# Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

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1 Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize to the case where R is a field, denoted F. In this chapter and the next, we will see that much of what works for the ring of integers also works for polynomials over a field including a division algorithm, factorization into irreducible elements, and even a form of modular arithmetic. Theorem 4.1. Let R be a ring. There exists a ring R[x] that contains an element x not in R and has the properties: (1) R is a subring of R[x]. (2) xr = rx for all r R (i.e., x lies in the center of R[x]). (3) Every nonzero element of R[x] can be uniquely written in the form a 0 + a 1 x + + a n x n, for some n 0, a i R, and a n 0. Definitions. The elements of R[x] are called polynomials with coefficients in R. The elements a i are called coefficients and x is called an indeterminate. The element a n is called the leading coefficient of the polynomial and n is called the degree. Elements of R are called constant polynomials, and have degree 0 if they are nonzero. The zero polynomial does not have a defined degree. The choice of the letter x is, of course, irrelevant. The construction of the ring R[x] is done in appendix G, pp. 542+, and has no x. See page 543. Here x = (0, 1, 0, 0, 0,... ). If R has an identity, then it also works as the identity of R[x] and has the formal description (1, 0, 0, 0,... ). More generally, the element r R corresponds to the element (r, 0, 0, 0,... ). From a more advanced viewpoint, there is a way to characterize the polynomial ring in terms of homomorphisms. R[x] is the unique ring containing R with the property that for any ring homomorphism f : R S and any element s in the center of S, there exists a unique homomorphism φ: R[x] S such that φ(r) = f(r) for all r R and φ(x) = s. This is the way to think of polynomials as functions, as it gives a meaning to evaluation of the polynomial at x = s. For example, take R = S = R, s = 2. φ(p(x)) is the element p( 2) R. For the next few sections, we ignore the idea of a polynomial as a function and just think of it as an element of the ring R[x]. Addition and multiplication are determined by the associative and distributive laws, so are exactly as you have always done them (unless you lack commutativity for multiplication, in which case you must be careful). If R is commutative, so is R[x] because x commutes with everything. If R has an identity, it also works as the identity in R[x]. 1

2 2 Theorem 4.2 (extended). Let f(x), g(x) be nonzero polynomials over a ring R. f(x)g(x) 0, then deg[f(x)g(x)] deg f(x) + deg g(x). If R has no zero divisors, then we always have If deg[f(x)g(x)] = deg f(x) + deg g(x). In particular, this applies if R is an integral domain. Proof. Write f(x) = a a n x n and g(x) = b b m x m, with a n, b m 0. Then f(x)g(x) = a 0 b 0 + (a 0 b 1 + a 1 b 0 )x + + a n b m x n+m. Thus the largest power of x that can occur is x n+m, so deg[f(x)g(x)] deg f(x)+deg g(x). If R has no zero divisors, then a n b m 0 and n+m is the degree, giving us deg[f(x)g(x)] = deg f(x) + deg g(x). Corollary 4.3. If R is an integral domain, so is R[x]. Proof. The theorem shows that R[x] has no zero divisors. R provides the identity. The converse of the corollary also holds. If R has a zero divisor r, say with rs = 0, r 0, s 0, then this still holds in R[x]. The corollary can also be iterated using induction to show that R[x, y], meaning (R[x])[y], is an integral domain when R is, and so is the n variable ring R[x 1, x 2,... x n ]. Discuss degrees of sums of polynomials. We have seen a division algorithm in Z and from high school work with polynomials, you expect one for polynomials with coefficients in a field. In Chapter 9 we will generalize both of these to a Euclidean domain, where we add an axiom that says there is a division algorithm (and hence a Euclidean algorithm for gcd s). Theorem 4.4 (Division Algorithm). Let F be a field, f(x), g(x) F [x] with g(x) 0. Then there exist unique polynomials q(x) and r(x) such that f(x) = g(x)q(x) + r(x) and either r(x) = 0 or deg r(x) < deg g(x). When we give an axiom for a division algorithm, we will need a special function to say r is in some sense smaller than g. For the integers we have their natural ordering and

