Lecture Notes on Polynomial Rings

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore to determine the prime elements, it suffices to determine the irreducible elements. We start with some basic facts about polynomial rings. Lemma 21.1. Let R be an integral domain. Then the units in R[x] are precisely the units in R. Proof. One direction is clear. A unit in R is a unit in R[x]. Now suppose that f (x) is a unit in R[x]. Given a polynomial g, denote by d(g) the degree of g(x) (note that we are not claiming that R[x] is a Euclidean domain). Now f (x)g(x) = 1. Thus 0 = d(1) = d(f g) ≥ d(f ) + d(g). Thus both of f and g must have degree zero. It follows that f (x) = f0 and that f0 is a unit in R[x]. D Lemma 21.2. Let R be a ring. The natural inclusion R −→ R[x] which just sends an element r ∈ R to the constant polynomial r, is a ring homomorphism. Proof. Easy. D The following universal property of polynomial rings, is very useful. Lemma 21.3. Let φ : R −→ S be any ring homomorphism and let s ∈ S be any element of S. Then there is a unique ring homomorphism ψ : R[x] −→ S, such that φ(x) = s and which makes the following diagram commute φ R S / / f ψ R[x] 1 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Proof. Note that any ring homomorphism ψ : R[x] −→ S that sends x to s and acts as φ on the coefficients, must send an xn + an−1 xn−1 + · · · + a0 to φ(an )sn + φ(an−1 )sn−1 + · · · + φ(a0 ). Thus it suffices to check that the given map is a ring homomorphism, which is left as an exercise to the reader. D Definition 21.4. Let R be a ring and let α be an element of R. The natural ring homomorphism φ : R[x] −→ R, which acts as the identity on R and which sends x to α, is called eval­ uation at α and is often denoted evα . We say that α is a zero (aka root) of f (x), if f (x) is in the kernel of evα . Lemma 21.5. Let K be a field and let α be an element of K. Then the kernel of evα is the ideal (x − α). Proof. Denote by I the kernel of evα . Clearly x−α is in I. On the other hand, K[x] is a Euclidean domain, and so it is certainly a PID. Thus I is principal. Suppose it is generated by f , so that I = (f ). Then f divides x − α. If f has degree one, then x − α must be an associate of f and the result follows. If f has degree zero, then it must be a constant. As f has a root at α, in fact this constant must be zero, a contradiction. D Lemma 21.6. Let K be a field and let f (x) be a polynomial in K[x]. Then we can write f (x) = g(x)h(x) where g(x) is a linear polynomial if and only if f (x) has a root in K. Proof. First note that a linear polynomial always has a root in K. Indeed any linear polynomial is of the form ax + b, where a = 0. Then it is easy to see that α = − ab is a root of ax + b. On the other hand, the kernel of the evaluation map is an ideal, so that if g(x) has a root α, then in fact so does f (x) = g(x)h(x). Thus if we can write f (x) = g(x)h(x), where g(x) is linear, then it follows that f (x) must have a root. Now suppose that f (x) has a root at α. Consider the linear polyno­ mial g(x) = x − α. Then the kernel of evα is equal to (x − α). As f is in the kernel, f (x) = g(x)h(x), for some h(x) ∈ R[x]. D 2 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Lemma 21.7. Let K be a field and let f (x) be a polynomial of degree two or three. Then f (x) is irreducible if and only if it has no roots in K. Proof. If f (x) has a root in K, then f (x) = g(x)h(x), where g(x) has degree one, by (21.6). As the degree of f is at least two, it follows that h(x) has degree at least one. Thus f (x) is not irreducible. Now suppose that f (x) is not irreducible. Then f (x) = g(x)h(x), where neither g nor h is a unit. Thus both g and h have degree at least one. As the sum of the degrees of g and h is at most three, the degree of f , it follows that one of g and h has degree one. Now apply (21.6). D Definition 21.8. Let p be a prime. Fp denotes the unique field