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Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic under addition?

And how could I prove this ?

What about $\mathbb{Q}$ and $\mathbb{Q}[i]$ ?

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$\bf C$ and $\bf R$ are isomorphic as vector spaces over $\bf Q$ (take uncountably infinite transcendence bases) so they are isomorphic as additive groups. The latter are not isomorphic as $\bf Q$ vector spaces so they are not isomorphic as groups either - alternatively, the two elements $1$ and $i$ in ${\bf Q}(i)$ do not generate a cyclic subgroup, whereas every finitely generated subgroup of $\bf Q$ is cyclic. –  anon Mar 21 '13 at 17:31
But don't try to write down an explicit additive isomorphism of $\mathbb C$ and $\mathbb R$ (you can't). This isomorphism cannot be proved in ZF, it requires choice. –  GEdgar Mar 21 '13 at 17:34
The first one is a duplicate: math.stackexchange.com/questions/302514/… –  Asaf Karagila Mar 21 '13 at 17:49
@anon:You want Hamel bases, not transcendence bases. –  Andreas Blass Mar 21 '13 at 17:59
Err, yes, that's right. –  anon Mar 21 '13 at 18:09

4 Answers 4

In fact, you can also show that : $$\frac{\mathbb C^+}{\mathbb R^+}\cong\mathbb R^+$$ by setting the following surjective homomorphism: $$f:\mathbb C^+\to\mathbb R^+,~~~(a+ib)\to a$$

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$\mathbb Q$ and $\mathbb Q[i]$ are not isomorphic as additive groups. Any two elements of $\mathbb Q$ have a "common divisor" ...

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I'm not quite sure how this proves that they're not isomorphic. –  Kasper Mar 21 '13 at 17:50
In $\mathbb Q[i]$, the two elements $i$ and $1$ have no commonn divisor: there is no $u$ so that both of them belong to the subgroup generated by $u$. –  GEdgar Mar 21 '13 at 17:53

Yes, they are isomorphic as additive groups.

They are in fact isomorphic as vector spaces over $\mathbb{Q}$ as they have the same dimension.

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The groups in question are all additive groups of fields of characteristic zero. In general the additive group $(F,+)$ of a field of characteristic zero is a uniquely divisible group: for all positive integers $n$, the map $[n]: F \rightarrow F$ given by $x \mapsto nx$ is an isomorphism. Indeed, it's a homomorphism for any commutative group and its inverse is $x \mapsto \left(\frac{1}{n}\right) x$.

I claim that any uniquely divisible commutative group $M$ admits the structure of a $\mathbb{Q}$-vector space in a unique way. For any nonzero rational number $\frac{p}{q}$ and any $x \in M$, we must define $\frac{p}{q} x$ to be the unique element $y$ of $M$ such that $qy = px$. It is easy to check that this works.

The only invariant of a vector space $V$ over any field $K$ is its dimension. Further, when the cardinality of $V$ is greater than the cardinality of $K$, the dimension of $V$ is equal to the cardinality of $V$. Thus:

$\mathbb{R}$ and $\mathbb{C}$ are both $\mathbb{Q}$-vector spaces of continuum cardinality; since $\mathbb{Q}$ is countable, they must have continuum dimension. Therefore their additive groups are isomorphic.

$\mathbb{Q}$ is a one-dimensional $\mathbb{Q}$-vector space whereas $\mathbb{Q}[i]$ is a two-dimensional $\mathbb{Q}$-vector space, so their additive groups are not isomorphic. (Note that the dimension of a $\mathbb{Q}$-vector space is the maximum cardinality of a $\mathbb{Z}$-linearly independent set. This leas to GEdgar's answer.)

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