3 for polynomials we have degrees. There is also a version of Theorem 4.4 when F is only a commutative ring with 1, but you end up with a power of the leading coefficient of g multiplying f since you can t divide by it. The book does a very careful discussion showing how the proof is similar to the long division algorithm for polynomials. If you have trouble understanding the proof, read the book s discussion. Proof. We begin with existence of q and r. The trivial cases: if f = 0 or deg f < deg g, then q = 0, r = f works. So we can assume that f 0 and deg f deg g. We use induction on n = deg f(x). The induction hypothesis is that the theorem works for any polynomial of smaller degree (not just degree n 1); this is an equivalent form of mathematical induction. The starting case is n = 0. In this case, f and g are constant polynomials, so lie in F. A solution is q = f/g, r = 0. In general, write f(x) = a n x n + a 1 x + a 0 and g(x) = b m x m + b 1 x + b 0, where n m, a n b m 0. The first term in the quotient using the high school algorithm is a n b m x n m, so we form f 1 (x) = f(x) a n b m x n m g(x) = (a n 1 a n b m b m 1 )x n 1 +, a polynomial of lower degree than f(x) (or possibly 0). Apply the induction hypothesis to f 1 and g, to obtain polynomials q 1 (x) and r(x) satisfying f 1 (x) = g(x)q 1 (x) + r(x), where r(x) = 0 or deg r(x) < deg g(x). Putting these together and solving for f gives [ ] an f(x) = g(x) x n m + q 1 (x) + r(x) b m as desired, the q(x) being a n b m x n m + q 1 (x). To show uniqueness, we assume there are other polynomials q 2 (x), r 2 (x) satisfying the same conditions. Then g(x)q(x) + r(x) = f(x) = g(x)q 2 (x) + r 2 (x), so that g(x)[q(x) q 2 (x)] = r 2 (x) r(x). The left hand side is either zero or has degree at least equal to deg g by Theorem 4.2. The right hand side has degree less than deg g if it is nonzero. Therefore both must be zero, hence q = q 2, r = r 2. Example: page 88, exercise 5(c), 14(b). Definition, p. 90. The definitions for divides and factor are the same as in the integers, and will be repeated later for any ring. Some things for F [x] are slightly different than for the integers. For example, (3x + 6) (x + 2) because x + 2 = 1 3 (3x + 6). In general, multiplication by a nonzero constant does not affect divisibility; this is because they are units in F [x]. This is different than working in Z[x], where there is no element 1 3! 3

4 4 By Theorem 4.2, if f(x) g(x), then deg f(x) deg g(x). To make greatest common divisors unique, we need a new condition (analogous to assuming they were positive in Z): Definition p. 91. Let f(x), g(x) F [x], not both 0. The greatest common divisor of f(x) and g(x) is the monic polynomial of highest degree that divides them both. The book gives no notation; I will continue to use gcd(f(x), g(x)). As before, we say f(x) and g(x) are relatively prime if their gcd is 1. The next theorem is analogous to Theorem 1.3 for the integers. It can be proved as we did there, by first checking that the Euclidean algorithm still works in F [x]. (In using it, the last nonzero remainder has to be multiplied by a constant to make it monic.) For variety, we shall follow the book s proof this time. Theorem 4.5. Let F be a field, f(x), g(x) F [x], not both 0. Then there exists a unique gcd d(x) of f(x) and g(x). Furthermore, there exist polynomials u(x) and v(x) such that d(x) = f(x)u(x) + g(x)v(x). Proof. Let t(x) be a (soon to be shown unique) monic polynomial of lowest degree in the set { f(x)m(x) + g(x)n(x) m(x), n(x) F [x] }. So by definition, t(x) = f(x)u(x) + g(x)v(x) for some polynomials u(x), v(x). Use the division algorithm to find q, r such that f(x) = t(x)q(x) + r(x), with r = 0 or deg r < deg t. Then r(x) = f tq = f (fu + gv)q = f(1 uq) + g( vq) S. By the minimality of the degree of t, we must have r = 0 and so t f. Similarly, t g so is a common divisor. For any other common divisor s(x), write f(x) = s(x)f 1 (x), g(x) = s(x)g 1 (x). Then t = fu + gv = sf 1 u + sg 1 v is divisible by s, so deg s deg t. Therefore, t is a greatest common divisor. Now let d(x) be another one. By the argument for s above, d t. Reversing their roles, we also have that t d. Thus they differ by a factor which is an invertible element of F [x], that is, an element of F. Since they are both monic, they must be equal. (Details are homework, p. 93.) Theorem 4.7. Let F be a field with f, g, h F [x]. If f(x) g(x)h(x) and gcd(f(x), g(x)) = 1, then f(x) h(x). The proof is identical to that for elements of Z.