with p elements. Of course, Fp is isomorphic to Zp . However, as we will see later, it is useful to replace Z by F . Example 21.9. First consider the polynomial x2 + 1. Over the real numbers this is irreducible. Indeed, if we replace x by any real number a, then a2 is non-negative and so a2 + 1 cannot equal zero. On the other hand ±i is a root of x2 + 1, as i2 + 1 = 0. Thus x2 + 1 is reducible over the complex numbers. Indeed x2 +1 = (x+i)(x−i). Thus an irreducible polynomial might well become reducible over a larger field. Consider the polynomial x2 + x + 1. We consider this over various fields. As observed in (21.7) this is reducible iff it has a root in the given field. Suppose we work over the field F5 . We need to check if the five elements of F5 are roots or not. We have 12 + 1 + 1 = 3 22 + 2 + 1 = 2 32 + 3 + 13 42 + 4 = 1 Thus x2 + x + 1 is irreducible over F5 . Now consider what happens over the field with three elements F3 . Then 1 is a root of this polynomial. As neither 0 nor 2 are roots, we must have x2 + x + 1 = (x − 1)2 = (x + 2)2 , which is easy to check. Now let us determine all irreducible polynomials of degree at most four over F2 . Any linear polynomial is irreducible. There are two such x and x + 1. A general quadratic has the form f (x) = x2 + ax + b. 6 0, else x divides f (x). Thus b = 1. If a = 0, then f (x) = x2 + 1, b= which has 1 as a zero. Thus f (x) = x2 + x + 1 is the only irreducible quadratic. 3 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Now suppose that we have an irreducible cubic f (x) = x3 +ax+bx+1. This is irreducible iff f (1) 6= 0, which is the same as to say that there are an odd number of terms. Thus the irreducible cubics are f (x) = x3 + x2 + 1 and x3 + x + 1. Finally suppose that f (x) is a quartic polynomial. The general irre­ ducible is of the form x4 + ax3 + bx2 + cx + 1. f (1) 6= 0 is the same as to say that either two of a, b and c is equal to zero or they are all equal to one. Suppose that f (x) = g(x)h(x). If f (x) does not have a root, then both g and h must have degree two. If either g or h were reducible, then again f would have a linear factor, and therefore a root. Thus the only possibilty is that both g and h are the unique irreducible quadratic polynomials. In this case f (x) = (x2 + x + 1)2 = x4 + x2 + 1. Thus x4 + x3 + x2 + x + 1, x4 + x3 + 1, and x4 + x + 1 are the three irreducible quartics. Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beau­ tiful and due to Gauss. The basic idea is a follows. Suppose we are given a polynomial with integer coefficients. Then it is natural to also consider this polynomial over the rationals. Note that it is much easier to prove that this polynomial is irreducible over the integers than it is to prove that it is irreducible over the rationals. For example it is clear that x2 − 2 is irreducible over √the integers. In fact it is irreducible over the rationals as well, that is, 2 is not a rational number. First some definitions. Definition 21.10. Let R be a commutative ring and let a1 , a2 , . . . , ak be a sequence of elements of R. The gcd of a1 , a2 , . . . , ak is an element d ∈ R such that (1) d|ai , for all 1 ≤ i ≤ k. (2) If d' |ai , for all 1 ≤ i ≤ k, then d' |d. Lemma 21.11. Let R be a UFD. Then the gcd of any sequence a1 , a2 , . . . , ak of non-zero elements of R exists. 