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.Originally Posted by suedenation
(b)No. Because $\displaystyle \mathbb{C}$ is a commutative ring, while $\displaystyle \mathbb{H}$ the quaternion numbers are not. Because for example, $\displaystyle i\times j=k$ while $\displaystyle j\times i=-k$
(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.
I am thinking about (d)
(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 $\displaystyle F_1$ and $\displaystyle F_2$ 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 $\displaystyle (a,b)\in F_1\times F_2$. Now, $\displaystyle (0,0')\in F_1\times F_2$ acts like a zero. Thus, we need to show that for all $\displaystyle (a,b)\in F_1\times F_2$ has an inverse $\displaystyle (a^{-1},b^{-1})$ where not both $\displaystyle a,b$ are zero. BUT $\displaystyle (0,a)$ is a nonzero element where $\displaystyle a\not= 0$ has no multiplicative inverse. Because $\displaystyle (0,a)(x,y)=(0,ay)$ thus, there is no way how it can produce $\displaystyle (1,1')$ which acts like a multiplicative inverse in $\displaystyle F_1\times F_2$
Thus, $\displaystyle \mathbb{Q}$ is a field. And $\displaystyle \mathbb{Z}_2$ is also a field. But $\displaystyle \mathbb{Q}\times \mathbb{Z}_2$ cannot be field. Thus, there is no isomorphism between $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{Q}\times\mathbb{Z}_2$ because it is not a field.
Note: By $\displaystyle 0,0'$ I mean that $\displaystyle 0\in F_1$ which is the additive identity and $\displaystyle 0'\in F_2$ which is the additive identity. Similary $\displaystyle 1,1'$.
Q.E.D.