5 5 Irreducibles and unique factorization. Theorem 4.8 (Units in R[x]). Let R be an integral domain. f(x) R[x] is a unit iff f(x) is a constant polynomial that is a unit in R. Proof. Clearly such functions are units in R[x]. On the other hand, suppose that f(x)g(x) = 1. Then deg f + deg g = deg 1 = 0, so deg f = deg g = 0. Hence they are in R, and units because their product is 1. If R has zero divisors, the theorem may fail. For example, over Z 9 we have (3x+1)(6x+ 1) = 1, so 3x + 1 and 6x + 1 are units in Z 9 [x]. Corollary 4.9. If F is a field, then f(x) F [x] is a unit iff it is a nonzero constant polynomial. Examples: Discuss units in Z[x], R[x], Z p [x]. Definition, p. 95. Two elements a, b in a commutative ring R are associates if there exists a unit u R with a = ub (equivalently, b = u 1 a). [Note that it is an equivalence relation; discuss associates in Z, F, F [x].] A nonzero element p in an integral domain R is irreducible if it is not a unit and whenever one can write p = bc with b, c R, then b or c is a unit. In particular, a nonconstant polynomial p(x) with coefficients in an integral domain R is called irreducible if its only divisors are its associates and units in R (which by 4.8 are the units in R[x]). A nonunit which is not irreducible is called reducible. (As the book points out, the words prime and irreducible have the same meaning in this context (cf. Theorem 4.11 and Theorem 9.15 for unique factorization domains). Examples. Every polynomial of degree 1 over a field is irreducible (think of degrees). From high school you might recall that every polynomial over R can be factored into linear and irreducible quadratic factors. This is a special property of the real numbers and will not be proved in this course. We will later see that there are irreducible polynomials of every degree in Q[x]. For polynomials of degree larger than 3, they may be reducible even though they have no roots in the field: x 4 + 2x = (x 2 + 1) 2 over R. But x is reducible as an element of C[x], so irreducibility depends on which ring you are thinking of it being an element of. Theorem Let F be a field. A nonzero polynomial is reducible in F [x] iff it can be written as a product of two polynomials of lower degree.

6 6 This is not true for F an integral domain: in Z[x], 2x + 2 = 2(x + 1) and neither 2 nor x + 1 is a unit. Theorem Let F be a field and p(x) F [x], p(x) / F. TFAE: (1) p(x) is irreducible. (2) If b(x), c(x) F [x] with p(x) b(x)c(x), then p(x) b(x) or p(x) c(x). (3) If b(x), c(x) F [x] with p(x) = b(x)c(x), then b(x) or c(x) is a nonzero constant polynomial. Proof. (1) = (2) Same as Theorem 1.8 for Z: gcd(p, b) p, so is a unit or associate of p. If a unit, then Theorem 4.7 says p c. Otherwise, p b. (2) = (3) By (2), p b or p c; we may assume the former, say pr = b, so p = bc = prc. Hence rc = 1, since F [x] is an integral domain. But then c is a unit (nonzero element of F ) as desired. (3) = (1) If b is any divisor of p, then p = bc for some c. By (3), either b is a unit, and c = b 1 p is an associate of p, or c is a unit, and b = c 1 p is an associate of p. Corollary Let F be a field and p(x) F [x] be irreducible. If p divides a product of polynomials, then it divides at least one of them. This is proved by induction, the same as for the integers. Theorem 4.13 (Unique factorization). Let F be a field. Every nonconstant polynomial in F [x] can be written as a product of irreducible polynomials. The factorization is unique up to order and multiplication by units. This theorem does not actually require that F be a field, but it certainly needs F to be a ring in which one has unique factorization. This is, in fact, all that is required. (See exercise 36, p. 307.) In particular, one still has unique factorization for Z[x] and R[x, y], but we shall not give a detailed proof in this course. Proof. The proof is essentially the same as the proof of the Fundamental Theorem of Arithmetic. I will only sketch the proof here, as a reminder. Let S be the set of polynomials which fail and assume it is nonempty. Then it has a polynomial of smallest degree; it is not irreducible, so it factors into polynomials of smaller degree (so not in S). Their factorizations give one of the original polynomial, hence a contradiction. For uniqueness, take two factorizations f(x) = p 1 p r = q 1 q s into irreducibles. p 1 divides the product of q i s, so divides one of them, hence is an associate of it. Cancel and continue to see that r = s and all others pair up as associates.

