2 let r z10z mod 10 and s 0 2 4 6 8 then i check as to...

2. Let R = Z10(Z mod 10) and S = { 0, 2, 4, 6, 8 }. Then (i) check as to whether S is a subring of R which contains an identity or not. (ii) find all the idempotents which exist in R.

Transcript

00:01 In this problem we are provided with the ring r which equals to z10, that is, it consists of the elements 0, 1, 2, 3, so on, up to 9.

00:17 And we are given the set s which equals to 0, 2, 4, 6 and 8.

00:25 In the first subpart, we are asked to check if s is a subring of r.

00:36 So let us do so by checking the four conditions.

00:41 The first condition is if s is closed under addition.

00:53 We know that all the numbers in s are clearly even numbers and the sum of two even numbers is also even.

01:02 And when we divide this even number by 10, the remainder that we will get would be 0, 2, 4, 6 or 8 only, which implies that the set s is closed under addition.

01:19 Next, let us move towards the second condition.

01:22 The second condition is the 0 element.

01:24 It is clear that 0 belongs to the set s.

01:28 So therefore the 0 element is present in s.

01:31 Next, the third condition is the existence of negatives.

01:40 In other words, the existence of the inverses.

01:43 So we know that the inverse of true 2 is negative 2, but negative 2 is congruent to 8 mod 10 and 8 is present in the set s.

01:58 Likewise, negative 4 is congruent to 6 mod 10 and 6 is also present in the set s.

02:06 Negative 6 is congruent to 4 mod 10.

02:11 4 is present in the set s and lastly negative 8 is congruent to 2 mod 10.

02:20 And 2 is present in the set s.

02:22 Therefore we can say that the set s is closed under negatives.