# Math Help - Isomorphism

1. ## 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.

2. 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.
(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.

3. (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 $F_1$ 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.