3+4=2 5+6=3 7 4=4. a + b =(a + b) mod m

Transcription

1 Rings and fields The ring Z m -part2(z 5 and Z 8 examples) Suppose we are working in the ring Z 5, consisting of the set of congruence classes Z 5 := {[0] 5, [1] 5, [2] 5, [3] 5, [4] 5 } with the operations of addition and multiplication [a] m +[b] m =[a + b] m and we agree to do two things: [a] m [b] m =[a b] m State up front that we re in the universe of Z 5, and drop the boxes and the subscripts. It s assumed that everything in context is being worked with modulo 5. Agree to use the principal representatives of each of the classes. If we get a different representative, we immediately knock it back down mod 5 so that it s one of 0, 1, 2, 3, 4. Instead of writing [3] 5 +[4] 5 =[7] 5 =[2] 5 we ll just announce at the start, in Z 5! and write Compute in Z 8 the values of = =3 7 4=4 Under these conventions, we can describe the ring Z m as the set {0, 1, 2,.., m 1}, with operations of addition and multiplication defined by a + b =(a + b) mod m a b =(a b) mod m i.e., do the computations like you normally would off to the side, mod the result down, and come back with your answer.

2 Since Z 5 has a finite number of elements, we can organize the results of all possible combinations under the operation of addition easily on a Cayley table: We can also do a Cayley table for multiplication in Z 5. Finish filling in the table: More properties We have established that for any Z m, the commutative properties of addition and multiplication, the associative properties of addition and multiplication, and the distributive property of multiplication over addition all hold. Now we look at specific Z m to see what other properties hold. Closure under addition? Closure under multiplication? Additive identity? Multiplicative identity? Additive inverse? Multiplicative inverse?

3 Run down the list for Z 5.Dowehave Closure under addition? Closure under multiplication? Additive identity? Multiplicative identity? Additive inverse? Multiplicative inverse?

4 Do the whole thing again, this time with Z 8 : construct Cayley tables for both addition and multiplication, and run through all the properties.

5 Generalizations For any m, Z m will be closed under addition and multiplication. This follows from the definitions of those operations; we always reduce things modulo m to guarantee we ll never leave the set {0, 1, 2,..., m 1}. For any m, we will have 0 as the additive identity, and 1 as the multiplicative identity. This follows from the definitions again: a +0=(a +0) mod m = a mod m = a a 1=(a 1) mod m = a mod m = a (since a is already such that 0 a<m, a mod m = a). For any m, every element will have an additive inverse, and the additive inverse of a will always have the value m a: a +(m a) =(a +(m a)) mod m = m mod m =0 0 will never have a multiplicative inverse in any modulus; the congruence 0x 1(modm) can t have a solution. Whether or not every nonzero element will have a multiplicative inverse comes down to whether the modulus is prime or not. The general rule is that only the elements relatively prime to m will have multiplicative inverses. If m is prime, all nonzero elements will have multiplicative inverses. The justification for this is that it s equivalent to asking when there are solutions to the congruence ax 1(modm). We already know that the answer to that is when gcd(a, m) = 1. Terminology A nonzero element x is called a zero divisor if there is a nonzero element y such that the product x y =0. InZ 8, 2 and 4 are both zero divisors, since 2 4 = 0, but neither of those elements is itself zero. The multiplicative identity 1 is sometimes called unity. Which should not be confused with a unit. Any element that has a multiplicative inverse is called a unit. InZ 5,1,2,3,and4areallunits. InZ 8, 1, 3, 5, and 7 are units.

6 Algebra (solving equations) in Z m Rule 1: Since additive inverses exist for every element in any m, it is always safe to add an additive inverse to both sides of an equation to cancel a constant term. This result is analogous to the established result for congruences that ax + b c (mod m) ax c b (mod m) Solve x +5=4inZ 8 by adding the additive inverse of 5 to both sides of the equation. Be sure to perform all arithmetic in Z 8. Rule 2: Since multiplicative inverses exist for every element which is relatively prime to m, we can solve ax = b in Z m by multiplying both sides by the multiplicative inverse of a as long as gcd(a, m) =1. The corollary is that if m is prime, we can solve ax = b for any nonzero a. This result is analogous to the established result that a linear congruence ax b (mod m) hasa solution consisting of one equivalence class if gcd(a, m) = 1. Solve 3x =7inZ 8 by multiplying the multiplicative inverse of 3 to both sides of the equation. Be sure to perform all arithmetic in Z 8.

7 Rule 3a: Multiplicative inverses do not exist for elements a for which gcd(a, m) 1. Ifa and m are not relatively prime, we cannot solve ax = b using inverses. Rule 3b: Unfortunately, that s not always the same as having no solution. We can still solve ax = b in Z m if gcd(a, m) divides b. Rule 3c: Even worse, if d =gcd(a, m) divides b, wegetmultiple solutions... precisely d of them. This is analogous to our semi-cancellation rule for congruences: if we are solving ax b (mod m), and d = gcd(a, m) divides b, then d is a factor of both a and b and we can write d(px) dq (mod m). Cancellation gives px q (mod m d ). Solve 2x =4inZ 8 by solving the equivalent congruence. Finally, we can combine the steps as in solving linear equations to solve equations in the form ax + b = c in Z m.inz 5, solve the equation 3x +4=1.