Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.

Transcription

1 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 require a proof, you should still use sentences to say what you are doing. Be sure to introduce the variables that you use (aside from certain symbols that are fixed by convention, e.g. Z). Don t use the same symbol for two different things at the same time. Proofs should be logically sound and the logic should be clear to the reader. You can t introduce new assumptions unless they are clearly harmless, like reordering some symbols you introduced earlier, or unless the logic of the proof requires it, such as a in a proof by contradiction. Your proof should start with the assumptions and end with the conclusion, unless you re using some unusual proof structure. If you re doing something unusual, clearly say what you are doing. If you need to break into cases, it should be clear that you are not leaving any cases out. Please write your proofs as if your intended audience is someone who knows a little less than you. You should provide enough detail that a reader can follow your reasoning. You do not need to provide every detail if you are using techniques from a previous course. Please read your solutions for clarity, logic and consistency after you write them, and rewrite your solutions if you find problems with them. Please read my comments on your homework when I return it. Solutions and example write-up for first homework Prove for any given positive integer N there exist only finitely many integers n with φ(n) = N where φ denotes Euler s φ-function. Conclude in particular that φ(n) tends to infinity as n tends to infinity. Solution: Write n = p a 1 1 pa k k, where p 1,..., p k are the distinct primes dividing n and a 1,... a k are positive integers. To bound n, our goal is to bound k, each p i and each a i in terms of N. Since φ(p a i i ) = pa i 1 i (p i 1) and φ(ab) = φ(a)φ(b) when a and b are relatively prime integers, this means that N = φ(n) = p a (p 1 1)p a (p 2 1) p a k 1 k (p k 1). Without loss of generality, suppose that p k is the largest prime factor dividing n. Since p k 1 divides N, we know N p k 1. In particular, N + 1 is greater than or equal to every prime dividing n. Since there are only finitely many primes less than or equal to N + 1, this also bounds the number k in terms of N

2 Let c be the maximum power to which any prime appears in the prime factorization of N: c is the maximum l such that q l divides N, for a prime q. Then for any i from 1 to k, a i 1 c (this is clear for the expression for N in terms of the prime factors of n). In particular, each a i c + 1. So n = p a 1 1 pa k k, where k is at must the number of primes less than or equal to N + 1, each p i is less than or equal to N + 1, and each a i is an integer from 0 to c. Since there are finitely many choices of k, the {p i } i and the {a i } i, this means that there are only finitely many possibilities for n. Now we want to show that φ(n) tends to infinity as n goes to infinity. If φ(n) does not go to infinity as n goes to infinity, then there is some M > 0 such that for infinitely many n, we have φ(n) < M. Of course this implies that φ(n) = N for infinitely many n for some N < M, since there are only finitely many numbers less than M. This contradicts what we just showed Prove that the derivative D x of a polynomial satisfies D x (f(x)+g(x)) = D x (f(x))+d x (g(x)) and D x (f(x)g(x)) = D x (f(x))g(x) + D x (g(x))f(x) for any two polynomials f(x) and g(x). Solution. First we verify additivity. Suppose f(x) = a n x n + + a 1 x + a 0 and g(x) = b k x k + + b 1 x + b 0. Without loss of generality (since everything is commutative) we assume that n k. We define b k+1,..., b n all to be 0, so that g(x) = b n x n + + b 1 x + b 0 and f(x) + g(x) = (a n + b n )x n + + (a 1 + b 1 )x + (a 0 + b 0 ). Then D x (f(x) + g(x)) = n(a n + b n )x n i(a i + b i )x i (a 1 + b 1 ) Now we verify the product rule. f(x) = ax n and g(x) = bx k. Then = na n x n a 1 + nb n x n b 1 = D x (f(x)) + D x (g(x)). First we suppose that f and g are monomials, so that D x (f(x)g(x)) = D x (abx n+k ) = (n + k)abx n+k 1 = nax n 1 bx k + kbx k 1 ax n = D x (f(x))g(x) + D x (g(x))f(x). We note that the product rule is also true if either of f or g is zero. To prove the general product rule, we argue by induction on the maximum of the number of terms in f and g. We just proved the base case, where f and g are both monomials. Now we suppose that f(x) = f 1 (x) + f 2 (x) and g(x) = g 1 (x) + g 2 (x), where the number of terms in each of f 1, f 2, g 1, and g 2 is less than the maximum of the number of terms in f and g (this is possible since they are not both monomials). Then by the inductive hypothesis, D x (f i (x)g j (x)) = D x (f i (x))g j (x) + D x (g j (x))f i (x) for all choices of i, j in {1, 2}. Then D x (f(x)g(x)) = D x ((f 1 (x) + f 2 (x))(g 1 (x) + g 2 (x))) = D x (f 1 (x)g 1 (x) + f 1 (x)g 2 (x) + f 2 (x)g 1 (x) + f 2 (x)g 2 (x)) = D x (f 1 (x)g 1 (x)) + D x (f 1 (x)g 2 (x)) + D x (f 2 (x)g 1 (x)) + D x (f 2 (x)g 2 (x)) = D x (f 1 (x))g(x) + D x (g 1 (x))f(x) + D x (g 2 (x))f(x) + D x (f 2 (x))g(x) = D x (f 1 (x) + f 2 (x))g(x) + D x (g 1 (x) + g 2 (x))f(x) = D x (f(x))g(x) + D x (g(x))f(x) 2

