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MTH6128 Number Theory 2 Algebraic Numbers

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MTH6128 Number Theory 2 Algebraic Numbers MTH6128 Number Theory Notes 2 Spring 2020 \The only way to learn Mathematics is to do Mathematics." Paul Halmos 2 Algebraic numbers 2.1 Algebraic numbers and algebraic integers Pythagoras and his school discovered that the square root of 2 is not a rational number. (You saw the proof of this in Introduction to Algebra.) However, it is an easy number to describe geometrically: it is the ratio of the diagonal of a square to its side. The number π has a more complicated geometric description: it is the ratio of the circumference of a circle to its diameter, but there is no simple method to construct a straight line which is equal to the circumference of a given circle. (We know now, for example, that such a line cannot be constructed with the traditional geometric instruments of \ruler and compass".) We make a distinction between algebraic numbers (which are roots of polyno- mials with rational coefficients) and transcendental numbers (which are not): Definition Let α be a complex number. Then: • α is an algebraic number if there is a non-zero polynomial f(x) with rati- onal coefficients such that f(α) = 0; • α is a transcendental number if α is not an algebraic number. Moreover, • α is an algebraic integer if there is a non-zero monic polynomial f(x) with integer coefficients such that f(α) = 0. (A polynomial is monic if the coefficient of the highest power of x is equal to 1, n n−1 i.e., it is of the form f(x) = x +an−1x +:::+a1x+a0, with a0; : : : ; an−1 2 Z.) 1 Note that there are two extra conditions in the definition of an algebraic inte- ger. The polynomial should be monic, and it should have integer coefficients. Just asking one of these conditions would just define algebraic numbers over again. For, given any non-zero polynomial f(x), • we can make it monic by dividing by the coefficient of the highest power of x; and • we can make the coefficients integers by multiplying by the least common multiple of the denominators. Example p • Let α = 2. Then α2 − 2 = 0, so α is a root of the polynomial x2 − 2. This is monic and has integer coefficients. Thus α is an algebraic integer. p p 1 2 1 • Let α = 2 2 = 1= 2. Then α = 2 , so α is a root of the monic rational 2 1 polynomial x − 2 , or (equivalently) of the non-monic integer polynomial 2x2 −1. Thus, α is an algebraic number; but we haven't decided yet whether it is an algebraic integer. We will see soon that it is not. • π is transcendental (that is, it is not an algebraic number). This was proved by Lindemann about 150 years ago. Any integer is an algebraic integer; the integer n satisfies the polynomial x − n = 0. Similarly, any rational number is an algebraic number. In the other direction, we have: Proposition 2.1 A rational number is an algebraic integer if and only if it is an integer. For this reason, we sometimes call the ordinary integers \rational integers". Proof We have seen that integers are algebraic integers; we have to prove that a rational number which is an algebraic integer is an integer. Let q = a=b be a rational number (a 2 Z, b 2 Z, b ≥ 1) in its lowest terms (so that gcd(a; b) = 1). Suppose that q satisfies a monic polynomial with integer coefficients, say n n−1 f(q) = q + cn−1q + ··· + c1q + c0 = 0: Putting q = a=b, and multiplying this equation by bn, we obtain n n−1 n−1 n a + cn−1a b + ··· + c1ab + c0b = 0: 2 Now every term in this equation except the first is divisible by b, so b divides an. Now assume b > 1, and choose a prime factor p of b. Since b divides an, p divides an as well. Applying Lemma 1.2, we find that p j a. But also p j b, so p j gcd(a; b) = 1. This is a contradiction. Therefore b = 1, so that q is an integer. There is a result, which I will not prove, which makes things like this much easier. This is known as Gauss's Lemma. It can be stated in many different ways. But the following will do for our purposes. Definition Let α be an algebraic number. The minimal polynomial of α is the non-zero, monic polynomial f(x) of smallest possible degree with rational