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


 Angelina Shields
 1 years ago
 Views:
Transcription
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.
it is easy to see that α = a
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 UF. Therefore
More informationPROBLEM SET 6: POLYNOMIALS
PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other
More informationThe Division Algorithm for Polynomials Handout Monday March 5, 2012
The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,
More informationModern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)
Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)
More informationIntroduction to Finite Fields (cont.)
Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number
More informationCHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY
January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.
More informationUnique Factorization
Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon
More informationFactoring Polynomials
Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent
More informationPOLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS
POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).
More informationMOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu
Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing
More informationModule MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013
Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of
More information3 1. Note that all cubes solve it; therefore, there are no more
Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if
More informationQuotient Rings and Field Extensions
Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.
More informationMath 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field
Math 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will
More informationz = i ± 9 2 2 so z = 2i or z = i are the solutions. (c) z 4 + 2z 2 + 4 = 0. By the quadratic formula,
91 Homework 8 solutions Exercises.: 18. Show that Z[i] is an integral domain, describe its field of fractions and find the units. There are two ways to show it is an integral domain. The first is to observe:
More informationH/wk 13, Solutions to selected problems
H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important
More information9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.
9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n1 x n1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role
More information7. Some irreducible polynomials
7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of
More information3 Factorisation into irreducibles
3 Factorisation into irreducibles Consider the factorisation of a nonzero, noninvertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could
More informationcalculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,
Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warmup Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
More informationAppendix A. Appendix. A.1 Algebra. Fields and Rings
Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationSOLVING POLYNOMIAL EQUATIONS
C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra
More informationDie ganzen zahlen hat Gott gemacht
Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk.
More informationPrime Numbers and Irreducible Polynomials
Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More information1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain
Notes on realclosed fields These notes develop the algebraic background needed to understand the model theory of realclosed fields. To understand these notes, a standard graduate course in algebra is
More informationDiscrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof. Harper Langston New York University
Discrete Mathematics Lecture 3 Elementary Number Theory and Methods of Proof Harper Langston New York University Proof and Counterexample Discovery and proof Even and odd numbers number n from Z is called
More informationCopy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.
Algebra 2  Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers  {1,2,3,4,...}
More informationRESULTANT AND DISCRIMINANT OF POLYNOMIALS
RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results
More informationFinite Fields and ErrorCorrecting Codes
Lecture Notes in Mathematics Finite Fields and ErrorCorrecting Codes KarlGustav Andersson (Lund University) (version 1.01316 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents
More information(a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9
Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More informationminimal polyonomial Example
Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationIntroduction to Algebraic Geometry. Bézout s Theorem and Inflection Points
Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationComputer Algebra for Computer Engineers
p.1/14 Computer Algebra for Computer Engineers Preliminaries Priyank Kalla Department of Electrical and Computer Engineering University of Utah, Salt Lake City p.2/14 Notation R: Real Numbers Q: Fractions
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More informationCS 103X: Discrete Structures Homework Assignment 3 Solutions
CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On wellordering and induction: (a) Prove the induction principle from the wellordering principle. (b) Prove the wellordering
More informationFactorization Algorithms for Polynomials over Finite Fields
Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 20110503 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is
More information(x + a) n = x n + a Z n [x]. Proof. If n is prime then the map
22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find
More information1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]
1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not
More informationsome algebra prelim solutions
some algebra prelim solutions David Morawski August 19, 2012 Problem (Spring 2008, #5). Show that f(x) = x p x + a is irreducible over F p whenever a F p is not zero. Proof. First, note that f(x) has no
More informationInteger roots of quadratic and cubic polynomials with integer coefficients
Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street
More informationElementary Number Theory We begin with a bit of elementary number theory, which is concerned
CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,
More informationFunctions and Equations
Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c
More information15. Symmetric polynomials
15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things.
More informationAlgebra 3: algorithms in algebra
Algebra 3: algorithms in algebra Hans Sterk 20032004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems
More informationSome Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.
Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,
More informationFACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set
FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,
More informationReal Roots of Univariate Polynomials with Real Coefficients
Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials
More informationIntroduction to Modern Algebra
Introduction to Modern Algebra David Joyce Clark University Version 0.0.6, 3 Oct 2008 1 1 Copyright (C) 2008. ii I dedicate this book to my friend and colleague Arthur Chou. Arthur encouraged me to write
More information4. CLASSES OF RINGS 4.1. Classes of Rings class operator Aclosed Example 1: product Example 2:
4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets
More informationr + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.
Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in
More information11 Ideals. 11.1 Revisiting Z
11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(
More informationIrreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients
DOI: 10.2478/auom20140007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca
More informationALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY
ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY HENRY COHN, JOSHUA GREENE, JONATHAN HANKE 1. Introduction These notes are from a series of lectures given by Henry Cohn during MIT s Independent Activities
More informationNonunique factorization of polynomials over residue class rings of the integers
Comm. Algebra 39(4) 2011, pp 1482 1490 Nonunique factorization of polynomials over residue class rings of the integers Christopher Frei and Sophie Frisch Abstract. We investigate nonunique factorization
More informationMath Review. for the Quantitative Reasoning Measure of the GRE revised General Test
Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important
More informationSome facts about polynomials modulo m (Full proof of the Fingerprinting Theorem)
Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem) In order to understand the details of the Fingerprinting Theorem on fingerprints of different texts from Chapter 19 of the
More informationElementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.
Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole
More informationThe cyclotomic polynomials
The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) =
More informationBasics of Polynomial Theory
3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where
More informationField Fundamentals. Chapter 3. 3.1 Field Extensions. 3.1.1 Definitions. 3.1.2 Lemma
Chapter 3 Field Fundamentals 3.1 Field Extensions If F is a field and F [X] is the set of all polynomials over F, that is, polynomials with coefficients in F, we know that F [X] is a Euclidean domain,
More informationLagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.
Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method
More informationFactoring of Prime Ideals in Extensions
Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree
More informationPutnam Notes Polynomials and palindromes
Putnam Notes Polynomials and palindromes Polynomials show up one way or another in just about every area of math. You will hardly ever see any math competition without at least one problem explicitly concerning
More information1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).
.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational
More informationIntroduction to finite fields
Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at
More informationCyclotomic Extensions
Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate
More information4 Unique Factorization and Applications
Number Theory (part 4): Unique Factorization and Applications (by Evan Dummit, 2014, v. 1.00) Contents 4 Unique Factorization and Applications 1 4.1 Integral Domains...............................................
More informationThe Fundamental Theorem of Arithmetic
Chapter 1 The Fundamental Theorem of Arithmetic 1.1 Prime numbers If a, b Z we say that a divides b (or is a divisor of b) and we write a b, if for some c Z. Thus 2 0 but 0 2. b = ac Definition 1.1 The
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationMA651 Topology. Lecture 6. Separation Axioms.
MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples
More information5. Factoring by the QF method
5. Factoring by the QF method 5.0 Preliminaries 5.1 The QF view of factorability 5.2 Illustration of the QF view of factorability 5.3 The QF approach to factorization 5.4 Alternative factorization by the
More informationProofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.
Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t
More informationKevin James. MTHSC 412 Section 2.4 Prime Factors and Greatest Comm
MTHSC 412 Section 2.4 Prime Factors and Greatest Common Divisor Greatest Common Divisor Definition Suppose that a, b Z. Then we say that d Z is a greatest common divisor (gcd) of a and b if the following
More information6. Fields I. 1. Adjoining things
6. Fields I 6.1 Adjoining things 6.2 Fields of fractions, fields of rational functions 6.3 Characteristics, finite fields 6.4 Algebraic field extensions 6.5 Algebraic closures 1. Adjoining things The general
More information3.6 The Real Zeros of a Polynomial Function
SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix,
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationMODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.
MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on
More informationLecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic. Theoretical Underpinnings of Modern Cryptography
Lecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic Theoretical Underpinnings of Modern Cryptography Lecture Notes on Computer and Network Security by Avi Kak (kak@purdue.edu) January 29, 2015
More informationSubsets of Euclidean domains possessing a unique division algorithm
Subsets of Euclidean domains possessing a unique division algorithm Andrew D. Lewis 2009/03/16 Abstract Subsets of a Euclidean domain are characterised with the following objectives: (1) ensuring uniqueness
More information8 Divisibility and prime numbers
8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express
More informationQUADRATIC RECIPROCITY IN CHARACTERISTIC 2
QUADRATIC RECIPROCITY IN CHARACTERISTIC 2 KEITH CONRAD 1. Introduction Let F be a finite field. When F has odd characteristic, the quadratic reciprocity law in F[T ] (see [4, Section 3.2.2] or [5]) lets
More information1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style
Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with
More informationMathematics Review for MS Finance Students
Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,
More informationThe finite field with 2 elements The simplest finite field is
The finite field with 2 elements The simplest finite field is GF (2) = F 2 = {0, 1} = Z/2 It has addition and multiplication + and defined to be 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 0 0 = 0 0 1 = 0
More informationThe van Hoeij Algorithm for Factoring Polynomials
The van Hoeij Algorithm for Factoring Polynomials Jürgen Klüners Abstract In this survey we report about a new algorithm for factoring polynomials due to Mark van Hoeij. The main idea is that the combinatorial
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS
ON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for
More informationMATH10040 Chapter 2: Prime and relatively prime numbers
MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive
More information3.2 The Factor Theorem and The Remainder Theorem
3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial
More informationMath 4310 Handout  Quotient Vector Spaces
Math 4310 Handout  Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable
More information1 Lecture: Integration of rational functions by decomposition
Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.
More informationU.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra
U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory
More information