Consider the subset S 0 2 4 6 8 10 12 of Z14Z a Show that S

Consider the subset S = {0, 2, 4, 6, 8, 10, 12} of Z/14Z. (a) Show that S = {0, 2, 4, 6, 8, 10, 12} forms a subring of Z/14Z. (b) Show that S forms a unital ring. (c) Show that S is not a unital subring of Z/14Z.

Solution

(a) Notice that all elements of S can be written as 2k

Let 2k, 2j in S

Then 2k+2j = 2(k+j) is in S, so clearly sum is in S

Also (2k)(2j) = 4kj = 2(2kj) is in S and so product is also in S

We can also see that 0, the additive identity of Z/14Z is in S by definition

Finally if 2k is in S, then -2k is in S and so S is subring of Z/14Z

(b) The ring has identity 8

Because 2.8 = 16=2, 4.8= 32 = 4, 6.8 = 48 = 6, 8.8 = 64 = 4, 10.8 = 80 = 10, 12.8 = 96 = 12

Hence S is unital

(c) As we can see that we have {2,4,6,10,12} and here 8 is missing and so S is not unital subring of Z/14Z