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Let $(\mathbb{R},+)$ be the additive group of the reals and $(\mathbb{C},+)$ be the additive group of the complex numbers. Prove that those groups are isomorphic.

I think I got a solution using the fact that the real numbers have a Hamel Basis $B$, then proving that $B\times {0}\cup 0\times B$ is a Hamel basis for the complex numbers and noting that these sets have the same cardinality. However this result relies on Zorn's lemma and is thus not very elementary. Is there a simpler way to get it?

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    $\begingroup$ No, there isn't. $\endgroup$ – Mariano Suárez-Álvarez Jan 21 '15 at 22:46
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    $\begingroup$ @DiegoMath: That is not correct in the context here, where both $\mathbb R$ and $\mathbb C$ are being considered as rational vector spaces. $\endgroup$ – Jonas Meyer Jan 21 '15 at 22:48
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    $\begingroup$ @Zero, the result is certainly not wrong. It is a standard, well-known example of a consequence of Choice. $\endgroup$ – Mariano Suárez-Álvarez Jan 21 '15 at 22:52
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    $\begingroup$ (To answer my own comment-question: no, they're not necessarily. See, e.g., journals.cambridge.org/… ) $\endgroup$ – Steven Stadnicki Jan 21 '15 at 23:24
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    $\begingroup$ @StevenStadnicki: Since the current Question asks about whether the result "relies on Zorn's Lemma" in an essential way, it does raise an issue not resolved by the earlier post. But the article/link you found does clarify this nicely; I encourage you to summarize the link as an Answer here. $\endgroup$ – hardmath Jan 22 '15 at 0:01
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Unfortunately, there is no more elementary argument than going through some form of AC, because the result actually does depend on some amount of choice. As shown by e.g. C.J. Ash (see this 1973 J. Australian Math Society paper), an isomorphism between $(\mathbb{R},+)$ and $(\mathbb{C},+)$ implies the existence of a non-measurable set of reals. The paper has the full argument, but the short version is that (assuming that all sets of reals are measurable) one takes an isomorphism $f:\mathbb{R}\oplus\mathbb{R}\mapsto\mathbb{R}$, defines the sets $S_n=f[\mathbb{R}\oplus[n,n+1)]\cap(0,1)$ (that is, the image of $\mathbb{R}\oplus[n,n+1)$ under $f()$, intersected with the unit interval), and then shows that (a) the $S_n$ partition $(0,1)$ and (b) they all have the same measure. This is enough to contradict countable additivity.

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This two groups are vector spaces over $\mathbf{Q}$, and as these two groups have the same cardinal, any basis (over $\mathbf{Q}$) of one of them has the same cardinal than has any basis (over $\mathbf{Q}$) of the other one. This allows you to show that these two $\mathbf{Q}$-vector spaces are isomorphic as $\mathbf{Q}$-vector spaces. As any such isomorphism is also a group isomorphism, you're done.

Take for instance a Hamel basis $(e_i)_{i\in I}$, and a bijection $f : I \to I\coprod I$, then the $\mathbf{Q}$-linear map $F$ defined on basis vectors by $F(e_i)=(e_{f(i)},0)$ or $(0,e_{f(i)})$ claimed that $f(i)$ is in the first copy of $I$ or in the second is an isomorphism from $\mathbf{R}$ to $\mathbf{R}^2$, and there is a obvious group isomophism from $\mathbf{R}^2$ to $\mathbf{C}$, the one sending $(x,y)$ to $x+iy$...

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  • $\begingroup$ So $\mathbb{Q}^2$ is isomorphic to $\mathbb{Q}$? $\endgroup$ – Rob Arthan Jan 21 '15 at 23:30
  • $\begingroup$ "any basis (over Q) of one of them has the same cardinal than any basis (over Q) of the other one", I think that part is wrong, Rob Arthan pointed out a counterexample $\endgroup$ – Zero Jan 21 '15 at 23:35
  • $\begingroup$ @Robert Green: the point is that your argument requires the $\mathbb{Q}$-vector spaces to be infinite-dimensional. You didn't mention that important point. "As these two groups have the same cardinal" doesn't cut it. What you have to note is that the groups have the same uncountable cardinal, so they must be infinite-dimensional over $\mathbb{Q}$. $\endgroup$ – Rob Arthan Jan 22 '15 at 0:00
  • $\begingroup$ @RobArthan I get your point but your basis seems to not work and also since you put that part just after "and these two groups have the same cardinal", it seemed that the reason why those basis had the same cardinal was that the groups had the same cardinal and both $Q$ and $Q^2$ have the same cardinal, so I thought it was a counterexample $\endgroup$ – Zero Jan 22 '15 at 0:00
  • $\begingroup$ @Zero: I think you have the Robs and Roberts mixed up. :-) $\endgroup$ – Rob Arthan Jan 22 '15 at 0:02

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