7 Examples. 1. (2x+2)(x+2) = (x+1)(2x+4) = (3x+3)( 2 3 x+ 4 3 ) are various factorizations of 2x 2 + 6x + 4 into irreducible polynomials in Q[x] Exercise 23, page 99. x is irreducible in Z 5 [x]: if it factors, each factor must be linear, say (x + a)(x + b). But then a, b Z 5 are roots of x 2 + 2; but none of the five elements work. In Z 5 [x], x 4 4 = x has no roots (try the 5 possibilities), but does factor as (x 2 + 2)(x 2 2), so this is a factorization into irreducibles. 3. Exercise 21, p. 99. In finite fields we can using counting techniques. For Z p [x], there are p degree 1 monic polynomials x + a. And( so) to count the degree 2 monic reducible p p(p 1) p(p + 1) polynomials, we need all products of these: + p = + p =, the number with distinct factors plus the number of squares. Since there are p 2 monic degree 2 polynomials x 2 +ax+b, this leaves the number of irreducible monic degree 2 polynomials as p 2 p(p+1) 2 = p(p 1) 2.

8 8 Polynomials as functions. Let R be a commutative ring, f(x) = a n x n + + a 0 R[x]. The function f : R R induced by f(x) is called a polynomial function and is defined by f(r) = a n r n + a n 1 r n a 1 r + a 0. We must be very careful in writing f(x) to differentiate between the polynomial as an element of R[x] (where x is an indeterminate) and the polynomial function from R to R (where x is a variable). It is quite possible for two different polynomials to induce the same polynomial function (though not on the real numbers). Your homework for this section will ask you to find nonzero polynomials that induce the zero function. A simple example is given by the polynomials x 2 + 1, x Z 3 [x]. They are different polynomials but induce the same function 0 1, 1 2, 1 2. Indeed, there are only 3 3 = 27 possible functions from Z 3 to Z 3, but infinitely many polynomials. Definition, p Let R be a commutative ring, f(x) R[x]. And element a R is called a root of f(x) if f(a) = 0 R (meaning that the induced function f : R R maps a 0 R ). The standard high school question of solving f(x) = x 2 + 3x + 2 = 0 means finding the roots of f(x), or equivalently, the set of all r R which are mapped to zero by the induced function. You did not need to be so careful about the difference between the two meanings of f(x) in high school because two polynomials in R[x] are equal iff their induced functions are the same. To make use of the idea of roots in determining reducibility of polynomials, we generalize a couple of high school theorems from R to arbitrary fields. Theorem 4.14 (Remainder Theorem). Let F be a field, f(x) F [x] and a F. The remainder when f(x) is divided by x a is f(a). Proof. Use the division algorithm and the fact that the remainder must be an element of F. Corollary (Factor Theorem). Let F be a field, f(x) F [x] and a F. a is a root of f(x) iff x a is a factor of f(x) in F [x]. Corollary Let F be a field and f(x) a polynomial of degree n in F [x]. Then f(x) has at most n roots in F Examples: x has no roots in R. It has 2 roots in C. It has infinitely many roots in the real quaternions H, in particular, i, j and k. So commutativity is needed! 2(x 2 + x) has four roots in Z 6, namely 0, 2, 3, 5. So we must not have zero divisors either. In fact,

