a 2 + b 2) ( c 2 + d 2) = (ac bd) 2 + (ad + bc) 2 = (ac + bd) 2 + (ad bc) 2

Transcription

1 We will be talking about integers. We start with the rational integers. These are the ordinary integers 0, ±1, ±, ±3,.... The text denotes the set of rational integers by J. A lot of people use the letter Z instead. If a and b are integers, we say that a divides b and write a b, if a = bc for some integer c. (See page 3, just before Theorem 1.3.) So 3 divides 1 but 3 does not divide. Instead of saying that a divides b, we sometimes say that a is a factor of b. If an integer divides 1, it is said to be a unit, or to be invertible. The rational integers that are units are 1 and 1. The text defines a prime to be an integer p, that is not unit, with the property that if p = ab, then either a or b is a unit (page 6). For the rational integers, that s the same as saying that p ±1, and the only divisors of p are ±1 and ±p (page 1). Theorem 1.3 in the text is Euclid s Proposition 30 of Book VII. It says that if a rational integer p is a prime, and p ab, then p a or p b. The converse is also true: if p is a rational integer, and whenever p ab, then p a or p b, then p is a prime. You should be able to prove the converse. I like to take Theorem 1.3 as the definition of a prime. That s because the usual formulation of the fundamental theorem of arithmetic is hideous. (Actually, the text s version of the fundamental theorem of arithmetic, Theorem 1.5, is better than most.) If we take Theorem 1.3 to be the definition of a prime, then the fundamental theorem is essentially the statement that every nonzero integer is either a unit or a product of primes. The required uniqueness of the primes in the product comes from repeated application of Theorem 1.3. That s essentially what is going on in the proof of the fundamental theorem on pages 4 and 5. The Gaussian integers G = J + Ji = {a + bi : a, b J}. Define Nα, for α = a + bi G by Nα = a + b. This is a positive integer if α 0, and is 0 if α = 0. The two square identity (Fibonacci, Brahmagupta, Diophantus): ( a + b ) ( c + d ) = (ac bd) + (ad + bc) = (ac + bd) + (ad bc) So 5 = 1 + and 13 = + 3 so 65 = = In terms of N this says NαNβ = N (αβ) = N (ᾱβ). This is also true for Gaussian numbers, that is, numbers of the form a+bi where a, b Q. Nonzero Gaussian numbers 1

2 have inverses: (a + bi) 1 = α 1 = α Nα a a + b b a + b i The Gaussian integer α is a unit if and only if Nα = 1. Consequently, if Nα is a (rational) prime, then α is a Gaussian prime. So 1 + i is a Gaussian prime. However, 3 is a Gaussian prime, but N3 = 9 is not a rational prime. The division algorithm: If α, β G, and α 0, then there exist π, ρ G such that β = πα + ρ and Nρ < Nα. Pick π so that the distance from π to β/α is less than 1. So Nρ = N (β πα) = NαN (β/α π) < Nα. This is Theorem 1.6. From the division algorithm we get Bezout s theorem: If α and β are in G, then there exist σ and τ in G such that σα + τβ divides both α and β. We say that σα + τβ is an algebraic gcd of α and β. Here is the proof. If α = 0, take σ = 0 and τ = 1. Otherwise write β = πα + ρ with Nρ < α. By induction on Nα, we can find λ and µ so that λρ + µα divides both ρ and α, hence divides both α and β. But λρ + µα = λ (β πα) + µα = (µ λπ) α + λβ so we can set σ = µ λπ and τ = λ. Euclid s Proposition 30 (Theorem 1.7) follows from Bezout s theorem. Suppose α βγ. Let δ = σα + τβ divide both α and β. So α = εδ. Also α δγ (multiply δ = σα + τβ by γ). If ε is a unit, then α β. If δ is a unit, then α γ. The proof of Theorem 1.7 in the text is nonconstructive. The next theorem is similar to the rational root test in that it reduces the question of whether a polynomial with rational coeffi cients has a nontrivial factor to a finite number of cases. Theorem 1 (Kronecker) There is an algorithm for factoring a polynomial with rational coeffi cients. Proof. Gauss s theorem says that it suffi ces to show how to factor polynomials with (rational) integer coeffi cients. So let f (X) be a polynomial of

