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Unit 14: Rings
We define + and . in Z + iZ to be the usual addition and multiplication of complex numbers. Notes
Thus, foram + in and s + it in Z + iZ,
(m + in) + (s + it) = (m + s) + i(n + t), and
(m + in) . (s + it) = (p nt) + i(mt + ns).
Verify that Z + iZ is a ring under this addition and multiplication. (This ring is called the ring of
Gaussian integers, after the mathematician Carl Friedrich Gauss.)
Solution: Check that (Z + iZ, +) is a subgroup of (C, I). Thus, the axioms RI-R4 are satisfied. You
can also check that
((a + ib) . (c + id)) . (m + in) = (a + ib) . ((c + id) . (m + in))
a + ib, c + id, m + in Z + iZ.
This shows that R5 is also satisfied.
Finally, you can check that the right distributive law holds, i.e.,
((a + ib) + (c + id)) . (m + in) = (a + ib) . (m + in) + (c + id) . (m + in) for any a + ib, c + id, m +
in Z + iZ.
Similarly, you can check that the left distributive law holds. Thus, (Z + iZ, + , .) is a ring. The
operations that we consider in it are not the usual addition and multiplication.
Example: Let X be a non-empty set, (XI ) be the collection of all subsets of X and A
denote the symmetric difference operation. Show that ((X), A, n) is a ring.
Solution: For any two subsets A and B of X,
A B = (A\B) (B\A)
It is clear that ( (X), A) is an abelian group. You also know that is associative. Now let us see
if distributes over A.
Let A, B, C E (X). Then
A (B C) = A [(B\C) (C\B)]
= [A (B\C)][A (C\ B)], since n distributes over U.
= [(A B)\(A C)][(A C)\(A B)], since distributes over complementation.
= (A B) A (A C).
So, the left distributive law holds.
Also, (B C) A = A (B C), since is commutative.
= (A B) P (A C)
= (B A) A ( C A).
Therefore, the right distributive law holds also.
Therefore, ( (X), A, ) is a ring.
So far you have seen examples of rings in which both the operations defined on the ring have
been commutative. This is not so in the next example.
Example: Consider the set
a 11 a 12
M (R) = a a a, a , a , and a; are real numbers
12
21
2
21 22
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