# MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao

Save this PDF as:

Size: px
Start display at page:

Download "MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu"

## Transcription

1 Integer Polynomials June 9, 007 Yufei Zhao 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 with integer polynomials. Looking at the coefficients Bound the size of the coefficients Modulos reduction. In particular, a b P (a) P (b) whenever P (x) Z[x] and a, b are distinct integers. Looking at the roots Bound their location on the complex plane. Examine the algebraic degree of the roots, and consider field extensions. Minimal polynomials. Many problems deal with the irreducibility of polynomials. A polynomial is reducible if it can be written as the product of two nonconstant polynomials, both with rational coefficients. Fortunately, if the original polynomial has integer coefficients, then the concepts of (ir)reducibility over the integers and over the rationals are equivalent. This is due to Gauss Lemma. Theorem 1 (Gauss). If a polynomial with integer coefficients is reducible over Q, then it is reducible over Z. Thus, it is generally safe to talk about the reducibility of integer polynomials without being pedantic about whether we are dealing with Q or Z. Modulo Reduction It is often a good idea to look at the coefficients of the polynomial from a number theoretical standpoint. The general principle is that any polynomial equation can be reduced mod m to obtain another polynomial equation whose coefficients are the residue classes mod m. Many criterions exist for testing whether a polynomial is irreducible. Unfortunately, none are powerful enough to be universal. One of the most well-known criteria is Eisenstein s criterion. Theorem (Eisenstein). Let f(x) = a n x n + a n 1 x n a 1 x + a 0 be a polynomial with integer coefficients such that p a i for 0 i n 1, p a n and p a 0. Then f(x) is irreducible. Proof. Suppose that f = gh, where g and h are nonconstant integer polynomials. Consider the reduction mod p (i.e., apply the ring homomorphism Z[x] F p [x]), and let f, ḡ, h denote the residues of f, g, h (i.e. the coefficients are residues mod p). We have f(x) = a 0 x n. Since F p [x] is a unique factorization domain, we see that the only possibilities for ḡ and h are cx k for some integers c and k 1. Then, the constant terms of g and h are both divisible by p, so p a 0. Contradiction. The most typical example for the application of Eisenstein s criterion is to show that the cyclotomic polynomial Φ p (x) is irreducible for prime p: Problem 1. Let p be a prime number. Show that f(x) = x p 1 + x p + + x + 1 is irreducible. 1

2 Solution. The polynomial f(x) is irreducible if and only if f(x + 1) is irreducible. We have f(x + 1) = (x + ( ) ( ) ( ) 1)p 1 p p p (x + 1) 1 = xp 1 + x p + x p x + p. 1 p Note that f(x + 1) fails the Eisenstein criterion for the prime p. Therefore f(x) is irreducible. Note that the proof of Eisenstein s criterion extends to other rings with similar properties. For instance, to show that x 4 + 5x + 5 is irreducible over the Gaussian integers Z[i], we can simply apply Eisenstein with the Gaussian prime + i. The proof of Eisenstein s Criterion can be slightly generalized to the following. The proof is more or less the same, and so it s left as exercise. Theorem 3 (Extended Eisenstein). Let f(x) = a n x n + a n 1 x n a 1 x + a 0 be a polynomial with integer coefficients such that p a i for 0 i < k, p a k and p a 0. Then f(x) has an irreducible factor of degree greater than k. We give one more result that relates to looking at the modulo reduction of polynomials, known as Hensel s lemma. Theorem 4 (Hensel). Let a 0, a 1,..., a k be integers, and let P (x) = a n x k + a 1 x+a 0, and let P (x) denote the derivative of P (x). Suppose that x 1 is an integer such that P (x 1 ) 0 (mod p) and P (x 1 ) 0 (mod p). Then, for any positive integer k, there exists an unique residue x (mod p k ), such that P (x k ) 0 (mod p k ) and x x 1 (mod p). The proof of Hensel s lemma closely mimics Newton s method of finding roots. We work up the powers of p, and find the a zero of P (x) mod p k for k =, 3,.... The details of the proof are omitted here. Root Hunting When working with integer polynomials, it is often not enough to stay in Z. We have to think outside the box and move our scope to the complex numbers. A lot can be a said about a polynomial if we know something about its complex zeros. Many irreducibility problems hinge on placing bounds on the zeros of the polynomial in the complex plane. We begin with a familiar example. Problem. Let f(x) = a n x n + a n 1 x n + + a 1 x + a 0 be a polynomial with integer coefficients, such that a 0 is prime and a 0 > a 1 + a + + a n. Show that f(x) is irreducible. Solution. Let α be any complex zero of f. Suppose that α 1, then a 0 = a 1 α + + a n α n a a n, a contradiction. Therefore, all the zeros of f satisfies α > 1. Now, suppose that f(x) = g(x)h(x), where g and g are nonconstant integer polynomials. Then a 0 = f(0) = g(0)h(0). Since a 0 is prime, one of g(0), h(0) equals 1. Say g(0) = 1, and let b be the leading coefficient of g. If α 1,..., α k are the roots of g, then α 1 α α k = 1/ b 1. However, α 1,..., α k are also zeros of f, and so each has an magnitude greater than 1. Contradiction. Therefore, f is irreducible. Next, we present a Perron s criterion, which has a similar statement but a much more difficult proof compared with the previous result. Theorem 5 (Perron). Let P (x) = x n + a n 1 x n a 1 x + a 0 be a polynomial with a 0 0 and Then P (x) is irreducible. a n 1 > 1 + a n + + a 1 + a 0.

