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Please me some clues verifying the following groups are not isomorphic. I 've found this problem interesting.

  1. $(\mathbb Z,+)$ and $(\mathbb Q,+)$
  2. $(\mathbb Q,+)$ and $(\mathbb R,+)$
  3. $(\mathbb Z[x],+)$ and $(\mathbb Q[x],+)$
  4. $(\mathbb Q[x],+)$ and $(\mathbb Q,+)$

Appreciate your help.

  • 5
    Hints for the first two. 1) Is $(\mathbb{Q},+)$ cyclic?. 2) Is $\mathbb{R}$ countable?...2012-10-13

3 Answers 3

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  1. In $\Bbb Q$, $z+z=c$ is solvable for all $c$.
  2. There is no bijection $\Bbb Q\to\Bbb R$.
  3. same as 1. (for $c=1$, say)
  4. Consider the image of $1$ of a homomorphism $f:\Bbb Q\to \Bbb Q[x]$.
  • 0
    Ah I was having a dense moment.2012-10-13
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Although Berci's first nice hint is easy to understand, I found another way just for it. $\mathbb Q$ is a divisible group but $\mathbb Z$ is not.

  • 0
    Nice example, Babak2013-03-28
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For the 4th part:

$f : \mathbb{Q} \to \mathbb{Q}[X]$ is isomomorphism let $f(1)=a$ then $f(m)=f(1 + \dots + 1) = mf(1) = ma$ also $f(0)=0$ thus we get $f(1-1)=f(0)$ which implies $f(1)+f(-1)=0$ thus $f(-1)=-f(1)=-a$ then $f(p) = p*f(1)$ also $f(p)=f(q*\frac{p}{q})=q*f(\frac{p}{q})$ from this we can get $f(\frac{p}{q})=\frac{p}{q}*f(1)$ therefore for all $r \in \mathbb{Q}$, $f(r)=rf(1)$ then it can be said $f$ cannot be surjective.