An Introduction to Proof-based Mathematics Harvard/MIT ESP: Summer HSSP Isabel Vogt
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1 An Introduction to Proof-based Mathematics Harvard/MIT ESP: Summer HSSP Isabel Vogt Class Objectives Field Axioms Finite Fields Field Extensions Class 5: Fields and Field Extensions 1
2 1. Axioms for a field Loosely speaking, a field F is a set of elements for which the familiar operations of arithmetic are defined and behave in the usual way. Here is a set of axioms for a field. You can use them to prove theorems that are true for any field. Short version: (a) The elements of F form a Abelian group under addition. (b) The nonzero elements of F form a Abelian group under multiplication. (c) Distributive law: a(b + c) = ab + ac. Long version: (a) Addition is commutative: a + b = b + a. (b) Addition is associative: (a + b) + c = a + (b + c). (c) Additive identity: 0 such that a F, 0 + a = a + 0 = a. (d) Additive inverse: a F, a such that a + a = a + ( a) = 0. (e) Multiplication is associative: (ab)c = a(bc). (f) Multiplication is commutative: ab = ba. (g) Multiplicative identity: 1 such that a F, 1a = a. (h) Multiplicative inverse: a F {0}, a 1 such that a 1 a = 1. (i) Distributive law: a(b + c) = ab + ac. 2
3 2. Theorems about fields As opposed to the axioms for a group, this set of axioms for a field is not and independent set, meaning some things can be proven as theorems from the other axioms: namely the commutativity of addition. Some well-known laws of arithmetic are omitted from the list of axioms because they are easily proved as theorems. The most obvious omission is a F, 0a = 0. RTP: Given a field F, a F, 0a = 0 Do the integers Z form a field? 3
4 3. Finite fields Building upon what we know of modular arithmetic, we can make from the integers a finite field. Choose a prime number p. Break up the set of integers into p subsets. Each subset is named after the remainder when any of its elements is divided by p. [0] p = {m m = np, n Z} [1] p = {m m = np + 1, n Z} [a] p = {m m = np + a, n Z} Notice that [a + kp] p = [a] p for any k. There are only p sets, but each has many alternate names. These p infinite sets are the elements of the field Z p. Define addition by [a] p + [b] p = [a + b] p. Here a and b can be any names for the subsets, because the answer is independent of the choice of name. The rule is Add a and b, then divide by p and keep the remainder. What is the simplest name for [5] 7 + [4] 7? What is the simplest name for the additive inverse of [3] 7? Define multiplication by [a] p [b] p = [ab] p. Again a and b can be any names for the subsets, because the answer is independent of the choice of name. The rule is Multiply a and b, then divide by p and keep the remainder. What is the simplest name for [5] 7 [4] 7? Notice that here [12] 7 = 12 (mod 7), but the bracket notation merely specifies [12] 7 = [5] 7 as an element of the finite field Z 7 4
5 4. Multiplicative inverses in Z p For Z p, all the properties of a field follow directly from the corresponding properties of integer arithmetic except for the existence of a multiplicative inverse. RTP: a Z p where p is prime, a 1 Z p such that aa 1 = a 1 a = 1 There is an ingenious and efficient way to find the multiplicative inverse of [a] p based on Euclid s algorithm for finding the greatest common divisor of a and p, which we may have time to examine when we turn to number theory. In Z 3, [2] 3 [2] 3 = [4] 3 = [1] 3. In Z 5, [2] 5 [3] 5 = [6] 5 = [1] 5. In Z 5, [4] 5 [4] 5 = [16] 5 = [1] 5. Let s do a few more: In Z 7, what is the multiplicative inverse of [2] 7? In Z 7, what is the multiplicative inverse of [3] 7? In Z 7, what is the multiplicative inverse of [ 1] 7? (This is not the standard name, but it is a convenient one!) In Z 7, what is the multiplicative inverse of [6] 7? 5
6 5. Rational numbers The rational numbers Q form a field. You learned how to add and multiply them years ago! The multiplicative inverse of a is b as long as a 0. b a The rational numbers are not a big enough field for doing Euclidean geometry or calculus. Here are some irrational quantities: 2 π. most values of trig functions, exponentials, or logarithms. coordinates of most intersections of two circles. In cases such as these, it is necessary to extend a field in order to solve certain equations. For example, are there any solutions to the equation x 2 = 2 over the rational numbers? Does the equation x 2 = 2 have any solutions over the set {a+b 2 a, b Q}? 6