4 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Proof. There are two obvious ways to proceed. The first is to take a common factorisation of each ai into a product of powers of primes, as in the case k = 2. The second is to recursively construct the gcd, by setting di to be the gcd of di−1 and ai and taking d1 = a1 . In this case d = dk will be D a gcd for the whole sequence a1 , a2 , . . . , ak . Definition 21.12. Let R be a UFD and let f (x) be a polynomial with coefficients in R. The content of f (x), denoted c(f ), is the gcd of the coefficients of f. Example 21.13. Let f (x) = 24x3 − 60x + 40. Then the content of f is 4. Thus f (x) = 4(8x3 − 15x + 10). Lemma 21.14. Let R be a UFD. Then every element of R[x] has a factorisation of the form cf, where c ∈ R and the content of f is one. Proof. Obvious. D Here is the key result. Proposition 21.15. Let R be a UFD. Suppose that g and h ∈ R[x] and let f (x) = g(x)h(x). Then the content of f is equal to the content of g times the content of h. Proof. It is clear that the content of g divides the content of f . There­ fore we may assume that the content of g and h is one, and we only have to prove that the same is true for f . Suppose not. As R is a UFD, it follows that there is a prime p that divides the content of f . We may write g(x) = an xn +an−1 xn−1 +· · ·+a0 and h(x) = bn xn +bn−1 xn−1 +· · ·+b0 . As the content of g is one, at least one coefficient of g is not divisible by p. Let i be the first such, so that p divides ak , for k < i whilst p does not divide ai . Similarly pick j so that p divides bk , for k < j, whilst p does not divide bj . Consider the coefficient of xi+j in f . This is equal to a0 bi+j + a1 bi+j−1 + · · · + ai−1 bj+1 + ai bj + ai+1 bj+1 + · · · + ai+j b0 . Note that p divides every term of this sum, except the middle one ai bj . Thus p does not divide the coefficient of xi+j . But this contradicts the definition of the content. D 5 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Theorem 21.16. (Gauss’ Lemma) Let R be a UFD and let f (x) ∈ R[x]. Let F be the field of fractions of R. Suppose that the content of f is one and that we may write f (x) = u1 (x)v1 (x), where u1 (x) and v1 (x) are in F [x]. Then we may find u(x) and v(x) in R[x], such that f (x) = u(x)v(x) where the degree of u is the same as the degree of u1 and the degree of v is the same as the degree of v. In particular if f is irreducible in R[x] then it is irreducible in F [x]. Proof. We have f (x) = u1 (x)v1 (x). Now clear denominators. That is, multiply through by the product c of all the denominators in u1 (x) and v1 (x). In this way we get an expression of the form cf (x) = u2 (x)v2 (x), where now u2 and v2 belong to R[x]. Now write u2 (x) = au(x) and v2 (x) = bv(x). We get cf (x) = abu(x)v(x). By (21.15) we can c divides ab, ab = cα, where α ∈ R. Therefore, replacing u(x) by αu(x), we have f (x) = u(x)v(x). D Corollary 21.17. Let R be a UFD. Then R[x] is a UFD. Proof. It is clear that the Factorisation algorithm terminates, by in­ duction on the degree. Therefore it suffices to prove that irreducible implies prime. Suppose that f (x) ∈ R[x] is irreducible. If f has degree zero, then it is an irreducible element of R and hence a prime element of R and there is nothing to prove. Otherwise we may assume that the content of f is one. By Gauss’ Lemma, f is not only irreducible in R[x] but also in F [x]. But then f is a prime element of F [x] as F [x] is a UFD. Now suppose that f divides gh, where g and h ∈ R[x]. As f is prime in F [x], f divides g or h in F [x]. Suppose it divides g. Then we may write g = f k, 6 MIT OCW: 18.703 Modern Algebra Prof. James McKernan some k ∈ F [x]. As in the proof of Gauss’ Lemma, this means we may write g = f k', some k ' ∈ R[x]. But then f (x) divides g in R[x]. D Corollary 21.18. Z[x] is a UFD. Definition 21.19. Let R be a commutative ring and let x1 , x2 , . . . , xn be indeterminates. A monomial in x1 , x2 , . . . , xn is a product of powers of x1 , x2 , . . . , xn . If I = (d1 , d2 , . . . , dn ), then let xdi i . xI = The degreeed of a monomial is the sum of the degrees of the indi­ vidual terms, di . The polynomial ring R[x1 , x2 , . . . , xn ] is equal to the set of all finite formal sums aI x I , I with the obvious addition and multiplication. The degree of a poly­ nomial is the maximum degree of a monomial term that appears with non-zero coefficient. Example 21.20. Let x and y be indeterminates. A typical element of Q[x, y] might be x2 + y 2 − 1. This has degree 2. Note that xy also has degree two. A more compli­ cated example might be 2 3 x − 7xy + y 5 , 3 a polynomial of degree 5. Lemma 21.21. Let R be a commutative ring and let x1 , x2 , . . . , xn be indeterminates. Let S = R[x1 , x2 , . . . , xn−1 ]. Then there is a natural isomorphism R[x1 , x2 , . . . , xn ] � S[xn ]. Proof. Clear. D To illustrate how this proceeds, it will probably help to give an ex­ ample. Consider the polynomial 2 3 x − 7xy + y 5 . 