3 Again, the idea is to put bounds on the modulus of the roots of f. The key lies in the following lemma. Lemma 1. Let P (x) = x n + a n 1 x n a 1 x + a 0 be a polynomial with a n 1 > 1 + a n + + a 1 + a 0. Then exactly one zero of P satisfies z > 1, and the other n 1 zeros of P satisfy z < 1. Let us see how we can prove Perron s criterion if we have this lemma. Suppose that P (x) = f(x)g(x), where f and g are integer polynomials. Since P has only one zero with modulus not less than 1, one of the polynomials f, g, has all its zeros strictly inside the unit circle. Suppose that z 1,..., z k are the zeros of f, and z 1,..., z k < 1. Note that f(0) is a nonzero integer, and f(0) = z 1 z k < 1, contradiction. Therefore, f is irreducible. Now, let us prove Lemma 1. We offer two proofs. The first proof is an elementary proof that uses only the triangle inequality. The second proof invokes theorems from complex analysis, but it is much more intuitive and instructive. First proof of the Lemma 1. (due to Laurentiu Panaitopol) Let us suppose wolog that a 0 0 since we can remove any factors of the form x k. Let s first prove that there is no root α of P (x) with α = 1. Suppose otherwise, then we have that thus a n 1 α n 1 = α n + a n α n + + a 1 α + a 0, a n 1 = a n 1 α n 1 = α n + a n α n + + a 1 α + a 0 α n + a n α n + + a 1 α + a 0 = 1 + a n + + a 1 + a 0. This contradicts the given inequality. Therefore, no zero of f(x) lies on the unit circle. Let s denote with α 1, α,..., α n be the zeros of P. Since α 1 α α n = a 0, it follows that at least one of the roots is larger than 1 in absolute value. Suppose that α 1 > 1 and let be the polynomial with roots α, α 3,..., α n. Then, Q(x) = x n 1 + b n x n + + b 1 x + b 0 P (x) = (x α 1 )Q(x) = x n + (b n α 1 )x n 1 + (b n 3 b n α 1 )x n + + (b 0 b 1 α 1 )x b 0 α 1 It follows that b n 1 = 1, a 0 = b 0 α 1, and a k = b k 1 b k α 1 for all 1 k n 1. Then, using the given inequality, we have b n α 1 = a n 1 > 1 + a n + + a 1 + a 0 = 1 + b n 3 b n α b 0 α 1 On the other hand, b n α 1 b n + α 1, so 1 + b n α 1 b n 3 + b n 3 α 1 b n b 1 x 1 b 0 + b 0 x 1 = 1 + b n + ( α 1 1) ( b n + b n b 1 + b 0 ). b n + α 1 > 1 + b n + ( α 1 1) ( b n + b n b 1 + b 0 ) and therefore b n + b n b 1 + b 0 < 1. 3

