Algebra - Groups & Rings

Question

For n > 1, let Zn = {[ 0 ], [ 1 ], [ 2 ],....., [ n - 1 ]}, the congruence classes of the integers modulo n: Show that Zn is a ring with respect to the operations defined by [ a ] + [ b ] = [ a + b ] and [ a ].[ b ] = [ a . b ] for all [ a ], [ b ] belong to Zn

Answer 1

We need to go through all the properties of a ring and verify them for Z_n. First, we need to check that Z_n is an abelian group under addition. Z_n is obviously commutative and associative (a + b = b + a and (a+b) + c = a + (b + c)). We also have [0] as the additive identity element, as [a] + [0] = [a + 0] = [a]. Now, for the additive inverse, we have -[a] = [n-a], as [a] + [n-a] = [a + n - a] = [n] = [0]. By all of these properties, Z_n is an abelian group under addition.

Now, we need to analyze the multiplicative properties. We obviously have [1] as the multiplicative identity, as [a].[1] = [a.1] = a. We also see that Z_n is associative under multiplication, as

([a].[b]).[c] = ([a.b]).[c] = ([a.b.c]) = [a].([b.c]) = [a].([b].[c])

Finally, we need to check that multiplication is distributive under addition. We have:

[a].([b] + [c]) = [a].([b + c]) = [a.(b+c)] = [a.b + a.c] = [a.b] + [a.c] = [a].[b] + [a].[c]

([b] + [c]).[a] = [b+c].[a] = [(b+c).a] = [b.a + c.a] = [b.a] + [c.a] = [b].[a] + [c].[a]

That's all we need to check, and we have that Z_n is a ring.