd makes perfect sense in any commutative ring. The difficulty is that a gcd of two elements need not exist, as shown in Problem 11. However, if \(R\) is a UFD, then we can see that a gcd always exists. In fact, if \(b = \pi_1^{e_1} \cdots \pi_n^{e_n}\) and \(a = \pi_1^{f_1} \cdots \pi_n^{f_n}\) is the factorization of \(a\) and \(b\) into irreducibles, where \(e_i, f_i\) can be \(0\), and if \(g_i = \min\{e_i, f_i\}\), then a gcd of \(a\) and \(b\) is \(\pi_1^{g_1} \cdots \pi_n^{g_n}\). If \(1\) is a gcd of \(a\) and \(b\), we say that \(a\) and \(b\) are relatively prime. Unlike in the integers, in a UFD a gcd of two elements need not be a linear combination of the elements. An example of this appears in Problem 15. However, if \(R\) is a PID and \(d = \gcd(a,b)\), then \(dR = aR + bR\), so \(d\) is a linear combination of \(a\) and \(b\) (Problem 17).
The definition of gcd can be extended to any finite set of elements instead of just two elements. An element \(d \in R\) is a gcd of \(a_1, \ldots, a_n\) if \(d\) divides each \(a_i\), and any \(e\) that divides all \(a_i\) also divides \(d\). In a UFD, the gcd of any finite set of elements does exist. Moreover, a gcd can be calculated by recursion from the equation \(\gcd(a_1, \ldots, a_n) = \gcd(a_1, \gcd(a_2, \ldots, a_n))\) (see Problem 16).
3 Polynomials over a Field
Let \(R\) be a ring. We denote by \(R[x]\) the ring of polynomials in one variable over \(R\). Given a polynomial \(f(x) = \sum_{i=0}^n r_i x^i\) with \(r_n \neq 0\), the degree of \(f\) is defined to be \(n\). For convenience, we define the degree of the zero polynomial to be \(-\infty\). While our primary interest is in polynomials over a field, we state a number of results for polynomials over an integral domain.
Lemma 3.1 Let \(R\) be an integral domain. If \(f(x), g(x) \in R[x]\) are nonzero, then \(\deg(fg) = \deg(f) + \deg(g)\). Consequently, \(R[x]\) is an integral domain.
Proof. Let \(f(x) = \sum_{i=0}^n a_i x^i\) and \(g(x) = \sum_{i=0}^m b_i x^i\) with \(\deg(f) = n\) and \(\deg(g) = m\). Therefore, \(a_n, b_m \neq 0\). The product of \(f\) and \(g\) is \(f(x)g(x) = \sum_{i=0}^{n+m} (\sum_{j+k=i} a_j b_k) x^i\). Clearly, all coefficients past degree \(n+m\) are \(0\). The coefficient of \(x^{n+m}\) is \(a_n b_m\), which is nonzero, since both \(a_n\) and \(b_m\) are nonzero. This proves the degree formula. Moreover, it shows that \(fg\) cannot be the zero polynomial unless \(f = 0\) or \(g = 0\); hence, \(R[x]\) is an integral domain. □
If either \(f = 0\) or \(g = 0\), then the degree formula still holds with the convention that \(\deg(0) = -\infty\), given that addition is defined by \(n + (-\infty) = -\infty\) for all integers \(n\).
The main theorem for polynomials over a field is the division algorithm. Since this result holds in more generality and is useful in its full version, we give the full version here.
Theorem 3.2 (Division Algorithm) Let \(R\) be an integral domain. Let \(f(x), g(x) \in R[x]\) with \(g(x) \neq 0\), and suppose that the leading coefficient of \(g\) is a unit in \(R\). Then there are unique polynomials \(q(x), r(x) \in R[x]\) satisfying \(f(x) = q(x)g(x) + r(x)\) and \(\deg(r(x)) < \deg(g(x))\).
