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Notes on Algebraic Numbers Notes on Algebraic Numbers Robin Chapman January 20, 1995 (corrected November 3, 2002) 1 Introduction This is a summary of my 1994–1995 course on Algebraic Numbers. (Revised and improved on 1993– 1994!) The background assumed is standard elementary number theory—as found in my Level III course—and a little (Abelian) group theory. Corrections and suggestions for improvement are wel- come, and will be credited in future editions! I first learned algebraic number theory from Stewart & Tall’s book ([3]) and this is an excellent account. However it’s more abstract than the approach of this course and deals with general algebraic number theory while I deal mainly with the theory of quadratic fields. A book dealing mainly with quadratic fields is Cohn ([1]); I have incorporated many of the ideas in this book into this course, but this is a rather difficult book to read. I am grateful to Jeremy Bygott for corrections to and suggestions on a previous version, and to Paul Epstein for a further correction. 2 Algebraic Numbers and Integers An algebraic number is a root of a polynomial equation with rational coefficients, i.e., α is an algebraic number if and only if there exist a1, a2, . , an ∈ Q with n n−1 n−2 α + a1α + a2α + ··· + an−1α + an = 0. Similarly an algebraic integer is a root of a monic polynomial equation with integer coefficients, i.e., β is an algebraic number if and only if there exist b1, b2, . , bm ∈ Z with m m−1 m−2 β + b1β + b2β + ··· + bm−1β + bm = 0. Clearly every algebraic integer is also an algebraic number. (As a convention I shall use Greek letters for algebraic numbers, and Roman letters for rational numbers.) We think of algebraic integers and algebraic numbers as generalizations of integers and rationals. All integers are algebraic integers, and all rationals are algebraic numbers. Some examples of algebraic √ √ √ q integers are i, 2, 3 10, (1 + 5)/2 and 2 cos 2π/9. The algebraic numbers include (1 + i)/2, 1/5 and sin 2π/7. The numbers e and π are not algebraic—they are transcendental. Transcendence proofs are very difficult, and will not be covered in this course. (See [2] for proofs that e and π are transcendental.) A basic fact about algebraic integers is that a rational number which is also an algebraic integer is an (ordinary) integer: Proposition 1 If α is an algebraic integer and α ∈ Q, then α ∈ Z. 1 Proof Write α = a/b in lowest terms. Suppose that α isn’t an integer so b 6= ±1. As α is an algebraic integer, there exist c1, c2, . cn ∈ Z with n n−1 α + c1α + ··· + cn−1α + cn = 0. Writing α = a/b and clearing denominators gives n n−1 n−1 n a + c1a b + ··· + cn−1ab + cnb = 0, which implies that an ≡ 0 (mod b). As b 6= ±1 then b has a prime factor, p, say. Hence p | an which implies that p | a. This means that p is a factor of both a and b contradicting the assumption that a/b is in its lowest terms. This contradiction establishes the result. 2 This result is useful as it shows that some algebraic numbers aren’t algebraic integers. The most important result in this section is that sums and product of algebraic numbers (resp. algebraic integers) are also algebraic numbers (resp. integers). Before we can prove this we need an alternative description of algebraic numbers and integers by means of matrix theory. Lemma 1 A number α is an algebraic number if and only if it is an eigenvalue of a (square) matrix with rational entries, and it is an algebraic integer if and only if it is an eigenvalue of a (square) matrix with integer entries. Proof Suppose first that A is a square matrix with integer entries. The characteristic polynomial of A is χA(λ) = det(λI − A). As the entries in λI − A involve only integer coefficients, it follows that its determinant χA(λ) is a monic polynomial with integer coefficients. Now if α is an eigenvalue of A, then χA(α) = 0 and so α is an algebraic integer. If A has rational entries then an identical argument shows that its eigenvalues are algebraic numbers. For the converse suppose that α is an algebraic integer, so that there exist b1, b2 . , bn ∈ Z with n n−1 α + b1α + ··· + bn−1α + bn = 0. We can rewrite this as n n−1 α = −b1α − · · · − bn−1α − bn. (∗) Let v be the (column) vector (1 α α2 ··· αn−1)T , so α 0 1 0 ··· 0 1 2 α 0 0 1 ··· 0 α . . .. . αv = . = . . = Av αn−1 0 0 0 ··· 1 αn−2 n n−1 α −bn −bn−1 −bn−2 · · · −b1 α where A is a matrix with integer entries. (NB we have used (∗) here.) Hence α is an eigenvalue of A, with eigenvector v. Similarly if α is an algebraic number then we can find a matrix A with rational entries having α as an eigenvalue. 