3 7 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Consider this as a polynomial in y, whose coefficients lie in the ring R[x]. That is y 5 + (−7x)y + 2/3x3 ∈ R[x][y]. Corollary 21.22. Let R be a UFD. Then R[x1 , x2 , . . . , xn ] is a UFD. Proof. By induction on n. The case n = 1 is (21.17). Set S = R[x1 , x2 , . . . , xn−1 ]. By induction S is a UFD. But then D S[x] � R[x1 , x2 , . . . , xn ] is a UFD. Now we give a way to prove that polynomials with integer coefficients are irreducible. Lemma 21.23. Let φ : R −→ S be a ring homomorphism. Then there is a unique ring homomorphism ψ : R[x] −→ S[x] which makes the following diagram commute R φ / S ψ R[x] / S[x] and which sends x to x. Proof. Let f : R −→ S[x] be the composition of φ with the natural inclusion of S into S[x]. By the universal property of R[x], there is a unique ring homomorphism ψ : R[x] −→ S[x]. The rest is clear. D Theorem 21.24. (Eisenstein’s Criteria) Let f (x) = an xn + an−1 xn−1 + · · · + a0 be a polynomial with integer coefficients. Suppose that there is a prime p such that p divides ai , i ≤ n − 1, p does not divide an and p2 does not divide a0 . Then f (x) is irreducible in Q[x]. 8 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Proof. There is no harm in assuming that the content of f is one, so that by Gauss’ Lemma, it suffices to prove that f is irreducible over Z. Suppose not. Then we may find two polynomials g(x) and h(x), of positive degree, with integral coefficients, such that f (x) = g(x)h(x). Suppose that f (x) = an xn + an−1 xn−1 + · · · + a0 g(x) = bd xd + bd−1 xd−1 + · · · + b0 h(x) = ce xe + ce−1 xe−1 + · · · + c0 , for some n, d and e > 1. As an = bd ce and an is not divisible by p, then neither is bd nor ce . Consider the natural ring homomorphism Z −→ Fp . This induces a ring homomorphism Z[x] −→ Fp [x]. It is convenient to denote the image of a polynomial g(x) as ḡ(x). As we have a ring homomorphism, f¯(x) = ḡ(x)h̄(x). Since the leading coefficient of f is not divisible by p, f¯(x) has the same degree as f (x), and the same holds for g(x) and h(x). On the other hand, every other coefficient of f (x) is divisible by p, and so f¯(x) = ān xn . Since Fp is a field, Fp is a UFD and so ḡ = ¯bd xd and ḡ(x) = c̄e xe . It follows that every other coefficient of g(x) and h(x) is divisible by p. In particular b0 and c0 are both divisible by p, and so, as a0 = b0 c0 , a0 must be divisible by p2 , a contradiction. D Example 21.25. Let f (x) = 2x7 − 15x6 + 60x5 − 18x4 − 9x3 + 45x2 − 3x + 6. Then f (x) is irreducible over Q. We apply Eisenstein with p = 3. Then the top coefficient is not divisible by 3, the others are, and the smallest coefficient is not divisible by 9 = 32 . Lemma 21.26. Let p be a prime. Then f (x) = xp−1 + xp−2 + · · · + x + 1, is irreducible over Q. 9 MIT OCW: 18.703 Modern Algebra Prof. James McKernan Proof. By Gauss’ Lemma, it suffices to prove that f (x) is irreducible over Z. First note that xp − 1 f (x) = , x−1 as can be easily checked. Consider the change of variable y = x − 1. As this induces an automorphism Z[x] −→ Z[x] by sending x to x − 1, this will not alter whether or not f is irreducible. In this case (y + 1)p − 1 f (y) = y p p−2 p p−3 p = y p−1 + y + y + ... 1 2 p−1 p−1 p−2 =y + py + · · · + p. Note that pi is divisible by p, for all 1 ≤ i < p, so that we can apply Eisenstein to f (y), using the prime p. D 10 MIT OCW: 18.703 Modern Algebra Prof. James McKernan