Proof. This argument is almost the same as the proof of the division algorithm for polynomials with real number coefficients. We first show the existence of \(q\) and \(r\) with the desired properties, then we prove the uniqueness. Let
The set \(S\) is clearly nonempty. Let \(r(x) \in S\) be a polynomial of minimal degree in \(S\). Then \(f = qg + r\). If \(\deg(r) \geq \deg(g)\), then say \(r(x) = \sum_{i=0}^n a_i x^i\) and \(g(x) = \sum_{i=0}^m b_i x^i\) with \(a_n, b_m \neq 0\) and \(n \geq m\). If \(q_1(x) = q(x) - a_n b_m^{-1} x^{n-m}\), which makes sense since \(b_m\) is assumed to be a unit in \(R\), then
which has degree less than \(n\), since the coefficient of \(x^n\) is \(0\). Consequently, this polynomial is in \(S\) and has smaller degree than \(r(x)\). This is a contradiction, which forces \(n < m\).
For uniqueness, suppose that there are \(q(x), q_1(x)\) and \(r(x), r_1(x) \in R[x]\) with \(f = qg + r\) and \(f = q_1 g + r_1\), and with \(\deg(r), \deg(r_1) < \deg(g)\). Then \(g(q_1 - q) = r - r_1\). If \(q_1 \neq q\), then the degree of \(g(q_1 - q)\) is at least \(\deg(g)\), which is larger than \(\deg(r - r_1)\). This contradiction shows that \(q_1 = q\), which forces \(r = r_1\). This proves the uniqueness. □
We state the usual division algorithm separately for emphasis.
Corollary 3.3 If \(F\) is a field and if \(f(x), g(x) \in F[x]\) with \(g(x) \neq 0\), then there are unique polynomials \(q(x), r(x) \in F[x]\) satisfying \(f(x) = q(x)g(x) + r(x)\) and \(\deg(r(x)) < \deg(g(x))\).
The division algorithm yields the fact that \(F[x]\) is a PID. From this, we will see that \(F[x]\) is a UFD.
Corollary 3.4 If \(F\) is a field, then \(F[x]\) is a PID.
Proof. Let \(I\) be an ideal of \(F[x]\). If \(I = \{0\}\), then \(I\) is generated by \(0\). If \(I \neq 0\), take \(g(x) \in I - \{0\}\) of minimal degree. If \(f(x) \in I\), by the division algorithm there are polynomials \(q\) and \(r\) with \(f = qg + r\) and \(\deg(r) < \deg(g)\). Since \(I\) is an ideal, \(r = f - qg \in I\). Minimality of \(\deg(g)\) forces \(r = 0\), which shows that \(f\) is in the ideal generated by \(g\). Therefore, \(I = (g)\) is principal, so \(F[x]\) is a PID. □
We can now use Kaplansky's theorem to give an easy proof that \(F[x]\) is a UFD.
Lemma 3.5 If \(R\) is a PID, then \(R\) is a UFD. In particular, if \(F\) is a field, then \(F[x]\) is a PID.
Proof. Suppose that \(R\) is a PID. If \(P\) is a prime ideal of \(R\), then \(P\) is principal, say \(P = (\pi)\). Therefore, \(P\) contains the principal prime ideal \((\pi)\). By Theorem 2.3, \(R\) is a UFD. □
The following fact will be used early in Chapter I.
Corollary 3.6 Let \(F\) be a field. If \(p(x) \in F[x]\), then the principal ideal \((p(x))\) is a maximal ideal of \(F[x]\) if and only if \(p(x)\) is irreducible. Consequently, any prime ideal of \(F[x]\) is maximal.
