Does anyone know how to do this question?

In each of the following cases, decide whether or not the two rings are isomorphic.

(a) Z and Q.

(b) C and H.

(c) Z6 × R and R × Z6.

(d) Q and Q × Z2.

(e) R and C.

Thanks very much. :confused:

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- March 27th 2006, 12:17 PMsuedenationIsomorphism
Does anyone know how to do this question?

In each of the following cases, decide whether or not the two rings are isomorphic.

(a) Z and Q.

(b) C and H.

(c) Z6 × R and R × Z6.

(d) Q and Q × Z2.

(e) R and C.

Thanks very much. :confused: - March 27th 2006, 02:12 PMThePerfectHackerQuote:

Originally Posted by**suedenation**

(b)No. Because is a commutative ring, while the quaternion numbers are not. Because for example, while

(c)Yes. I really do not see how order is important in a direct product between rings.

(e)No. The reals are ordered but the complex numbers cannot be ordered.

I am thinking about (d) - March 28th 2006, 06:18 PMThePerfectHacker
(d) Nope.

Explanation

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It is known that the direct product between two groups is a group. And also the direct product between two rings is a ring. But the direct product between two fields IS NOT a field EVER.

Proof: Let and be fields. Then, by definition a field is a commutative division ring. Meaning that every non-zero element in a field has a multiplicative inverse. Thus, we have . Now, acts like a zero. Thus, we need to show that for all has an inverse where not both are zero. BUT is a nonzero element where has no multiplicative inverse. Because thus, there is no way how it can produce which acts like a multiplicative inverse in

Thus, is a field. And is also a field. But cannot be field. Thus, there is no isomorphism between and because it is not a field.

Note: By I mean that which is the additive identity and which is the additive identity. Similary .

Q.E.D.