coefficients such that f(α) = 0. (Any algebraic number satisfies a monic polynomial with rational coefficients, and we can certainly choose one of smallest degree. Why is it unique? Suppose that f1(x) and f2(x) were two different monic polynomials of the same (smallest) degree satisfied by α, and let g(x) = f1(x) − f2(x). Then f1(α) = f2(α) = 0, so g(α) = 0, but g has smaller degree than f1 and f2; we can make it monic by dividing by the leading coefficient. But this contradicts the minimality of the degree of f1 and f2.) Theorem 2.2 (A version of Gauss' Lemma) The algebraic number α is an algebraic integer if and only if its minimal polynomial has integer coefficients. Now let q be a rational number. It satisfies the polynomial x − q = 0, and clearly this is monic and has smallest possible degree, so it is the minimal poly- nomial of q. So q is an algebraic integer if and only if the coefficients 1 and −q of this polynomial are both integers, i.e. if andp only if q is an integer. 1 Let us now reconsider the number α = 2 2. We saw that it satisfies the 2 1 monic polynomial x − 2 . This polynomial is irreducible, and its coefficients are not all integers. So α is not an algebraic integer. This confirms what we claimed earlier. 2.2 Quadratic numbers and quadratic integers We now define a quadratic number or a quadratic integer, to be an algebraic num- ber or algebraic integer which is not rational but satisfies a quadratic equation. More formally: 3 Definition An algebraic number is a quadratic number if its minimal polyno- mial is of degree 2. An algebraic number is a quadratic integer if its minimal polynomial is of degree 2 and has integer coefficients. Remark A quadratic number is irrational. Therefore, we also call quadratic numbers quadratic irrationals. An algebraic number is a quadratic integer if and only if it is a quadratic number and an algebraic integer. Real quadratic numbers will be very important to us in this course, so let us 2 see what they lookp like. The solutions to the quadratic equation x + bx + c = 0 are x = (−b ± b2 − 4c)=2. We can write this in simpler form as follows. Definition An integer d is said to be squarefree if for every prime p, p j d ) p2 6 j d. Said otherwise, d is squarefree if no prime divides it more than once, so a squarefree number is - up to sign - a product of distinct primes. Every integer is of the form c2d for some c; d 2 , d squarefree. pZ p Now for any rational number q, we can write q = v d where v 2 Q and d is a squarefree integer. For example, r r r280 23 · 5 · 7 23 · 3 · 5 · 7 · 11 2 p = = = 2310: 297 33 · 11 34 · 112 99 Here 2310 = 2 · 3 · 5 · 7 · 11 is squarefree. The general procedure is as in this example: multiply top and bottom by all the primes occurring to odd powers in the denominator; take all even powers of primes out of the square root; and what is left is of the right form. So we conclude: Propositionp 2.3 A complex number α is a quadratic number if and only if α = u + v d where u; v 2 Q, v 6= 0 and d is a squarefree integer not equal to 1. Proof "(": f(x) = (x − u)2 − v2d satisfies f(α) = 0. q 2 p p2 ")": Let f(x) = x + px + q satisfy f(α) = 0. Then α = − 2 ± 4 − q. p2 a p2 a 1 1 2 Write 4 − q = b for some a; b 2 Z, b > 0, so that 4 − q = b = b2 ab = b2 c d for p c some c; d 2 Z, c > 0, d 6= 1, d squarefree. Set u = − 2 , v = ± b . 4 p Remarkp The expressionp u + v d for α is unique, in the sense that if we have u1 +v1 d1 = u2 +v2 d2 for some u1; u2; v1; v2 2 Q and squarefree integers d1; d2 not equal to 1, then u1 = u2, v1 = v2 and d1 = d2. p 2.3 The ring of integers of Q( d) p p Recall that Q( d) = fu + v d : u; v 2 Qg is a field. For d = 1 this is just the rational numbers, which contains Z, the ring of integers.p In this section we will introduce the analog of the integers for the field Q( d). p Definition Suppose d is -free. The ring of integers of ( d) denoted O p Q Q( d) is defined as p O p = fα 2 ( d): α is an algebraic integerg: Q( d) Q It turns out that O p is a ring, however this is not clear from the definition. Q( d) p Also, note that ⊂ O p since ⊂ ( d) and every integer is an algebraic Z Q( d) Z Q integer.
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