Proof. This really is a fact about PIDs, as the proof will show. Suppose that \(p(x)\) is irreducible. Let \(M\) be a maximal ideal of \(F[x]\) containing \(p(x)\). Since \(F[x]\) is a PID, \(M = (f(x))\) for some polynomial \(f(x)\). Then \(f\) divides \(p\) since \((p) \subseteq (f)\). But \(p\) is irreducible, so \(p\) has no divisors other than units or associates. Since \((f) = M \neq F[x]\), we see that \(f\) is not a unit; hence, \(f\) is an associate to \(p\). Therefore, \((f) = (p)\), so \((p)\) is maximal. Conversely, suppose that \((p)\) is maximal. If \(p\) is not irreducible, then \(p = fg\) with both \(f\) and \(g\) nonconstant polynomials. Then \((p) \subset (f) \subset R\). This contradicts maximality, so \(p\) is irreducible. If \(M\) is a prime ideal, then \(M = (p)\) for some irreducible polynomial by arguments similar to those just given; hence, \(M\) is maximal. □
4 Factorization in Polynomial Rings
The goal of this section is to show that \(R[x]\) is a UFD whenever \(R\) is a UFD. However, we have some work to do in order to prove this.
Definition 4.1 Let \(R\) be a UFD, and let \(f(x) \in R[x]\). The content \(c(f)\) of \(f\) is the gcd of the coefficients of \(f\). If the content of \(f\) is \(1\), then \(f\) is said to be primitive.
The following lemma is easy to prove, but we will use it in a number of places in this book. The proof follows immediately from the definition of addition and multiplication in polynomial rings and quotient rings; this will be left to the reader.
Lemma 4.2 Let \(R\) be a ring, and let \(I\) be an ideal of \(R\). Then the map \(\varphi : R[x] \to R/I[x]\) given by \(\varphi(\sum_i a_i x^i) = \sum_i (a_i + I) x^i\) is a surjective ring homomorphism.
In particular, if \(p\) is a prime number and \(\bar{a}\) represents the equivalence class of \(a\) modulo \(p\), then the map \(\mathbb{Z}[x] \to \mathbb{F}_p[x]\) given by \(\sum_i a_i x^i \mapsto \sum_i \bar{a}_i x^i\) is a ring homomorphism.
Proposition 4.3 Let \(R\) be a UFD, and let \(f, g \in R[x]\). Then \(c(fg) = c(f)c(g)\). In particular, if \(f\) and \(g\) are primitive, then \(fg\) is primitive.
Proof. We may write \(f(x) = c(f) f_1(x)\) and \(g(x) = c(g) g_1(x)\) for some primitive polynomials \(f_1\) and \(g_1\). So, \(fg = c(f)c(g) \cdot f_1 g_1\). If we prove that the product of primitive polynomials is primitive, then we will have proved the proposition. So, suppose that \(f\) and \(g\) are primitive. If \(fg\) is not primitive, then there is a prime element \(\pi\) that divides all of the coefficients of \(fg\). Consider the polynomial ring \(R/(\pi)[x]\) over \(R/(\pi)\). Since \(\pi\) is a prime element, \(R/(\pi)\) is an integral domain. Let \(\bar{f}\) and \(\bar{g}\) be the images of \(f\) and \(g\) in \(R/(\pi)[x]\). Since \(f\) and \(g\) are primitive, \(\pi\) does not divide all the coefficients of \(f\) or \(g\), so \(\bar{f} \neq 0\) and \(\bar{g} \neq 0\). Therefore, \(\bar{f} \cdot \bar{g} = \overline{fg}\) by Lemma 4.2, and \(\bar{f} \cdot \bar{g} \neq 0\), since \(R/(\pi)[x]\) is an integral domain. However, if \(\pi\) divides all the coefficients of \(fg\), then \(\overline{fg} = 0\), a contradiction. Therefore, \(fg\) is indeed primitive. □
The following theorem is perhaps the most important result about polynomials over a UFD.
Theorem 4.4 (Gauss' Lemma) Let \(R\) be a UFD, let \(F\) be its quotient field, and let \(f(x) \in R[x]\). Then \(f\) is irreducible over \(R\) if and only if \(f\) is primitive and irreducible over \(F\).