4 Then, for any complex number α with α 1, we have Q(α) = α n 1 + b n α n + b n 3 α n b 1 α + b 0 α n 1 b n α n b n 3 α n 3 b 1 α b 0 α n 1 α n 1 ( b n + b n b 1 + b 0 ) = α n 1 (1 b n b n 3 b 1 b 0 ) > 0 And so α cannot be a root. It follows that all the zeros of Q lie strictly inside the unit circle. This completes the proof of the lemma. In the polynomial P, the second term x n 1 is dominating, in the sense that the absolute value of its coefficient is greater than the sum of the absolute values of all the other coefficients. In the above proof, we managed to construct a new polynomial Q, whose leading term is dominating. While exactly one zero of P is outside the unit circle, none of the zeros of Q is outside the unit circle. This observation generalizes to the following result. Proposition 6. Let P (z) = a n z n + a n 1 z n a 1 z + a 0 be a polynomial with complex coefficients, and such that a k > a 0 + a a k 1 + a k a n for some 0 k n. Then exactly k zeros of P lie strictly inside the unit circle, and the other n k zeros of P lie strictly outside the unit circle. This is indeed true. The easiest way to prove this result is to invoke a well-known theorem in complex analysis, known as Rouché s theorem. Theorem 7 (Rouché). Let f and g be analytic functions on and inside a simple closed curve C. Suppose that f(z) > g(z) for all points z on C. Then f and f + g have the same number of zeros (counting multiplicities) interior to C. The proof of Rouché s theorem uses the argument principle. It can be found in any standard complex analysis textbook. In practice, for polynomials, Rouché s theorem is generally applied to some circle, and is useful when one term is very big compared to the other terms. Proposition 6 becomes very easy to prove with the aid of Rouché s theorem. Indeed, let us apply Rouché s theorem to the functions a k z k and P (z) a k z k with the curve being the unit circle. The given inequality implies that a k z k > P (z) a k z k for all z = 1. It follows that P has the same number of zeros as a k z k inside the unit circle. It follows that P has exactly k zeros inside the unit circle. Also, it is not hard to show that P has no zeros on the unit circle (c.f. first proof of Lemma 1). Thus we have proved Proposition 6. Second proof of Lemma 1. Apply Proposition 6 to k = n 1. While we re at it, let s look at couple of neat applications of Rouché s theorem, just for fun. These are not integer polynomial problems, but they contain useful ideas. Problem 3. (Romania??) Let f C[x] be a monic polynomial. Prove that we can find a z C such that z = 1 and f(z) 1. Solution. Let deg P = n. Suppose that f(z) < 1 for all z on the unit circle. Then f(z) < z n for all z on the unit circle. So, by Rouché s theorem, f(z) z n has n roots inside the unit circle, which is impossible, since f(z) z n has degree n 1. The Fundamental Theorem of Algebra is also an easy consequence of Rouché s theorem. Theorem 8 (Fundamental Theorem of Algebra). Any polynomial P (x) C[x] of degree n has exactly n complex zeros. 4

5 Proof. Let P (x) = a n x n + + a 1 x + a 0. For a sufficiently large real number R, we have a n R n > a n 1 R n a 1 R + a 0. Apply Rouché s theorem to the functions a n x n and P (x) a n x n on the circle z = R, we find that P (x) has exactly n zeros inside the circle. Also, since we may choose R arbitrarily large, so there are no additional zeros. Note that the above proof also gives a bound (although rather weak) for the zeros of a polynomial. This bound is attributed to Cauchy. Finally, the following result is a slightly stronger version of Rouché s theorem. Theorem 9 (Extended Rouché). Let f and g be analytic functions on and inside a simple closed curve C. Suppose that f(z) + g(z) < f(z) + g(z) for all points z on C. Then f and g have the same number of zeros (counting multiplicities) interior to C. There are many ways of bounding polynomial zeros on the complex plane. The following result is worth mentioning, as it has proven useful quite a few times. Proposition 10. Let P (x) = a 0 + a 1 x + + a n x n, where 0 < a 0 a 1 a n are real numbers, then any complex zero of the polynomial satisfies z 1. Proof. If z > 1, then, since z is a zero of (1 x)p (x), we get Thus, contradiction. Therefore, z 1. a 0 + (a 1 a 0 )z + + (a n a n 1 )z n a n z n = 0. a n z n = a 0 + (a 1 a 0 )z + + (a n a n 1 )z n a 0 + (a 1 a 0 ) z + + (a n a n 1 ) z n < a 0 z n + (a 1 a 0 ) z n + + (a n a n 1 ) z n = a 0 z n a 0 z n + a 1 z n a 1 z n + + a n z n = a n z n It follows as a simple corollary that for any polynomial with positive real coefficients, P (x) = a 0 + a 1 x + + a n x n, all its zeros lie in the annulus a k 1 a k 1 min z max 1 k n a k 1 k n a k Finally, we present one more irreducibility criterion, known as Cohn s criterion. Essentially, it says that if f(x) has nonnegative integer coefficients, and f(n) is prime for some n greater than all the coefficients, then f is irreducible. Theorem 11 (Cohn s Criterion). Let p be a prime number, and b an integer. Suppose that p n p n 1 p 1 p 0 is the base-b representation of p, with 0 p i < b for each i and p n 0, then the polynomial is irreducible. f(x) = p n x n + p n 1 x n + + p 1 x + p 0 The following proof is due to M. Ram Murty 1. As before, we begin with a lemma bounding the complex zeros of the polynomial. 1 M. Ram Murty, Prime Numbers and Irreducible Polynomials, Amer. Math. Monthly. 109 (00)

