abstract algebra Example 2.58However, Examples 2.55 and 2.56 do generalize suitably: it turns outthat for any prime p, the ring Z/pZ is a field (with p elements). Recallfrom the discussions in Examples 2.20 and 2.21 that the elements ofZ/pZ are equivalence classes of integers under the relation a ~bif andonly if a-b is divisible by p. The equivalence class [a], of an integer a isthus the set of integers of the form a + p, a ± 2p, a ±3p, .... Additionand multiplication in Z/pZ are defined by the rules1. [a]p + [b, = [a + blp2. [a), (b), = [a - b)p youtheExercise 2.58.1yoursShow that addition and multiplication are well-defined, that is, ifa~a' and b ~b', then a +b~d+b and a ·b~d.b.

Given: Addition and multiplication ℤ/pℤ is defined as, ap+bp=a+bp and ap·bp=a·bp respectively. To show: Addition and multiplication in ℤ/pℤ is well defined.To show addition and multiplication in ℤ/pℤ is well defined it is sufficient to show that a~a' and b~b' then a+b~a'+b' and a·b~a'·b' 1) Since we have given, a~b if and only if a-b is divisible by p. Suppose a~a' and b~b', ⇒a-a' and b-b' is divisible by p. Hence, there sum a-a'+b-b' is divisible by p. ⇒a+b-a'+b' is divisible by p. ⇒a+b~a'+b' 2) Since we have given, a~b if and only if a-b is divisible by p. Suppose a~a' and b~b', ⇒a-a' and b-b' is divisible by p. ⇒a-a'=k1p and b-b'=k2p Consider, a·b-a'·b'=ab-ab'+ab'-a'b' =ab-b'+a-a'b' =ak2p+k1pb' =ak2+k1b'p⇒a·b-a'·b'=k3p, where k3=ak2+k1b' Which shows a·b-a'·b' is divisible by p. ⇒a·b~a'b' Hence, addition and multiplication in ℤ/pℤ is well-defined.