Proof. Suppose that \(f\) is primitive and irreducible over \(F\). If \(f\) factors in \(R[x]\) as \(f = gh\) with neither \(g\) nor \(h\) a unit in \(R[x]\), then since \(f\) is irreducible over \(F\), either \(g\) or \(h\) must be a constant. But if \(g\) is a constant, \(g\) would divide all the coefficients of \(gh = f\). This is impossible, since \(f\) is primitive, so \(f\) is irreducible over \(R\).
Conversely, suppose that \(f\) is irreducible over \(R\). Since we can write \(f = c(f) \cdot f_1\) with \(f_1\) primitive, \(c(f)\) is a unit in \(R[x]\); hence, \(c(f)\) is a unit in \(R\). Therefore, \(f\) is primitive. If \(f\) is not irreducible over \(F\), then we can write \(f = gh\) with \(g, h \in F[x]\) both of degree at least \(1\). By using common denominators, we can write \(gh = a/b \cdot g_1 h_1\), where \(g_1, h_1 \in R[x]\) are primitive and \(a, b \in R\) are relatively prime. Then \(b f(x) = a g_1(x) h_1(x)\). By Proposition 4.3, we have \(b = c(bf) = a\), since \(g_1\) and \(h_1\) are primitive. But this contradicts \(\gcd(a,b) = 1\) unless \(a\) and \(b\) are both units in \(R\). If \(a\) and \(b\) are units in \(R\), then \(f = (a g_1)(1/b \cdot h_1)\) is a nontrivial factorization of \(f\) in \(R[x]\), which contradicts the assumption that \(f\) is irreducible over \(R\). Therefore, \(f\) is indeed irreducible over \(F\). □
We can now prove that \(R[x]\) is a UFD if \(R\) is.
Theorem 4.5 If \(R\) is a UFD, then \(R[x]\) is a UFD.
Proof. We give here a somewhat nonstandard proof of this theorem, making use of Theorem 2.3. This proof is easier to understand if the reader has some experience in localization. Some of the details in this proof are left to Problem 18. A more standard proof of this fact can be found in Hungerford [13, Thm. 3.11.1] or Herstein [12, Thm. 3.11.1].
Let \(Q\) be a nonzero prime ideal of \(R[x]\). By Theorem 2.3, we wish to show that \(Q\) contains a nonzero prime element. If \(P = Q \cap R\), then \(P\) is a prime ideal in \(R\). If \(P \neq 0\), then \(P\) contains a nonzero prime element \(\pi\) of \(R\), which is also a prime element of \(R[x]\). Therefore, \(Q\) contains a nonzero prime element of \(R[x]\). The more difficult case is if \(P = 0\), which we now consider. Let \(F\) be the quotient field of \(R\). Then \(F[x]\) is a UFD, as we have already seen. Let \(Q' = Q F[x]\), the ideal of \(F[x]\) generated by \(Q\). Then \(Q'\) is a prime ideal, since \(P = 0\) (see Problem 18). Because \(F[x]\) is a UFD, there is a polynomial \(f(x) \in Q'\) that is irreducible in \(F[x]\). We can write \(f(x) = \frac{a}{b} g(x)\) with \(g(x) \in R[x]\) a primitive polynomial and \(a, b \in R\). Since \(a/b\) is a unit in \(F\), we have \(g(x) \in Q'\). Furthermore, \(g(x)\) is irreducible in \(F[x]\), since \(g(x)\) is an associate to \(f(x)\). Therefore, \(g(x)\) is irreducible over \(R\) since \(g(x)\) is primitive, by Gauss' lemma. But \(g(x) \in Q' \cap R[x]\), so \(g(x) \in Q\) (see Problem 18 again). Finally, we need to show that \(g(x)\) is prime in \(R[x]\), which will finish the proof. We see that \(g(x) F[x] \cap R[x] = g(x) R[x]\), since \(g(x) \in R[x]\) (see Problem 18). However, the ideal \(g(x)F[x]\) is prime, since \(F[x]\) is a UFD and \(g(x)\) is irreducible. Thus, \(g(x) \in R[x]\) is prime, since the intersection of a prime ideal of \(F[x]\) with \(R[x]\) is a prime ideal of \(R[x]\). □
Corollary 4.6 If \(F\) is a field, then \(F[x_1, \ldots, x_n]\) is a UFD for any \(n\).