6 Lemma. Let f(x) = a n x n + a n 1 x n + + a 1 x + a 0 belong to Z[x]. Suppose that a n 1, a n 1 0, and a i H for i = 0, 1,..., n, where H is some positive constant. Then any complex zero α of f(x) either has nonpositive real part, or satisfies α < H Proof. If z > 1 and Re z > 0, we observe that f(z) z n an + a n 1 H z ( > Re a n + a n 1 z whenever 1 ) ( ) 1 z + 1 z n H z z H z z = z z H z z z H. It follows that α cannot be a zero of f if α H and Re α > 0. To prove Theorem 11 for the case b 3, we notice that Lemma implies if α is a zero of f(x), then b α > 1. Suppose that f(x) = g(x)h(x), where g and h are nonconstant integer polynomials. Since f(b) is prime, one of g(b), h(b) is 1. Say g(b) = 1, and the zeros of g are α 1,..., α k. We have g(b) = b α 1 b α k > 1, contradiction. Therefore, f is irreducible. The b = case is special, and requires more analysis. Lemma 3. Let f(x) = x n + a n 1 x n + + a 1 x + a 0 with a i {0, 1} for each i. Then all the zeros of f lie in the half plane Re z < 3. Proof. The cases n = 1 and can be verified by hand. Assume that n 3. Then, for z 0, we have f(z) 1 z n a m a ( m 1 z z z ) z m > 1 + a m 1 + a m 1 z z z ( z 1). If z satisfies arg z π/4, then we have Re(1/z ) 0, and we get f(z) z n > 1 1 z ( z 1). If z 3, then z ( z 1) ( 3 ( 3 ) 1) = 9 8 > 1, and so f(z) 0. On the other hand, if z is a zero of f with arg z > π/4, and suppose that Re z > 0, then from Lemma we have z < 1+ 5, and thus Re z < 1+ 5 < 3. It follows that all zeros of f lie in the half-plane Re z < 3. To finish off the proof, suppose that f(x) = g(x)h(x), where g and h are integer polynomials. Since f() is prime, one of g(), h() is 1. Say g() = 1. By Lemma 3, all the zeros of f lie in the half plane Re z < 3, which means that they satisfy z > z 1. Thus, if α 1,..., α k are the zeros of g, we have g() = α 1 α k > 1 α 1 1 α k = g(1) 1. So g() > 1, contradiction. 0 6