3 degree n with integer coeffi cients (not necessarily monic). We show how to determine all the factors of f (X) in J [X]. Look at the integers f (0), f (1),..., f (n) If f (i) = 0, then X i is a factor of f (X), so we can pass to f (X) / (X i) and we are done by induction. So we may assume that f (i) 0 for i = 0, 1,..., n. If f (X) is a multiple of g (X) J [X], then g (i) divides f (i) for i = 0, 1,..., n. Let S i be the finite set of divisors of f (i). Then we know that g (i) = s i S i for i = 0, 1,..., n. Moreover, since deg g (X) n, the values s i determine g (X). In fact, we can write a formula for g (X) in terms of these values (Lagrange interpolation) g (X) = n s i i=0 j i X j i j Of course the right hand side generally has rational coeffi cients. If it has integer coeffi cients, then it is a candidate for a factor of f (X). So we just see if it works. Theorem (6.9) Every algebraic number field has an integral basis. Proof. Suppose ω 1,..., ω n is a basis of integers. Where would we look for an integer that was not in Jω Jω n? Suppose we have an integer ω = a 1 ω a n ω n where some a i is not in J. We may assume that pa i J for some prime p and all i. We may further assume that 0 a i < 1, and that 0 < a 1. So a 1 = r = m/p with 0 < m < p. Let ω 1 = ω = rω 1 + a ω + + a n ω n, and ω i = ω i for i > 1. The determinant of this matrix is r, so [ω 1,..., ω n] = r [ω 1,..., ω n ], whence p divides [ω 1,..., ω n ]. Thus there are finitely many possibilities for p, and a i = m i /p with 0 m i < p, so there are finitely many candidates for ω. We can test to see if any of them are integers. If not, then ω 1,..., ω n is an integral basis. If so, then ω 1,..., ω n is a basis of integers with smaller discriminant. Lemma 3 Let p be an odd prime, and ζ a primitive p-th root of 1. λ = 1 ζ and a J. If a/λ is an algebraic integer, then p divides a. Let 3

4 Proof. Let x = a/λ = a/ (1 ζ). Then ζ = 1 a/x so 1 = (1 a/x) p. Multiplying this equation by x p gives x p = (x a) p. So x = a/λ satisfies the polynomial (x a) p x p = pax p 1 p (p 1) + a x p 1 + a p ( ) = a px p 1 p (p 1) ax p a p 1 Thus a/λ satisfies a polynomial of the form g (x) = px p 1 + p ( ) + a p 1 Since p does not divide a, the polynomial x p 1 g (1/x) is irreducible by Eisenstein. Hence g is irreducible, so g/p is the minimum polynomial of a/λ over R. As a/λ is an integer, the constant term of g/p must lie in J. But that constant term is a p 1 /p, so p divides a. Theorem 4 (6.13) Let p be an odd prime, and ζ a primitive p-th root of 1. Then 1, ζ,..., ζ p is an integral basis for R (ζ). Proof. It is a basis of integers because ζ satisfies the irreducible polynomial 1 + x + + x p 1 = 0. If λ = 1 ζ, then [ 1, ζ,..., ζ p ] = [ 1, λ,..., λ p ] because J + Jζ + + Jζ p = J + Jλ + + Jλ p by equations (6.7). Moreover, Theorem 5.1 says D (ζ) = ( 1) (p 1)/ p p. Let ω 1,..., ω p 1 be an integral basis for R (ζ). Then λ j = p 1 i=1 c ijω i for j = 0,..., p where the c ij J. So [ 1, λ,..., λ p ] = c ij [ω 1,..., ω p 1 ] whence c ij = ±p k for some k 0. So we can write where a (i) j J. ω i = a(i) 0 + a (i) 1 λ + + a (i) p λ p p k 4