Proof. This follows by induction on \(n\) and the previous theorem, the \(n = 1\) case having been proven earlier. □
More generally, if \(F\) is a field and \(X\) is any set of variables, possibly infinite, then the polynomial ring \(F[X]\) is a UFD. A proof of this fact can be obtained from the following two points. First, any element of \(F[X]\) is a polynomial in finitely many of the variables, and unique factorization holds for polynomials in finitely many variables, and second, adding more variables does not affect whether an element is irreducible.
5 Irreducibility Tests
It is hard in general to determine if a polynomial is irreducible over a field \(F\). However, if \(F\) is the quotient field of a UFD, there are some simple tests that can determine when a polynomial is irreducible over \(F\). While these tests may seem somewhat specialized, nonetheless they can be quite useful.
The first test is actually a test for roots, but it is also an irreducibility test for polynomials of degree 2 and 3. Let \(R\) be a UFD, and let \(F\) be its quotient field. Suppose that \(f(x) = a_0 + \cdots + a_n x^n \in R[x]\). If \(\alpha/\beta \in F\) is a root of \(f(x)\) with \(\gcd(\alpha,\beta) = 1\), then by multiplying the equation \(f(\alpha/\beta) = 0\) by \(\beta^n\), we obtain the equation
Therefore, \(a_0 \beta^n = -\alpha (a_1 \beta^{n-1} + \cdots + a_{n-1} \alpha^{n-2} \beta + a_n \alpha^{n-1})\). Since \(\alpha\) is relatively prime to \(\beta\), it follows that \(\alpha\) divides \(a_0\). By a similar manipulation, we see that \(\beta\) divides \(a_n\). If \(f\) has degree 2 or 3, then \(f\) has a linear factor if and only if it is reducible over \(F\). We record these observations as the first irreducibility test.
Proposition 5.1 (Rational Root Test) Let \(R\) be a UFD with quotient field \(F\). Suppose that \(f(x) = a_0 + \cdots + a_n x^n \in R[x]\) has a root \(\alpha/\beta \in F\) with \(\gcd(\alpha,\beta) = 1\). Then \(\alpha\) divides \(a_0\), and \(\beta\) divides \(a_n\). If \(\deg(f) \leq 3\), then \(f\) is irreducible over \(F\) if and only if \(f\) has no roots in \(F\).
Example 5.2 The polynomial \(x^2 - p\) is irreducible over \(\mathbb{Q}\) if \(p\) is a prime, as is \(x^3 + 3x + 1\), by the rational root test. The cubic \(x^3 + 2x^2 - 4x + 1\) factors as \((x - 1)(x^2 + 3x - 1)\). The first factor could have been easily found by the rational root test; since \(x^3 + 2x^2 - 4x + 1\) is monic, any rational root of it is in \(\mathbb{Z}\) and must divide \(1\), so \(\pm 1\) are the only possibilities. The fourth degree polynomial \(x^4 - 4\) factors as \((x^2 - 2)(x^2 + 2)\), but it has no rational roots. Thus for polynomials of degree 4 or larger, the existence of a rational root is not necessary for a polynomial to factor over \(\mathbb{Q}\).
The next irreducibility test is the one we use the most in this book.
Proposition 5.3 (Eisenstein Criterion) Let \(R\) be a UFD with quotient field \(F\), and let \(f(x) = a_0 + \cdots + a_n x^n \in R[x]\).