3 The third line and the fifth line use the additivity of the formal derivative, which we just proved Find all irreducible polynomials of degrees 1, 2 and 4 over F 2 and prove that their product is x 16 x. Solution. Working over F 2, we can enumerate all polynomials of a given degree by going through all possibilities for coefficients. There will be 2 n polynomials of degree n for each n. All linear polynomials are irreducible, so we enumerate the linear polynomials: x, x + 1. Since we have enumerated the linear polynomials, we can enumerate the reducible quadratic polynomials by enumerating their products: x 2, x = (x + 1) 2, x 2 + x. The remaining quadratic polynomial is x 2 + x + 1 (there are four total) and this polynomial is therefore the only irreducible quadratic polynomial. There are 16 quartic polynomials. To find the irreducibles, first we find the reducible ones. The ones with x as a factor are exactly the ones with constant coefficient 0. The ones with x + 1 as a factor but not with x as a factor are products of x + 1 with cubic polynomials not divisible by x. The four cubic polynomials not divisible by x are the four with nonzero constant coefficient: Their products with x + 1 are x 3 + 1, x 3 + x + 1, x 3 + x 2 + 1, x 3 + x 2 + x + 1. x 4 + x 3 + x + 1, x 4 + x 3 + x 2 + 1, x 4 + x 2 + x + 1, x The only reducible quartic polynomial without x or x + 1 as a factor is (x 2 + x + 1) 2 = x 4 + x We have described the 13 reducible quartic polynomials. The remaining three quartic polynomials are irreducible: To finish the problem, we take the product x 4 + x + 1, x 4 + x 3 + 1, x 4 + x 3 + x 2 + x + 1. x(x + 1)(x 2 + x + 1)(x 4 + x + 1)(x 4 + x 3 + 1)(x 4 + x 3 + x 2 + x + 1). It is straightforward to verify that this is x 16 + x For any prime p and any nonzero a F p prove that x p x + a is irreducible and separable over F p. 3

