4. Let M2 (Z) be the ring of all 2 x 2 matrices over the...

4. Let M_2 (Z) be the ring of all 2 x 2 matrices over the integers and let S = {[x x - y], [x - y y] : x, y ? Z}. Prove or disprove that S is a subring of M_2 (Z). Consider the ring D as follows D = {a + ib + cj + dk : a, b, c, d ? R} where i^2 = j^2 = k^2 = -1 and ij = k = -ji, jk = i = -kj, ik = j = -ki. Prove or disprove that D forms a field. Find a zero-divisor and a non-zero element a satisfying a^2 = a (other than 1) in the ring Z_5[i], where Z_5[i] = {x + iy : x, y ? Z_5, i^2 = -1}.

Transcript

00:03 Hello, in this question we are, in the first part we are given this subset of the ring of integers, a ring of matrices, 2 by 2 matrices with entries in integers, with integer entries.

00:19 This is a subset of this ring and if we take two elements in this subset and we multiply them, the product of these two elements is like this and we can see that if we open this we will get ax, this term is equal to ax minus bx plus bx minus ay, so that this term cancels out and we will get ax minus by it should be.

00:52 We get ax minus by and if we take the difference of these two terms we can see that this product cancels out, this product term cancels out and we get ax minus by.

01:05 So that this thing is the difference of these two.

01:08 Similarly here if we open this we will get ax minus ay plus ya minus by and this minus ay plus ya cancels out, we will get ax minus by which is the difference of these two.

01:25 So we see that the product of any two terms in s is also a term, any two elements in s is again an element in s.

01:34 It is also so s is closed under multiplication.

01:38 Now if we take any two elements of s it is easy to see that it is closed under addition.

01:46 Hence s is a subset of m2g which is closed under multiplication which is closed under addition.

01:53 Also it contains if we take x equal to 1 y equal to 1 you can see that this the element will be 1 1 0 0 which is identity element of m2g.

02:04 So s is a subset of m2g which is closed under multiplication addition and contains identity.

02:11 Hence it will be a subring of m2g.

02:15 S is a subring of m2g.

02:21 Note that other properties to be a subring like associativity, distributivity those things will be inherited because s is a subset in m2g which is a ring.

02:34 So this is the first part of the question.

02:39 In the second part we are given this ring which is called the quaternion ring and in this ring we are asked whether this ring is a field.

02:49 It is not because in the definition itself it is given that j and k are two elements that j times k is equal to i which is equal to minus of k times j.

03:02 This implies that k times j is equal to or minus of k times j is equal to j times k and it is not equal to 0 which means that j k is negative of k times j and hence j and k do not commute.

03:23 Hence j k do not commute.

03:31 It is not equal to 0 is important because if j k were equal to 0 then this condition would be satisfied and they would still commute.

03:38 But as it is not equal to 0 then we have that this implies that j k is not equal to k j.

03:47 J k is not equal to k j.

03:49 Hence we have that j and k do not commute.

03:51 So we have so we see that d is not a commutative ring...