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Both additive groups have the same cardinality and are divisible. Thus, my initial guess would be that they are isomorphic as abelian groups. But I'm not sure how to prove this. They have different dimensions as vector spaces over $\mathbb Q$, so I can't prove it that way...

Nishant
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    Nope. The $\mathbb Q$-vector space structure can be recovered from the structure as an abelian group, so if they're not isomorphic as vector spaces, then they're not isomorphic as abelian groups. – Dustan Levenstein May 18 '14 at 03:30
  • Wait, why isn't it possible for a vector space to have different dimensions over a field defending on how you define the action of the field on the abelian group? – Nishant May 18 '14 at 04:14
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    @Nishant: this is possible in general, but not for $\mathbb{Q}$. The point is that the axioms uniquely specify what the action of $\mathbb{Z}$ on any abelian group has to be, and then if there exists any extension of this action to an action of $\mathbb{Q}$ then the group is necessarily uniquely divisible and this uniquely determines the extension. – Qiaochu Yuan May 18 '14 at 04:52
  • Any finite subset of the rationals generates a cyclic group. But this is not true of the algebraics. – Gerry Myerson May 18 '14 at 12:34
  • Out of curiosity, what's an example of a vector space with different dimensions over a field depending on the chosen action of the field? – Nishant May 18 '14 at 16:18
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    $\mathbb R$ and $\mathbb C$ have the same dimension over $\mathbb Q$, so the additive group $\mathbb C$ can be equipped with the structure of a 1-dimensional $\mathbb R$-vector space. – Dustan Levenstein May 18 '14 at 17:00

2 Answers2

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Of course this question has been answered in the comments already, but here a more explicit proof that such isomorphism indeed does not exist:

Let $\Phi: \mathbb{Q} \rightarrow \bar{\mathbb{Q}}$ a homomorphism of additive groups.

It is $\Phi(1) \cdot \sqrt{2} \in \bar{\mathbb{Q}}$. We claim it has no preimage. Assume $a, b \in \mathbb{Z}$ such that $\Phi(\frac{a}{b})= \Phi(1) \cdot \sqrt{2}$.

Adding both sides $b$ times implies $a \cdot \Phi(1) = b \cdot \Phi(1) \cdot \sqrt{2}$. Hence $\frac{a}{b} = \sqrt{2}$. Contradiction.

Louis
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Fact. Any $\mathbb{Z}$-linear map between $\mathbb{Q}$-vector spaces is already $\mathbb{Q}$-linear.

Hence, two $\mathbb{Q}$-vector spaces are already isomorphic when their underlying abelian groups are isomorphic. Put differently, any two $\mathbb{Q}$-vector spaces with different $\mathbb{Q}$-dimensions are not isomorphic as abelian groups. The $\mathbb{Q}$-dimension of $\overline{\mathbb{Q}}$ is even infinite.

Martin Brandenburg
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