If (S, +) is an abelian group, is it possible to construct two non-isomorphic rings from (S, +)?

Question

Suppose (S, +, *) and (S,+, #) are two rings. Then since both rings have the same abelian group (S, +), must (S, +, *) and
(S, +, #) necessarily be isomorphic?

More generally, if (R, +, *) and (S, @, $) are two rings, and the abelian groups (R, +) and

(S, @) are isomorphic, then must the two rings (R, +, *) and (S, @, $) necessarily be isomorphic to each other?

Inquisitively,

Edwin

Answer 1

The answer is no. Here are two nonisomorphic rings that have the same underlying abelian group:

EXAMPLE 1: The rational integers modulo 8, with multiplication over addition. The underlying group has representatives {1,3,5,7}, and is the direct product of two cyclic groups of order 2,
namely {1,3} and {1,5}, since 3 X 5 = 15 is congruent to 7 modulo 8.

EXAMPLE 2: The set is {a + bi: a and b are viewed as rational integers modulo 2}, and again, the operations are multiplication over addition. This is also known as the ring of Gaussian integers modulo 2. The underlying group is represented by {0,1,i,1+i}. Again, this group is the direct product of the two cyclic groups {0,1} and {0,i}, both under addition modulo 2.

To prove that these rings are not isomorphic, we observe that the square of (1 + i) is 2i, which is equivalent to 0 in the second ring, whereas in the first ring, there is no odd integer modulo 8 whose square is congruent to 0 modulo 8.

The above example contradicts the second part of this query as well.

Answer 2

There is also no rational justification for believing inductive reasoning (that is the past will resemble the future), because we use the past resembling the future as a basis for saying the past will resemble the future (which is circular). However, living based on experience is of utmost importance to survival and it works (think science and technology). There is a logical reason for believing deductive reasoning, because if the premises of the the argument are correct then the conclusion must be correct by definition. E.g. All bachelors are unmarried. Tom is a bachelor, therefore Tom is not married.

Answer 3

I guess the usual product and the convolution product give you a counter example on the space of bounded and integrable real functions. But of course you could argue that the Fourier transform actually gives an isomorphism except that it is not a one to one mapping from that space onto itself...