7 RTP: The set {a + b 2 a, b Q} forms a field. This is a field extension. We have a Q, and then we extend that set by some rational factor of an element not in Q. Similarly we could have extended the rationals by a factor of 3 to solve the equation x 2 = 3 and had another field extension. 7
8 6. Complex numbers as a field The real numbers (represented, for example, by infinite decimal numbers) also form a field. However, there are still polynomial equations that have no solutions in this field, notably i = 0 We can extend the field of real numbers by introducing a symbol i that represents a solution to this equation. Then, when we compute with the resulting complex numbers, we just replace i 2 by -1. The set notation is again a real number, plus a real number multiple of this element i: {a + bi a, b R} (a) Prove that the field of complex numbers is closed under multiplication. (b) Prove that every nonzero complex number has a multiplicative inverse. 8
9 7. Extending a finite field We have seen that we can: Extend the field of rational numbers by choosing an equation of the form x 2 k = 0, where k is a positive integer that is not a perfect square. The extended field has elements of the form a + b k, where a and b are in the unextended field of rational numbers and k is a positive number. We calculate in the extended field by using the replacement rule k 2 = k Extend the field of real numbers by choosing the equation i = 0. The extended field has elements of the form a + bi, where a and b are in the unextended field of real numbers and i is any solution to i = 0 we cannot tell one from the other. We calculate in the extended field by using the replacement rule i 2 = 1. In the finite field Z 2, the elements are [0] 2 and [1] 2 What is the additive inverse of [1] 2? We can use Z 2 to solve the following polynomials. What are the solutions? u 2 = 0 u = 0 u 2 + u = 0 Does the following polynomial have a solution in Z 2? u 2 + u + 1 = 0 To get around this, we need to extend our finite field! 9
10 We define u to be a solution of u 2 + u + 1 = 0. Given that u must have a multiplicative inverse, what other element must be in this field? We can find it by saying u 2 + u = 1, then factoring to get u(u + 1) = 1 Thus u and u + 1 are multiplicative inverses of each other. What are the additive inverses of u and u + 1? The resulting finite field, called F 4, has four elements, 0, 1, u, and u + 1. The replacement rule is u 2 = u 1 = u + 1. (a) Use the replacement rule to calculate (u + 1) 2. (b) Show that u + 1 is also a solution to the same quadratic equation. Just as we cannot distinguish i from i, we cannot distinguish u from u
11 (c) Make a table of addition facts for F 4 (d) Make a table of multiplication facts for F 4. 11
12 8. Multistage field extensions (a) The set of numbers of the form a + b 2 with rational coefficients, is a field. Show that, in this field, is a perfect square but is not. (b) Let u = What polynomial equation does u satisfy? (c) How can you use u to create an extended field? (d) Another way to make an extended field is to use two independent square roots. For example, use elements like a + b 2 + c 3 where a, b, and c are rational numbers. A representative element is x = What polynomial equation with integer coefficients does x satisfy? (e) In general, when you eliminate a square root by squaring both sides of an equation, you double the degree of the equation. If you write a number that involves n independent square roots, what degree of polynomial equation is it guaranteed to satisfy? 12
13 9. Theorems from field axioms RTP: Given a field F, a, b F, a + b = b + a using only the other axioms of a field and considering (1 + 1)(a + b) RTP: a F, a = ( 1)a. Remenber: a means the additive inverse of a, while 1 means the additive inverse of 1 13
14 RTP: a, b F, ( a)( b) = ab. RTP: If n is not prime, n = ab where neither a nor b is 1. Use this fact to show that Z n is not a field in this case. 14
15 10. Order of Finite Fields We can now see that all finite fields must be either of order p, where p is prime, or of some power of a prime p. We get the fields of order p from the sets Z p. We get the fields of order p n, n Z from field extensions to solve certain polynomial equations. For instance, we derived F 4 from solving a quadratic polynomial (degree 2) with coefficients from Z 2. (2) 2 = 4. We derived F 9 from solving a quadratic with coefficients from Z 3. (3) 3 = 9. We derive F 8 from solving a cubic (degree 3) over Z 2, and so on. 15
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