5 If p k divides a (i) j for all i and j, then we are done. Otherwise, we may assume that the displayed equation holds for some i, some k > 0, and a (i) j J, where p does not divide a (i) m for some m. Thus p k 1 ω i is an integer in R (ζ) of the form a (i) 0 + a (i) 1 λ + + a (i) p λ p p where p does not divide a (i) m for some m. We will show that this can t happen. We can choose m to be the smallest index for which this happens, and by deleting the first m terms in right-hand sum, we get an integer of the form a (i) m λ m + + a (i) p λ p p where p does not divide a (i) m. Now p = (1 ζ) ( 1 ζ ) (1 ζ p 1) because 1 + x + + x p 1 = (x ζ) ( x ζ ) (1 ζ p 1). So p = λ p 1 κ where κ is an integer, hence p = λ m+1 κ where κ is an integer. Thus a (i) m λ m + + a (i) p λ p λ m+1 is an integer, so a (i) m /λ is an integer. Thus p divides a (i) m, by the Lemma, a contradiction. Dedekind on algebraic integers. Let θ = 5. Choose β 1 so that β 1 = + θ and set β = 3/β 1. The β = 9/ ( + θ) = θ, so β is an integer. Also (β 1 + β ) = + θ + θ + 6 =. What about (1 + θ) / (β 1 + β )? If we square it, we get (1 + θ 5) / = + θ so it s an integer. Is it β 1? The question is whether 1 + θ = (β 1 + β ) β 1 = + θ + 3. Well, yes. So we have = β 1 and 1 + θ = (β 1 + β ) β 1. Can we get a linear combination with integer coeffi cients of β 1 and β 1 + β equal to 1? Well, β 1 β = 3, and (β 1 + β ) = so 1 = β 1 β (β 1 + β ) which is a linear combination of β 1 and β 1 + β. Revised notation. Choose λ in first quadrant so λ = + θ. Let µ = 3/λ = λ, so µ = θ. Note that (λ + µ) = = so λ + µ =. Also (λ µ) = 4 6 =, so λ µ =. Thus λ = + 5

6 We have θλ = λ 3µ and θµ = 3λ + µ. Is 1, θ, λ, µ an integral basis? which has determinant = 4 5, thank you Scientific WorkPlace. So = 8 5, which is not very square free (which would do it). What is D (λ)? There are 6 terms ( + ) ( ) ( ) = 40 The square of this is 40 = ( 4 3 5) = 8 3 5, so 1, λ, λ, λ 3 is not an integral basis. Can we show that 1, θ, λ, µ is an integral basis by extending the principal ideal domain Z [ ] into Q (, 5 )? The question is when is a + b 5 integral over Z [ ] when a, b Q ( ). For b 0, the criterion is that a +5b and a are in Z [ ]. Let c = a Z [ ], so c /4+5b Z [ ] so c + 5 (b) Z [ ] whence d = b Z [ ] because 5 is prime in Z [ ]. Theorem 5 A finite commutative ring in which every proper ideal is a product of maximal ideals is a principal ideal ring. Proof. Call the ring R. We will induct on the number n of maximal ideals of R. We may assume that R is nontrivial. First suppose n = 1, and the maximal ideal is P. Then zero is a product of maximal ideals, so P m = 0 for some m. Let I be a nonzero proper ideal. Then I = P i for some positive inreger i. Note that if P i = P i+1, then I = P i = P m = 0, so we can choose a P i with a / P i+1. Then Ra, which is a power of P, must be P i. For the induction step, where n > 1, divide the maximal ideals into two nonempty sets. Then we can write the zero ideal as a product of two ideals I and J such that I + J = R and each of I and J is contained in fewer than n maximal ideals. The natural map R R I R J 6

7 is onto because I + J = R, and one-to-one because I J = IJ = 0. By induction, each of R/I and R/J is a principal ideal ring. It follows that the product is a principal ideal ring. Note that an arbitrary finite commutative ring need not be a principal ideal ring. For example Z [x, y] (x, xy, y ) in which the ideal (x, y) is not principal. A corollary to this theorem is that the ring of integers in a finite-dimensional extension field of the rational numbers has the property that each ideal is generated by two-elements, and that one of those elements can be taken to be an arbitrary nonzero element of the ideal. 7