# Isomorphism

• Mar 27th 2006, 12:17 PM
suedenation
Isomorphism
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:
• Mar 27th 2006, 02:12 PM
ThePerfectHacker
Quote:

Originally Posted by suedenation
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:

(a) No. The integers are not dense. And the rationals are dense.

(b)No. Because $\mathbb{C}$ is a commutative ring, while $\mathbb{H}$ the quaternion numbers are not. Because for example, $i\times j=k$ while $j\times i=-k$

(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.

• Mar 28th 2006, 06:18 PM
ThePerfectHacker
(d) Nope.

Explanation
-------------
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 $F_1$ and $F_2$ 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 $(a,b)\in F_1\times F_2$. Now, $(0,0')\in F_1\times F_2$ acts like a zero. Thus, we need to show that for all $(a,b)\in F_1\times F_2$ has an inverse $(a^{-1},b^{-1})$ where not both $a,b$ are zero. BUT $(0,a)$ is a nonzero element where $a\not= 0$ has no multiplicative inverse. Because $(0,a)(x,y)=(0,ay)$ thus, there is no way how it can produce $(1,1')$ which acts like a multiplicative inverse in $F_1\times F_2$

Thus, $\mathbb{Q}$ is a field. And $\mathbb{Z}_2$ is also a field. But $\mathbb{Q}\times \mathbb{Z}_2$ cannot be field. Thus, there is no isomorphism between $\mathbb{Q}$ and $\mathbb{Q}\times\mathbb{Z}_2$ because it is not a field.

Note: By $0,0'$ I mean that $0\in F_1$ which is the additive identity and $0'\in F_2$ which is the additive identity. Similary $1,1'$.
Q.E.D.