- Suppose there is a prime element \(\pi \in R\) such that \(\pi\) divides \(a_0, \ldots, a_{n-1}\) but not \(a_n\) and that \(\pi^2\) does not divide \(a_0\). Then \(f\) is irreducible over \(F\).
- Suppose there is a prime element \(\pi \in R\) such that \(\pi\) divides \(a_1, \ldots, a_n\) but not \(a_0\) and that \(\pi^2\) does not divide \(a_n\). Then \(f\) is irreducible over \(F\).
Proof. We prove statement 1; the proof of statement 2 is similar. By factoring out the content of \(f\), we may suppose that \(f\) is primitive. Consider the ring \(R/(\pi)[x]\), and let \(\bar{h}\) denote the image of \(h \in R[x]\) obtained by reducing coefficients modulo \(\pi\). The condition on the coefficients of \(f\) shows that \(\bar{f} = \overline{a_n} x^n \neq 0\). Suppose that \(f\) factors over \(F\). Then by Gauss' lemma, \(f\) also factors over \(R\). Say \(f = gh\) with \(g, h \in R[x]\) both nonconstant polynomials. Then \(\bar{f} = \bar{g} \cdot \bar{h}\) in \(R/(\pi)[x]\) by Lemma 4.2. Since \(\bar{f} = \overline{a_n} x^n\), then \(\bar{g}\) and \(\bar{h}\) each must be of the form \(c x^k\), since the only monic irreducible factors of \(x^n\) are powers of \(x\). If \(\deg(\bar{g})\) and \(\deg(\bar{h})\) are both positive, then \(\pi\) divides the constant term of both \(g\) and \(h\). But \(a_0\) is the product of these constant terms, which would force \(\pi^2\) to divide \(a_0\), which is false. Therefore, either \(\bar{g}\) or \(\bar{h}\) is a constant. Since \(g\) and \(h\) are nonconstant, \(\pi\) divides the leading coefficient of \(g\) or \(h\). But \(a_n\) is the product of these leading coefficients, so \(\pi\) divides \(a_n\), which is false. The only possibility left is that \(f\) is irreducible over \(F\), which proves the criterion. □
Example 5.4 The polynomial \(x^5 - 12x^3 + 2x + 2\) is irreducible over \(\mathbb{Q}\) by an application of the Eisenstein criterion with \(\pi = 2\). Similarly, with \(\pi = 3\), the polynomial \(x^3 - 3x + 3\) is irreducible. The Eisenstein criterion does not tell anything about \(x^4 + 2x + 4\), since there is no prime that satisfies all the needed conditions.
Let \(p\) be a prime. The polynomial \(x^p - 1\) factors as
The Eisenstein criterion would appear to be useless to determine whether \(x^{p-1} + x^{p-2} + \cdots + x + 1\) is irreducible. However, this is not the case. By a change of variables, we can determine that this polynomial is irreducible over \(\mathbb{Q}\). Before doing so, however, we formalize the idea in the following lemma. The proof is straightforward and is left for the reader (Problem 12).
Lemma 5.5 Let \(f(x) \in R[x]\) and \(a \in R\). Then the map \(f(x) \mapsto f(x+a)\) is a ring isomorphism. Therefore, if \(f(x+a)\) is irreducible, then \(f(x)\) is irreducible.
Example 5.6 Let \(f(x) = x^p - 1\) with \(p\) a prime. Then
Since \(x^p - 1\) factors as \((x-1)(x^{p-1} + \cdots + x + 1)\), replacing \(x\) by \(x+1\) yields \(f(x+1) = x g(x)\) with \(g(x)\) the image of \(x^{p-1} + \cdots + x + 1\) after substituting \(x+1\) for \(x\). Therefore, \(x^{p-1} + \cdots + x + 1\) is irreducible if \(g(x)\) is irreducible. However, the coefficients of \(g(x) = x^{p-1} + p x^{p-2} + \cdots + p\) are all binomial coefficients of the form \(\binom{p}{i}\), which are all divisible by \(p\) (see Problem 13), except for the leading coefficient \(\binom{p}{p} = 1\). Since the constant term is \(p\), Eisenstein's criterion with \(\pi = p\) shows that \(g(x)\) is irreducible; hence, \(x^{p-1} + \cdots + x + 1\) is irreducible.