9 the theorem holds if F is only an integral domain, with a similar proof to the one we are about to give, where one first extends the division algorithm as we mentioned earlier to monic polynomials. A proof can also be based on Corollary 4.16 together with Theorem 9.30 to be done later, which says that any integral domain can be embedded in a field (like Z in Q). 9 Proof of Corollary Induct on n. If n = 0, f(x) is a constant in F, so has no roots. Assume n > 0 and the theorem holds for polynomials of degree n 1. If f(x) has no roots, we are done. If f(x) has a root a F, then f(x) = (x a)g(x) for some g(x) F [x] of degree n 1. By the induction hypothesis, g(x) has at most n 1 roots in F. Any root b a of f(x) satisfies 0 = f(b) = (b a)g(b), so b is also a root of g. Thus the number of roots of f(x) is at most 1 + (n 1) = n. Notice where we used the fact that F has no zero divisors. Another corollary of the factor theorem is that if f(x) F [x] is irreducible, then it has no roots in F. This gives us a test to see if a polynomial is reducible, but does not tell if it is irreducible (recall x 4 + 2x = (x 2 + 1) 2 R[x]). As we noted in an earlier example, we do have the converse if the degree of f(x) is 2 or 3. Corollary If F is a field and f(x) F [x] has degree 2 or 3, then f(x) is irreducible iff f(x) has no roots in F. We also see that what happened for finite fields with polynomial functions not being the same as polynomials does not happen with infinite fields. Corollary Let F be an infinite field and f(x), g(x) F [x]. f(x) and g(x) induce the same function on F iff f(x) = g(x) in F [x]. Proof. ( =) is clear. (= ) Assume that f(a) = g(a) for every a F. Then every element of F is a root of the polynomial f(x) g(x). Since F is infinite, the only way Corollary 4.16 can be true is if f(x) g(x) is the zero polynomial. That is, we have f(x) = g(x). Examples: page 105: 11, 16. Irreducibility in Q[x]. We shall reduce the problem of irreducibility in Q[x] to looking at Z[x]. Indeed, if f(x) Q[x], we can multiply by an integer to clear denominators: cf(x) Z[x] for some

10 10 c Z. It is clear that f(x) factors into polynomials of degree at least one iff cf(x) does at least with rational coefficients. We shall see that only integer coefficients are needed. There exist algorithms for completely factoring polynomials in Z[x], but they are beyond the scope of this course. We shall look at some standard theorems regarding polynomials over Z. The first is commonly done in high school: Theorem 4.20 (Rational root test). Let f(x) = a n x n + + a 1 x + a 0 Z[x]. If the rational number r/s 0 is a root of f(x) and is in lowest terms, then r a 0 and s a n. Proof. Since r s is a root, we have ( r ) n ( r ) n 1 ( r a n + an a1 + a 0 = 0. s s s) Multiply by s n to clear denominators and rearrange the terms to get a 0 s n = r[ a n r n 1 a 1 s n 1 ] a n r n = s[ a n 1 r n 1 a 0 s n 1 ]. and Since r s is in lowest terms, we have gcd(r, s) = 1, whence r a 0 and s a n. Examples. 1. f(x) = 2x 4 + x 3 2x 2 4x 3 has possible roots ±1, ±3, ± 1 2, ± 3 2. Check and find that f( 1) = 0 = f(3/2). Factor out x + 1 and 2x 3 to get f(x) = (x + 1)(2x 3)(x 2 + x + 1). The last factor has no rational (or even real) roots, so we have a complete factorization into irreducibles over Q. 2. x 3 + 2x 2 4x + 3 is irreducible over Q (but not over R!). The main issue in reducing the factoring problem to Z is the question of whether a polynomial with integer coefficients might factor with rational coefficients, but not with integer coefficients. In fact, this cannot happen. Lemma Let f, g, h Z[x] with f = gh and assume that p Z is a prime which divides every coefficient of f. Then either p divides every coefficient of g or p divides every coefficient of h. Proof. Write f(x) = a i x i, g(x) = b i x i and h(x) = c i x i. Assume the claim is false. Then there is a first coefficient b r not divisible by p and a first coefficient c t not divisible by p. From f = gh, we see that the coefficient a r+t of f is just r+t i=0 b ic r+t i. By choice of b r and c t, p divides every term in the sum for a r+t except b r c t since every other term has a factor from b 0,..., b r 1 or from c 0,..., c t 1. Since p does not divide b r c t, it cannot divide the sum, a contradiction of the fact that p divides all coefficients of f.

