MTH6128 Number Theory 2 Algebraic Numbers

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 polynomials 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 rational 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 ∈ Z.)

1 Note that there are two extra conditions in the definition of an algebraic integer. 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 √ • 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. √ √ 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 ∈ Z, b ∈ 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 | a. But also p | b, so p | 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 polynomial 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 and√ 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 number 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 polynomial 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 look√ 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 | d ⇒ p2 6 | 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 ∈ , d squarefree. √Z √ Now for any rational number q, we can write q = v d where v ∈ Q and d is a squarefree integer. For example, r r r280 23 · 5 · 7 23 · 3 · 5 · 7 · 11 2 √ = = = 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:

Proposition√ 2.3 A complex number α is a quadratic number if and only if α = u + v d where u, v ∈ 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 ∈ Z, b > 0, so that 4 − q = b = b2 ab = b2 c d for p c some c, d ∈ Z, c > 0, d 6= 1, d squarefree. Set u = − 2 , v = ± b .

4 √ Remark√ The expression√ 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 ∈ Q and squarefree integers d1, d2 not equal to 1, then u1 = u2, v1 = v2 and d1 = d2. √ 2.3 The ring of integers of Q( d) √ √ Recall that Q( d) = {u + v d : u, v ∈ Q} is a field. For d = 1 this is just the rational numbers, which contains Z, the ring of integers.√ In this section we will introduce the analog of the integers for the field Q( d). √ Definition Suppose d is -free. The ring of integers of ( d) denoted O √ Q Q( d) is defined as √ O √ = {α ∈ ( d): α is an algebraic integer}. Q( d) Q It turns out that O √ is a ring, however this is not clear from the definition. Q( d) √ Also, note that ⊂ O √ since ⊂ ( d) and every integer is an algebraic Z Q( d) Z Q integer. In the following proposition we will characterize the quadratic integers.

Proposition√ 2.4 A quadratic number α is a quadratic integer if and only if α = u + v d, where d is a squarefree integer not equal to 1 and either

• u, v ∈ Z, v 6= 0, or

1 1 • u − 2 , v − 2 ∈ Z and d ≡ 1 (mod 4). √ Example The golden ratio φ = (1 + 5)/2 is a quadratic integer, with u = 1 2 v = 2 and d = 5. It is a root of the polynomial x − x − 1. √ Example The complex number ω = (−1 + −3)/2 is a quadratic integer. Indeed, it is an algebraic integer since it is a cube root of unity; that is, it is a root of x3 − 1. Now x3 − 1 = (x − 1)(x2 + x + 1), so ω has minimal polynomial x2 + x + 1. √ √ Example 1/ 2 = 1 2 is an algebraic number but not an algebraic integer, 2 √ √ 1 since it has the form u+v d with u = 0, v = 2 and d = 2. Similarly, (3+ 7)/2 is an algebraic number but not an algebraic integer; we have u = 3/2, v = 1/2, and d = 7 6≡ 1 (mod 4).

5 The proof of Proposition 2.4 is not diﬃcult, but for sake of brevity has been moved to an appendix. I won’t expect you to know√ the proof. A naive guess would be that O √ = {m + n d : m, n ∈ }, but if d = 5 Q( d) √ Z in the example above we saw that φ = (1 + 5)/2 ∈ O √ so the guess is not Q( 5) quite correct. Using the proposition we can now describe O √ . Q( d) Proposition 2.5 For d square-free we have √ √ O √ = {m + n d : m, n ∈ } =: [ d] if d ≡ 2, 3 (mod 4) Q( d) Z Z and √   " √ !#  1 + d  1 + d O √ = m + n : m, n ∈ =: if d ≡ 1 (mod 4) Q( d) Z Z  2  2

√ Example For d = −1 we get OQ( −1) = Z[i], which is referred to as the Gaussian integers.