2 A slightly more elaborate argument proves that sums and products of algebraic numbers or integers are also algebraic numbers or integers. Theorem 1 If β and γ are algebraic numbers (resp. algebraic integers), then β + γ and βγ are also algebraic numbers (resp. algebraic integers). 2 Proof I’ll just give the proof for algebraic integers, as that for algebraic numbers follows mutatis mutandis as in Lemma 1. Suppose β and γ are algebraic integers. The idea of the proof is to find a non-zero vector w and two integer matrices B and C with Bw = βw and Cw = γw. This will imply that β + γ and βγ are eigenvalues of B + C and BC respectively. Suppose that n n−1 β + b1β + ··· + bn−1β + bn = 0 and m m−1 γ + c1γ + ··· + cm−1γ + cm = 0 where the bjs and cks are integers. Let w be the mn by 1 vector (1 β β2 ··· βn−1 γ βγ β2γ ··· βn−1γ γ2 βγ2 ··· βn−1γm−1)T . Consider βw. The entries in this vector all have the form βj+1γk where 0 ≤ j < n and 0 ≤ k < m. If j + 1 < n this is already an entry in w; if j + 1 = n it equals n k n−1 k k k β γ = −b1β γ − · · · − bn−1βγ − bnγ . In all cases βj+1γk is a linear combination, with integer coefficients, of entries in w. Hence there is an integer matrix B with Bw = βw. Similarly there is an integer matrix C with Cw = γw. (Writing down general expressions for B and C is fairly horrible so I won’t do it, but from the numerical examples I’ll give in lectures I hope it’s plain that this construction is a lot easier than it looks.) Now (B + C)w = Bw + Cw = βw + γw = (β + γ)w and (BC)w = B(Cw) = B(γw) = γBw = (βγ)w so β + γ and βγ are eigenvalues of B + C and BC respectively. As B + C and BC have integer coefficients, then by Lemma 1 β + γ and βγ are algebraic integers. 2 3 Quadratic Fields From now on we abandon the general study of algebraic numbers to concentrate on a specific class of such numbers, quadratic numbers, i.e., those which are solutions of quadratic equations with√ rational coefficients. By the solution formula for quadratics each such number has the form a + b d where a, b and d are rational numbers. We won’t even look at all quadratic numbers at once;√ if we fix a non-square rational number d we shall consider the set of all numbers of√ the form a + b d with √a and b ∈ Q. This set is called a quadratic field and is denoted by Q(√d). (NB if d >√0 then d is by convention the positive√ square root, and if d = −D < 0 then d will denote i D.)√ If d > 0 then every element of Q( d) is real, and we say is a real quadratic field; if d < 0 then Q( d) contains non-real numbers—it is an imaginary quadratic field. The structure of real quadratic fields is radically different from that of imaginary quadratic fields, as we shall see on many subsequent occasions. We should perhaps pause to justify our terminology; a field is by definition an algebraic structure which admits addition, subtraction, multiplication and division (except by 0) where these have all the familiar properties. This is indeed the case. √ Lemma 2 Let K = Q( d) be a quadratic field, where d is a non-square rational. Then if α, β ∈ K then α + β, α − β, αβ and, provided β 6= 0, α/β ∈ K. 3 Proof The proofs for addition, subtraction and multiplication are easy so I’ll skip them. For division it’s OK to treat the special case√ of α = 1, for if 1/β ∈ K then by the multiplication rule α/β = α(1/β) ∈ K. Now if β = a + b d then we might try to write √ √ 1 1 a − b d a − b d = √ = √ √ = . β a + b d (a + b d)(a − b d) a2 − b2d But remembering the commandment, “thou shalt not divide by zero”, we must check that the denominator a2 − b2d is non-zero. But if a2 − b2d = 0 then either b 6= 0 and then d = (a/b)2 is a square contrary to hypothesis, or b = 0 and then a = 0 implying that β = 0, again contrary to hypothesis.
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