11 Theorem 4.22 (Gauss s Lemma see page 325). Let f(x) Z[x]. f(x) factors as a product of polynomials of degrees m and n in Q[x] iff f(x) factors as a product of polynomials of degrees m and n in Z[x]. 11 Proof. ( =) is obvious. For ( = ), suppose f(x) = g(x)h(x) in Q[x]. Clear denominators by finding integers c, d > 0 such that cg(x), dh(x) Z[x] (i.e., find a common denominator and multiply by it). Then cdf(x) = [cg(x)][dh(x)] is a factorization of cdf(x) in Z[x] into polynomials of the same degrees. We will be done if we can show that cd = 1. If not, it has some positive prime divisor p, which then divides every coefficient of cdf(x), hence every coefficient of cg(x) or of dh(x) by the lemma; we may assume it is cg(x). Write cd = pt and factor out p to write cg(x) = pk(x) with k(x) Z[x]. Therefore ptf(x) = p[k(x)][dh(x)]. Cancel p. Continuing in this way, we can cancel every prime factor of the original cd, leaving a factorization of f(x) into polynomials of the same degrees as g and h. Exercise 10, p. 114: If f(x) = g(x)h(x) Z[x] and f is monic, then the leading coefficients of g and h must multiply to be 1, hence are either both +1 or both 1. In the latter case, write f(x) = [ g(x)][ h(x)], so we get a factorization into monic polynomials in either case. Example, p Show x 4 5x is irreducible over Q. Theorem 4.23 (Eisenstein s Criterion). Let f(x) = a n x n + +a 1 x+a 0 be a nonconstant polynomial with integer coefficients. Assume there is a prime p such that (1) p divides a 0, a 1,..., a n 1 (2) p does not divide a n (3) p 2 does not divide a 0. Then f is irreducible in Q[x]. Proof. Assume that f is reducible. By Gauss s Lemma, we can factor f over Z, say f(x) = (b 0 + b 1 x + + b r x r )(c 0 + c 1 x + + c s x s ), with r, s 1. Since p a 0 = b 0 c 0, we know p divides one of them, say b 0. Since p 2 a 0, p cannot also divide c 0. Since a n = b r c s is not divisible by p, neither is b r, so there is some minimum index k so that b k is not divisible by p. (So p does divide b 0,..., b k 1.) Now we have a k = b 0 c k + b 1 c k b k 1 c 1 + b k c 0 in which p divides every term except the last, a contradiction. This is a very useful theorem, though it certainly doesn t cover all cases. It is easy to construct numerous irreducible polynomials, and in particular, polynomials of every degree. One of the most interesting standard applications of the theorem is shown in Exercise 20, p. 114: for p prime, f(x) = x p 1 + x p x + 1 is irreducible in Q[x]. If we multiply f(x) by x 1, we get x p 1, whose roots are the p roots of 1 on the unit circle in the complex plane, namely e 2πik/p, k = 0, 1,..., p 1. Note that we can factor

12 12 f(x) iff we can factor f(x + c) for any c Q. To apply the Eisenstein criterion, we look at f(x + 1) = (x + 1)p 1 (x + 1) 1 = (x + 1)p 1 [( x ( ) ( ) ) p p /x = x p + x p x + 1 1] p 1 1 ( ) ( ) p p = x p 1 + x p 2 + +, p 1 1 in which we see that the three conditions of Eisenstein s criterion hold for the coefficients. Therefore f(x) is irreducible. There is one other very useful theorem; it can be used to show that a polynomial is irreducible, but not directly that it is reducible. Theorem Let f(x) = n i=0 a ix i Z[x] and assume that p is a positive prime not dividing a n. Let f(x) be the image of f(x) in Z p [x] under the homomorphism that reduces all coefficients modulo p. If f(x) is irreducible in Z p [x], then f(x) is irreducible in Q[x]. Proof. By Gauss s lemma, it suffices to show that f(x) is irreducible in Z[x]. Assume it is reducible: f(x) = g(x)h(x). Modulo p we have f(x) = ḡ(x) h(x). Furthermore, since p a n, it also does not divide the leading coefficients of g(x) or h(x), so we have a genuine factorization of f(x) whenever we have one of f(x) (that is, the degrees do not change). Since f(x) is irreducible in Z p [x], the original polynomial f(x) is irreducible also. As an example, consider x 2 2x 1 Z[x]. Modulo 2, it equals x 2 1 = (x + 1) 2 so tells us nothing. Modulo 3, we have x 2 2x 1 which is seen to be irreducible by trying the 3 possible roots and finding that none works. In general, there are only finitely many polynomials of a given degree in Z p [x] so all possibilities can be checked; this would be very time consuming, but in practice, there are special algorithms that work quite well by computer. That is why symbolic algebra programs such as Derive can easily factor polynomials over the integers or modulo p (or modulo n also). Example. Page 114, #11. 30x n 91 has no roots in Q for any n > 1. By Eisenstein s criterion with p = 7, the polynomial is irreducible in Q[x] and therefore it has no roots by Corollary We have mentioned the results of Section 4.6 on complex numbers and will do no more with it at this time.

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