√ h √ i Example We saw the golden ratio φ = (1 + 5)/2 ∈ O √ = 1+ 5 . Q( 5) Z 2

Remark Using Proposition 2.4 it is a straightforward exercise to prove that O √ is a ring. Q( d) √ Proof If d ≡ 2, 3 (mod 4) then by Proposition 2.4 α = u + v d ∈ O √ iﬀ √ Q( d) u, v ∈ Z iﬀ α ∈ Z[ d]. The case d ≡ 1 (mod 4) is slightly harder. h √ i Step 1. Show that 1+ d ⊂ O √ (Homework). Z 2 Q( d) h √ i Step 2. Show that O √ ⊂ 1+ d . If α ∈ O √ and α ∈ then Q( d) Z 2 Q( d) Z h √ i √ 1+ d √ clearly α ∈ Z . By Proposition 2.4 if α = u + v d ∈ O ( d) and is not 2 √ Q 1 1 in Z then α = u − 2 + (v − 2 ) d where u, v ∈ Z. Let a = 2u − 1 and b = 2v − 1 so a, b are odd and √ ! " √ !# a b √ b − a 1 + d 1 + d α = + d = + b ∈ . 2 2 2 2 Z 2

6 2.4 Units in O √ and Pell’s equation Q( d) By analogy with the integers number theorists are also interested in the arithmetic structure of O √ . One of the most fundamental questions one can ask is: Q( d) ”What are the units in O √ ?”. Q( d) Recall that in a ring R an element α ∈ R is a unit provided that α · β = 1R for some β ∈ R.

Example The units in Z are ±1.

Example In Z[i], ±1, ±i are units. In fact, these are all the units. √ √ Deﬁnition Given α = u + v d ∈ Q( d) the conjugate of α, denoted by α is √ α = u − v d. √ √ √ √For example the conjugate of 1 + 2 is 1 − 2, and the conjugate of 5 is − 5. Also, α is rational if and only if α = α (why?). Here are some properties of conjugates. √ Lemma 2.6 Suppose α, β ∈ Q( d) where d is squarefree then a) α = β if and only if α = β √ √ b) αα ∈ Z if α = u + v d ∈ Z[ d] c) αβ = αβ √ √ Proof Write√ α = u +√v d and β = r + s d. To prove a) notice that if α = β then u + v d = r + s d so √ (u − r) = (v − s) d since the left hand side is a rational number we must have v − s = 0 (otherwise the right√ hand side√ would be irrational). So u = r and v = s. This implies u − v d = r − s d as desired. Now consider √ √ 2 2 αα = (u + v d)(u − v d) = u − dv ∈ Z whenever u, v ∈ Z. This proves b). Finally, √ √ √ √ √ αβ = (u + v d)(r + s d) = ur+vsd−(us+vr) d = (u−v d)(r−s d) = αβ, which proves c).

7 Proposition√ 2.7 Suppose that d ≡ 2, 3 (mod 4) is squarefree. An element α = u + v d ∈ O √ is a unit if and only if Q( d)

u2 − dv2 = ±1.

Remark The equation x2 − dy2 = ±1 is known as Pell’s equation. Finding the solutions of this equation will be a major topic of this course. The above proposition shows that ﬁnding solutions to Pell’s equation for squarefree d,√ with d ≡ 2, 3 (mod 4) corresponds to ﬁnding units in the ring of integers of Q( d).

Example In Z[i], α = u + vi is a unit iﬀ u2 + v2 = ±1. The only way this can happen is if u = 0, v = ±1 or u = ±1 and v = 0. Hence the units in Z[i] are precisely ±1, ±i. √ Remark√ It turns out that Z[ 3] has inﬁnitely many units! We’ll see that (2 + 3)n is a unit for each n ∈ N . √ √ Proof√(⇒) Suppose that α = u + v d ∈ Z[ d] is a unit. Then there exists β ∈ Z[ d] such that αβ = 1 so by the lemma, part a), αβ = 1. Combining these equations gives 1 = αβαβ = (αα)(ββ) by part c) of the lemma. By part b) we know ββ ∈ Z and also αα = u2 − dv2 so we get 1 = (u2 − dv2) · ” an integer”. Hence u2 − dv2 is a unit in Z so u2 − dv2 = ±1. (⇐) Suppose that u2 − dv2 = ±1. Then

(αα)2 = (u2 − dv2)2 = 1 √ 2 so α(α · α ) = 1 so that α is a unit in Z[ d].