4 Solution: Separability is easy. Defining f(x) = x p x+a, we find that D x (f(x)) = 1, which has no roots in any extension of F p. Therefore f and D x (f(x)) have no common roots in any extension of F p, and therefore f is separable. Now for irreducibility. Suppose K is a splitting field for f and α K is a root of f. Then α p α + a = 0. If α F p, then α p 1 = 1, since the multiplicative group of F p = Z/pZ is cyclic of order p 1. Then α p = α, and α p α = 0. This would imply that a = 0, counter to our hypothesis. So α is not in F p. Now we consider α + b, where b F p. We know (α + b) p = α p + b p = α a + b. This implies that (α + b) p (α + b) + a = 0, in other words that α + b is a root of f(x). So the roots of f(x) are exactly α, α + 1,..., α + (p 1). (This also proves that f(x) is separable.) Suppose f factors as f(x) = g 1 (x) g k (x) for some k > 1, where g 1 (x),..., g k (x) are monic irreducible polynomials over f(x). Let i and j be distinct indices from 1,..., k. Suppose α is a root of g i (x) and α is a root of g j (x). Then g i (x) is the minimal polynomial of α and g j (x) is the minimal polynomial of α. Since α and α are roots of f(x), we know α = α + b and α = α + c for some b, c F p. Then α = α + b c, and therefore g i (x + b c) is an irreducible monic polynomial over F p with α as a root. Then g i (x + b c) and g j (x) are both minimal polynomials for α, and therefore they are equal. In particular, this implies that deg g i = deg g j. Since i and j were arbitrary, this implies that deg g 1 = = deg g k, and therefore that p = deg f(x) = deg g 1 (x) + + deg g k (x) = k deg g 1 (x). Since p is prime, this implies that either k = 1 or k = p. If k = p, then f(x) splits completely over F p and therefore α F p, a contradiction. So k = 1, meaning that f(x) is irreducible Suppose m and n are relatively prime positive integers. Let ζ m be a primitive mth root of unity and let ζ n be a primitive nth root of unity. Prove that ζ m ζ n is a primitive mnth root of unity. Solution: Suppose ζ = ζ m ζ n. By commutativity, ζ mn = (ζ m ζ n ) mn = (ζ m m) n (ζ n n) m = 1 n 1 m = 1. Of course, this implies that ζ is an mnth root of unity (although not necessarily a primitive one). To show it is a primitive mnth root of unity, it is enough to show that if ζ k = 1, then mn k. This would imply that the order of ζ is mn, and therefore that the (multiplicative) cyclic group it generates contains mn distinct elements (the set of mnth roots of unity). So suppose k is an integer and ζ k = 1. Then (ζ n ζ m ) k = 1, meaning that ζ k n = ζ k m. First we raise both sides of this expression to the n power, to get 1 = ζ nk n = ζ nk m. Since ζ m is a primitive mth root of unity, we can only have ζm nk relatively prime to m, this implies that m k. Similarly, we know 1 = ζ mk m = ζ mk n, = 1 if m nk. Since n is and therefore n mk. Since m is relatively prime to m, this implies n k. Since m and n both divide k and n is relatively prime to m, then mn k. This implies that ζ is a primitive mnth root of unity. 4

5 Prove there are only a finite number of roots of unity in any finite extension K of Q. Solution: Suppose K is an extension of Q of degree n. Suppose ζ m is a primitive mth root of unity and ζ m K. This means that Q(ζ m ) is a subfield of K, and in particular, φ(m) = [Q(ζ m ) : Q] [K : Q] = n. From Problem , we know that there are only finitely many m with φ(m) n. Further, for any given m such that ζ m K, there are only m distinct mth roots of unity. So there are only finitely many m with any mth root of unity in K, and for each of those, there are only finitely many mth roots of unity in K. So there are only finitely many roots of unity in K Prove that for n odd, n > 1, Φ 2n (x) = Φ n ( x). Solution: Suppose that ζ is a primitive nth root of unity. Then ( ζ) 2n = ( 1) 2n ζ 2n = 1, so ζ is a 2nth root of unity. Now suppose ( ζ) k = 1. Then ( 1) k ζ k = 1. If k is even, then ζ k = 1, implying that n k, since ζ is a primitive nth root of unity. Then 2n k, since n is odd. If k is odd, then ζ k = 1, implying that ζ 2k = 1, implying that n 2k. But since k and n are odd, this means that n k, implying that ζ k = 1, a contradiction. So 2n k, and therefore ζ is a primitive 2nth root of unity. So for every primitive nth root of unity ζ, we have that ζ is a primitive 2nth root of unity. (Alternatively, since n and 2 are relatively prime and 1 is the primitive square root of unity, this is an application of problem ) Let S n denote the set of primitive nth roots of unity and S 2n the set of primitive 2nth roots of unity. Let S n = { ζ ζ S n }. We have shown that S n S 2n. Since n is odd and φ(2) = 1, it follows from the properties of φ that φ(2n) = φ(n), and therefore S n = S 2n. So S n = S 2n. The set of roots of Φ n ( x) is S n and the set of roots of Φ 2n (x) is S 2n. Both polynomials are separable since they are irreducible polynomials over Q. Also, they are monic. A separable monic polynomial is determined by its set of roots. Since S n = S 2n, this implies that Φ 2n (x) and Φ n ( x) are equal. 5