Question

5. Show that the ring Z[7] = {a+bV7 | a, b e Z} is an integral domain.

Answer 1

For, a = 1 and b = 0, the element 1 belongs to Z[√7] is the identity of the Ring.

Let, x = a + b√7 and y = c + d√7 be 2 elements in the ring.

Then, x•y = (a+b√7)•(c+d√7) = ac + 7bd + √7(ad+bc)

&, y•x = (c+d√7)•(a+b√7) = ca + 7db + √7(cb+da) = ac + 7bd + √7(ad+bc) = x•y

So, x•y = y•x implies, the ring is commutative.

So, Z[√7] is a commutative ring with identity.

Next, we are to show that, the ring has no zero divisor.

Let, x & y be as above and, x•y = 0

So, ac + 7bd + √7(ad+bc) = 0

So, ac + 7bd = 0 = ad + bc

If we put, a = 0 then, bd = 0 = bc

If, b is non-zero then both c & d are zero & consequently y is 0.

If, b = 0 then consequently, x = 0

So, x•y = 0 implies either x = 0 or y = 0

So, Z[√7] has no zero divisor.

So, Z[√7] is an integral domain.