2.5 Appendix: The proof of Proposition 2.4

Proof Proof of Proposition 2.4√ Let α be a quadratic number. By the Proposi- tion 2.3, we can write α = u + v d, where u, v ∈ Q and d is a squarefree integer not equal to 1.

8 2 Let f(x) = x √− ax + b be the minimal polynomial√ of α. So one root of f(x) is α =√u + v d, the other√ one has to be = u − v d. Therefore, f(x) = (x − (u + v d))(x − (u − v d)). We conclude that

• a = 2u, and

• b = u2 − dv2.

So α is an algebraic integer if and only if 2u, u2 − dv2 ∈ Z. In particular, 2u ∈ Z 1 means that either u ∈ Z, or u − 2 ∈ Z, depending on whether 2u is an even or an odd integer.

Case 1: u ∈ Z. Then dv2 ∈ Z. Since d is squarefree, this implies that v ∈ Z. a Suppose not. Write v = b for a, b ∈ Z, b > 1, gcd(a, b) = 1. By assumption, 2 da2 2 2 dv = b2 ∈ Z. So b | da . Let p be a prime with p | b. Such a prime exists if b > 1. Then p2 divides b2. Thus p2 | da2. But we also have p 6 | a as gcd(a, b) = 1. It follows that p2 | d. This is a contradiction since d is squarefree.

1 1 Case 2: u − 2 ∈ Z. Say u = s + 2 where s ∈ Z. We have 1 s2 + s + − dv2 ∈ , 4 Z

2 1 2 so dv − 4 ∈ Z, and in particular 4v d − 1 ∈ 4Z. The last condition shows that v∈ / Z; for if it is, then 4v2d − 1 would be odd. a Again, write v = b for a, b ∈ Z, b > 0, gcd(a, b) = 1. By assumption, 2 4da2 2 2 n 4dv = b2 ∈ Z. So b | 4da . Let p be a prime, n ≥ 1 an integer with p | b. Then p2n divides b2. Thus p2n | 4da2. But we also have p 6 | a as gcd(a, b) = 1. It follows that p2n | 4d. As d is squarefree, this implies p2n−1 | 4. Hence p = 2 and 1 n = 1. So the denominator of v must be 2, and v = t + 2 for some t ∈ Z. Then

2 2 2 1 2 1 u − dv = s + s + 4 − d(t + t + 4 ), so (1 − d)/4 ∈ Z, whence d ≡ 1 (mod 4).

9 Progress Check 1. What is a monic polynomial? Give some examples. It is sug- gested 2. What is an algebraic number? Give some examples. you discuss 3. What is an algebraic integer? Give some examples. these questions with 4. What is a transcendental number? Give an example. your √ collea- 5. Show that 3 is an algebraic integer. gues to reinforce √ your 6. Show that ( −3 + 1)/2 is an algebraic integer. (Hint: call this number under- 2 2 standing x, calculate x and see if you can subtract a multiple of x to x to get of the an integer.) lesson. You √ should 7. Show that ( 3 + 1)/2 is not an algebraic integer. (Hint: do as in the try to previous exercise to ﬁnd an quadratic polynomial that has this number respond √ these as a root and then use Theorem 2.2). Do the same for −3/2. questions √ without 8. What is a ﬁeld? What is ( d)? your Q lecture notes to 9. What does O √ mean? Q( d) check whether 10. What is a quadratic number? Give a couple of examples. you have learnt the 11. What is a quadratic integer? Give a couple of examples. material.

12. What is a square-free integer? Give an example of a square-free integer and an example of an integer that is not square-free.