7 Problems 1. If q is a rational number and cos qπ is also rational, show that cos qπ {0, ± 1, ±1}.. Let P (x) be a monic polynomial with integer coefficients such that all its zeros lie on the unit circle. Show that all the zeros of P (x) are roots of unity, i.e., P (x) (x n 1) k for some n, k N. 3. If P (x) is a polynomial that such that P (n) is an integer for every integer n, then show that ( ) ( ) ( ) x x x P (x) = c n + c n c 0, n n 1 0 for some integers c n,..., c 0. (Note that the coefficients of P are not necessarily integers.) 4. Let f be an irreducible polynomial in Z[x], show that f has no multiple roots. 5. Player A and B play the following game. Player A thinks of a polynomial, P (x), with non-negative integer coefficients. Player B may pick a number a, and ask player A to return the value of P (a), and then player B may choose another number b and ask player A to return the value of P (b). After the two questions, player B must guess P (x). Does player B have a winning strategy? 6. Determine all pairs of polynomials f, g Z[x], such that f(g(x)) = x x (a) (USAMO 1974) Let a, b, c be three distinct integers, and let P be a polynomial with integer coefficients. Show that in this case the conditions P (a) = b, P (b) = c, P (c) = a cannot be satisfied simultaneously. (b) Let P (x) be a polynomial with integer coefficients, and let n be an odd positive integer. Suppose that x 1, x,..., x n is a sequence of integers such that x = P (x 1 ), x 3 = P (x ),..., x n = P (x n 1 ), and x 1 = P (x n ). Prove that all the x i s are equal. (c) (Putnam 000) Let f(x) be a polynomial with integer coefficients. Define a sequence a 0, a 1,... of integers such that a 0 = 0 and a n+1 = f(a n ) for all n 0. Prove that if there exists a positive integer m for which a m = 0 then either a 1 = 0 or a = 0. (d) (IMO 006) Let P (x) be a polynomial of degree n > 1 with integer coefficients and let k be a positive integer. Consider the polynomial Q(x) = P (P (... (P (x)... )) } {{ } k P s Prove that there are at most n integers t such that Q(t) = t. 8. (IMO Shortlist 1997) Find all positive integers k for which the following statement is true: if P (x) is a polynomial with integer coefficients satisfying the condition 0 P (c) k for c = 0, 1,..., k + 1, then F (0) = F (1) = = F (k + 1). 9. Let f(x) = x 4 + 6x + 1. Show that for any prime p, f(x) is reducible over F p, but f(x) is irreducible over Z. 10. Let m, n, and a be positive integers and p a prime number less than a 1. Prove that the polynomial f(x) = x m (x a) n + p is irreducible. 11. Let p be prime. Show that f(x) = x p 1 + x p + 3x p (p 1)x + p is irreducible. 1. (IMO 1993) Let f(x) = x n +5x n 1 +3, where n > 1 is an integer. Prove that f(x) cannot be expressed as the product of two nonconstant polynomials with integer coefficients. 13. (Romania TST 003) Let f(x) Z[x] be an irreducible monic polynomial with integer coefficients. Suppose that f(0) is not a perfect square. Show that f(x ) is also irreducible. This problem appeared in Reid Barton s handout in 005. Compare with the IMO 006 problem. 7

8 14. Let z 1, z,..., z n Z[i] be Gaussian integers (i.e., complex numbers of the form a + bi, where a, b Z) such that z i z 1 > for all i > 1. Prove that the polynomial (x z 1 )(x z ) (x z n ) + 1 is irreducible over Z[i]. 15. (Brazil 006) Let f(x) be an irreducible polynomial, and suppose that it has two roots whose product is 1. Show that the degree of f is even. 16. (MathLinks Contest) Let a be a nonzero integer, and n 3 be another integer. Prove that the the following polynomial is irreducible over the integers: P (x) = x n + ax n 1 + ax n + + ax Let a 1 a a n > 0 be positive integers. Show that the following polynomial is irreducible: P (x) = x n a 1 x n 1 a x n a n 18. (MOP 007) Let p(x) be a polynomial with integer coefficients. Determine if there always exists a positive integer k such that p(x) k is irreducible. 19. (Iran TST 007) Does there exist a sequence a 0, a 1, a,... in N, such that for each i j, gcd(a i, a j ) = 1, and for each n, the polynomial n i=0 a ix i is irreducible in Z[x]? 0. (China TST Quizzes 006) Let n be a positive integer, and let A 1, A,..., A k be a partition of the set of positive integers. Show that for some i {1,,..., k}, there are infinitely many irreducible polynomials of degree n and whose coefficients are distinct elements from A i. 1. Prove that x n x 1 is irreducible over the integers for all n.. (Iran 003) Let f 1, f,..., f n be polynomials with integer coefficients. Show that there exists a reducible polynomial g(x) Z[x] such that f i (x) + g(x) is irreducible for i = 1,,..., n. 3. (IMO Shortlist 1997) Let f be a polynomial with integer coefficients and let p be a prime such that f(0) = 0, f(1) = 1, and f(k) 0 or 1 (mod p) for all positive integers k. Show that deg f p (IMO Shortlist 005) Find all monic integer polynomials p(x) of degree two for which there exists an integer polynomial q(x) such that p(x)q(x) is a polynomial having all coefficients ±1. 5. (IMO Shortlist 005) Let a, b, c, d, e and f be positive integers. Suppose that the sum S = a + b + c + d + e + f divides both abc + def and ab + bc + ca de ef fd. Prove that S is composite. 6. (IMO 00) Find all pairs of integers m, n 3 such that there exist infinitely many positive integers a for which a m + a 1 a n + a 1 is an integer. 7. (IMO Shortlist 00) Let P (x) be a cubic polynomial with integer coefficients. Suppose that xp (x) = yp (y) for infinitely many pairs x, y of integers with x y. Prove that the equation P (x) = 0 has an integer root. 8. (IMO Shortlist 1996) For each positive integer n, show that there exists a positive integer k such that k = f(x)(x + 1) n + g(x)(x n + 1) for some polynomials f, g with integer coefficients, and find the smallest such k as a function of n. 9. (Romania TST 1998) show that for any n N, the polynomial P (x) = (x + x) n + 1 is irreducible over the integers. 8