The following result is our last irreducibility test.
Proposition 5.7 Let \(R\) be a UFD with quotient field \(F\), and let \(f(x) \in R[x]\) be monic. If \(\pi \in R\) is a prime element and if \(\bar{f} \in R/(\pi)[x]\) is irreducible, then \(f\) is irreducible over \(F\).
Proof. The polynomial \(f\) is primitive since it is monic. If \(f\) factors over \(F\), then \(f\) factors over \(R\) by Gauss' lemma. If \(f = gh\) with \(g, h \in R[x]\) nonconstant polynomials, then \(\bar{f} = \bar{g} \cdot \bar{h}\) in \(R/(\pi)[x]\) by Lemma 4.2. If \(\bar{f}\) is irreducible, this means \(\bar{g}\) or \(\bar{h}\) is a unit in \(R/(\pi)\). Since \(g\) and \(h\) are nonconstant, this would force \(\pi\) to divide the leading coefficient of either \(g\) or \(h\), which cannot happen since \(f\) is monic. Therefore, \(f\) is irreducible over \(R\), so \(f\) is also irreducible over \(F\). □
The converse of this proposition is false, since \(x^2 + x + 1\) is irreducible over \(\mathbb{Z}\), but \(x^2 + x + 1 = (x+2)(x+2)\) over \(\mathbb{F}_3\). Also, if \(f\) is not monic, then the result is false, since \(2x^2 + 3x + 1 = (2x+1)(x+1)\) factors over \(\mathbb{Z}\), but its image in \(\mathbb{F}_2[x]\) is \(x+1\), which is irreducible.
Example 5.8 Over \(\mathbb{F}_2\), the polynomials \(1 + x^3 + x^4\) and \(1 + x^3 + x^6\) can be seen to be irreducible by trial and error. Therefore, \(1 + x^3 + x^4\) and \(1 + x^3 + x^6\) are irreducible over \(\mathbb{Q}\), as are \(3 + 5x^3 + 7x^4\) and \(-1 + 11x^3 + x^6\).
Example 5.9 Let \(p\) be a prime, and consider \(x^p - x - 1\). This polynomial has no roots in \(\mathbb{F}_p\), since every element of \(\mathbb{F}_p\) is a root of \(x^p - x\). While a polynomial can have no roots but be reducible, in this case this does not happen. Problem 3 of Section 10 shows that a prime degree polynomial that has no roots in a field \(F\) is irreducible over \(F\) under the following hypothesis: For any field \(K\) containing \(F\), if the polynomial has a root in \(K\), then it factors into linear factors in \(K\). Using the result of this exercise, we show that the hypothesis holds for \(x^p - x - 1\), which then implies that it is irreducible over \(\mathbb{F}_p\), and so \(x^p - x - 1\) is irreducible over \(\mathbb{Q}\).
Suppose that \(K\) is a field containing \(\mathbb{F}_p\) for which \(x^p - x - 1\) has a root \(a\). So \(a^p - a = 1\). We claim that \(a+1, a+2, \ldots, a+p-1\) are also roots of \(x^p - x - 1\) in \(K\). To see this, if \(1 \leq i \leq p-1\), then \((a+i)^p - (a+i) - 1 = a^p + i^p - a - i - 1 = a^p - a - 1 = 0\), since \(i^p = i \pmod{p}\) by Fermat's little theorem. Therefore, we have \(p\) roots of \(x^p - x - 1\) in \(K\), so \(x^p - x - 1\) factors into linear factors in \(K\). Therefore, Problem 3 of Section 10 shows that \(x^p - x - 1\) is irreducible over \(\mathbb{F}_p\), since it has no root in \(\mathbb{F}_p\).