13. Assume d is square free. What are the elements of O √ in the case Q( d) d ≡ 2, 3 (mod 4)?

14. Assume d is square free. What are the elements of O √ in the case Q( d) d ≡ 1 (mod 4)?

15. What are the Gaussian integers?

16. What is the Golden ratio? Is it an algebraic integer? Why? √ √ 17. Calculate the norm of 2 + 3 2 and of 1 − −5.

10 18. What are the Gaussian integers of norm 1?

19. What is Pell’s equation?

20. How is Pell’s equation related to the norm of a quadratic integer?

More interesting facts This is 1. Note that the norm of a quadratic number not exa- √ minable N(x + y d) = x2 − dy2, x, y ∈ content Q but will is related but not exactly the same as the norm of a real or complex help your under- number. standing √ of the (a) In the case d is positive then x + y d is a real number√ and it is course 2 2 clear that its norm as an algebraic number√N(x+y d) = x −dy is not the same as the real number |x + y d|. 2 2 Note that the√ norm x − dy is not even guaranteed to be positive, e.g. N(1 − 2) = 12 − 2 · 12 = −1. √ √ (b) In the case d is negative then x + y d = x + y −di is a complex√ number and again its norm as an algebraic number N(x + y d) = 2 2 x −dy is similar√ but not the same as the norm of it as a complex number |x + y −di| = px2 − dy2. √ 2. The Golden ratio (1 + 5)/2 has been known since antiquity and is meant to reﬂect some ’ideal’ aesthetic proportion. It turns up in archi- tecture, music, biology, painting, ﬁnance etc.

11 See https: // en. wikipedia. org/ wiki/ Golden_ ratio for more information.

3. Incidentally note that the A4, A5, . . . system of paper sizes is not based on the golden ratio but on the fact that when you fold an A4 sheet in half, the resulting A5 sheet has the same proportions. Can you ﬁnd from this the ratio between the sides of an A4 sheet? (Hint: it is an algebraic integer).

4. Note that the equation deﬁning the golden ratio x2 − x − 1 = 0 can also be written as x−1 = x − 1, i.e. the golden ratio, x = 1.61803398875 ..., is a number such that when you invert it you get the initial number minus one: x−1 = 0.61803398875 .... There is only one more number with this same property, calculate it. At some point the value in euro of 1 sterling pound was equal to the golden ratio, GBP EUR = 1.61803398875 ..., so that the reciprocal ex- change rate, pounds per euro was the same minus one: EURGBP = 0.61803398875 .... The following R code will display when this happe- ned:

5. The name Pell equation is a miss-attribution by none other than Le- onard Euler. It seems that John Pell had nothing to do with what we call the Pell equation.

12 6. The famous cattle problem of Archimedes (287-212 BC): Compute, O friend, the number of the cattle of the sun which once grazed upon the plains of Sicily, divided according to color into four herds, one milk-white, one black, one dappled and one yellow. The number of bulls is greater than the number of cows, and the relations between them are as follows: 1 1 • White bulls = 2 + 3 black bulls + yellow bulls, 1 1 • Black bulls = 4 + 5 dappled bulls + yellow bulls, 1 1 • Dappled bulls = 6 + 7 white bulls + yellow bulls, 1 1 • White cows = 3 + 4 black herd, 1 1 • Black cows = 4 + 5 dappled herd, 1 1 • Dappled cows = 5 + 6 yellow herd, 1 1 • Yellow cows = 6 + 7 white herd. If thou canst give, O friend, the number of each kind of bulls and cows, thou art no novice in numbers, yet can not be regarded as of high skill. Consider, however, the following additional relations between the bulls of the sun: • White bulls + black bulls = a square number, • Dappled bulls + yellow bulls = a triangular number. If thou hast computed these also, O friend, and found the total number of cattle, then exult as a conqueror, for thou hast proved thyself most skilled in numbers.

(https: // en. wikipedia. org/ wiki/ Archimedes% 27s_ cattle_ problem ) can be reduced and solved by using the Pell equation. The exact solution took around 2000 years to calculate and was found in 1965. The answer contains 206,545 digits.