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up 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)).

### 3 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

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

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

### PUTNAM 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

### 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

### Module 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

### CHAPTER 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.

### Modern 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)

### PROBLEM 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

### Prime 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.

### 7. 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

### Unique 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

### Factoring 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

### Die 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.

### Introduction 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

### H/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.

### POLYNOMIAL 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).

### March 29, 2011. 171S4.4 Theorems about Zeros of Polynomial Functions

MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 4: Polynomial and Rational Functions 4.1 Polynomial Functions and Models 4.2 Graphing Polynomial Functions 4.3 Polynomial

### calculating 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

### The 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,

### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 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

### THE 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

### a 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

### Quotient 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.

### Alex, I will take congruent numbers for one million dollars please

Alex, I will take congruent numbers for one million dollars please Jim L. Brown The Ohio State University Columbus, OH 4310 jimlb@math.ohio-state.edu One of the most alluring aspectives of number theory

### FACTORING 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

### FACTORING SPARSE POLYNOMIALS

FACTORING SPARSE POLYNOMIALS Theorem 1 (Schinzel): Let r be a positive integer, and fix non-zero integers a 0,..., a r. Let F (x 1,..., x r ) = a r x r + + a 1 x 1 + a 0. Then there exist finite sets S

### Non-unique factorization of polynomials over residue class rings of the integers

Comm. Algebra 39(4) 2011, pp 1482 1490 Non-unique factorization of polynomials over residue class rings of the integers Christopher Frei and Sophie Frisch Abstract. We investigate non-unique factorization

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE 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

### MA651 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

### Factorization 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

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### The 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 ) =

### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients

DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca

### Integer 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

### INTRODUCTORY SET THEORY

M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H-1088 Budapest, Múzeum krt. 6-8. CONTENTS 1. SETS Set, equal sets, subset,

### Putnam 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

### 1 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

### 3 Factorisation into irreducibles

3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could

### Some 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,

### FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.

FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is

### EXERCISES FOR THE COURSE MATH 570, FALL 2010

EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime

### 1 = (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 real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

### Roots of Polynomials

Roots of Polynomials (Com S 477/577 Notes) Yan-Bin Jia Sep 24, 2015 A direct corollary of the fundamental theorem of algebra is that p(x) can be factorized over the complex domain into a product a n (x

### Introduction 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

### 11 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(

### The Mean Value Theorem

The Mean Value Theorem THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers

### Zeros of a Polynomial Function

Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we

### The 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

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### 9. 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 n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### Real 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

### Some applications of LLL

Some applications of LLL a. Factorization of polynomials As the title Factoring polynomials with rational coefficients of the original paper in which the LLL algorithm was first published (Mathematische

### COMMUTATIVE RINGS. Definition: A domain is a commutative ring R that satisfies the cancellation law for multiplication:

COMMUTATIVE RINGS Definition: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative

### QUADRATIC 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

### ALGEBRAIC 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

### Math 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

### BX in ( u, v) basis in two ways. On the one hand, AN = u+

1. Let f(x) = 1 x +1. Find f (6) () (the value of the sixth derivative of the function f(x) at zero). Answer: 7. We expand the given function into a Taylor series at the point x = : f(x) = 1 x + x 4 x

### U.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

### WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly

### Factoring 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

### Collinear Points in Permutations

Collinear Points in Permutations Joshua N. Cooper Courant Institute of Mathematics New York University, New York, NY József Solymosi Department of Mathematics University of British Columbia, Vancouver,

### Factorization Algorithms for Polynomials over Finite Fields

Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is

### Galois Theory III. 3.1. Splitting fields.

Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where

### PROOFS BY DESCENT KEITH CONRAD

PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### DEFINITION 5.1.1 A complex number is a matrix of the form. x y. , y x

Chapter 5 COMPLEX NUMBERS 5.1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of matrices. DEFINITION 5.1.1 A complex number is a matrix of

### FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22 RAVI VAKIL CONTENTS 1. Discrete valuation rings: Dimension 1 Noetherian regular local rings 1 Last day, we discussed the Zariski tangent space, and saw that it

### minimal 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

### SOLVING 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

### Copy 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,...}

### Algebra 3: algorithms in algebra

Algebra 3: algorithms in algebra Hans Sterk 2003-2004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems

### The Factor Theorem and a corollary of the Fundamental Theorem of Algebra

Math 421 Fall 2010 The Factor Theorem and a corollary of the Fundamental Theorem of Algebra 27 August 2010 Copyright 2006 2010 by Murray Eisenberg. All rights reserved. Prerequisites Mathematica Aside

### 3. INNER PRODUCT SPACES

. INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### OSTROWSKI FOR NUMBER FIELDS

OSTROWSKI FOR NUMBER FIELDS KEITH CONRAD Ostrowski classified the nontrivial absolute values on Q: up to equivalence, they are the usual (archimedean) absolute value and the p-adic absolute values for

### The Chinese Remainder Theorem

The Chinese Remainder Theorem Evan Chen evanchen@mit.edu February 3, 2015 The Chinese Remainder Theorem is a theorem only in that it is useful and requires proof. When you ask a capable 15-year-old why

### SMT 2014 Algebra Test Solutions February 15, 2014

1. Alice and Bob are painting a house. If Alice and Bob do not take any breaks, they will finish painting the house in 20 hours. If, however, Bob stops painting once the house is half-finished, then the

### Congruence properties of binary partition functions

Congruence properties of binary partition functions Katherine Anders, Melissa Dennison, Jennifer Weber Lansing and Bruce Reznick Abstract. Let A be a finite subset of N containing 0, and let f(n) denote

### GROUPS ACTING ON A SET

GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for

### FACTORING AFTER DEDEKIND

FACTORING AFTER DEDEKIND KEITH CONRAD Let K be a number field and p be a prime number. When we factor (p) = po K into prime ideals, say (p) = p e 1 1 peg g, we refer to the data of the e i s, the exponents

### Subsets 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

### CONTRIBUTIONS TO ZERO SUM PROBLEMS

CONTRIBUTIONS TO ZERO SUM PROBLEMS S. D. ADHIKARI, Y. G. CHEN, J. B. FRIEDLANDER, S. V. KONYAGIN AND F. PAPPALARDI Abstract. A prototype of zero sum theorems, the well known theorem of Erdős, Ginzburg

### RESULTANT 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

### ON HENSEL S ROOTS AND A FACTORIZATION FORMULA IN Z[[X]]

#A47 INTEGERS 4 (204) ON HENSEL S ROOTS AND A FACTORIZATION FORMULA IN Z[[X]] Daniel Birmaer Department of Mathematics, Nazareth College, Rochester, New Yor abirma6@naz.edu Juan B. Gil Penn State Altoona,

### FACTORING 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,

### SUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by

SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples

### Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

### MATH10040 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

### 1 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

### Number Theory Hungarian Style. Cameron Byerley s interpretation of Csaba Szabó s lectures

Number Theory Hungarian Style Cameron Byerley s interpretation of Csaba Szabó s lectures August 20, 2005 2 0.1 introduction Number theory is a beautiful subject and even cooler when you learn about it

### Ideal Class Group and Units

Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals

### The degree of a polynomial function is equal to the highest exponent found on the independent variables.

DETAILED SOLUTIONS AND CONCEPTS - POLYNOMIAL FUNCTIONS Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. Thank you! PLEASE NOTE

### Continued 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

### Numerical Analysis Lecture Notes

Numerical Analysis Lecture Notes Peter J. Olver 5. Inner Products and Norms The norm of a vector is a measure of its size. Besides the familiar Euclidean norm based on the dot product, there are a number

### Baltic Way 1995. Västerås (Sweden), November 12, 1995. Problems and solutions

Baltic Way 995 Västerås (Sweden), November, 995 Problems and solutions. Find all triples (x, y, z) of positive integers satisfying the system of equations { x = (y + z) x 6 = y 6 + z 6 + 3(y + z ). Solution.