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How we can get automorphism $Aut(C/Q)$ which fixes $Q$ ? I know that since transcendence basis for C/Q is infinite from this how can we say order of $Aut (C)$ is infinite because if this is the case then $Aut (R)$ is also infinite

maths student
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1 Answers1

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Fix a transcendence basis $\mathscr{B}$ of $\mathbb{C}$ over $\mathbb{Q}$. Then $\mathscr{B}$ is infinite (actually of cardinality $2^{\aleph_0}$).

If $\alpha\colon\mathscr{B}\to\mathscr{B}$ is any bijection, this extends to an automorphism of $\mathbb{C}$ because of the properties of transcendence bases and the fact $\mathbb{C}$ is algebraically closed. If $\hat{\alpha}\colon\mathbb{C}\to\mathbb{C}$ is the endomorphism defined by extending $\alpha$, then $\mathbb{C}$ is algebraic over the image of $\hat{\alpha}$ hence equal to it. Hence $\hat{\alpha}$ is an automorphism.

Any infinite set has infinitely many permutations.

However, this has no consequence on the group of automorphisms of the reals, which is trivial: the only field automorphism of $\mathbb{R}$ is the identity. If you follow the same path as before, you cannot conclude that the image of $\hat{\alpha}$ is $\mathbb{R}$. See Is an automorphism of the field of real numbers the identity map? for more on this topic.

egreg
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  • @Johncena The key fact is that $\mathbb{C}$ is algebraically closed, whereas $\mathbb{R}$ isn't. – egreg Oct 31 '17 at 08:49
  • @reuns See Is an automorphism of the field of real numbers the identity map?. $\mathbb{R}$ has many *endomorphisms*, but just one automorphism. – egreg Oct 31 '17 at 08:55
  • @dragonballz Look at automorphisms of $\mathbb{Q}(\pi)$ (sending $\pi$ to $\frac{a\pi+b}{c\pi+d}, ad-bc \ne 0$) and of $\mathbb{Q}(\sqrt{2})$ (sending $\sqrt{2} \to -\sqrt{2}$) and think about how those can extend to $\mathbb{C}$ – reuns Oct 31 '17 at 08:56
  • @egreg Right those extend only to $\mathbb{C}$ but not to $\mathbb{R}$ – reuns Oct 31 '17 at 09:00
  • Thanks for adding an explanation as to why $\hat\alpha$ is surjective as an endomorphism of $\Bbb{C}$ (+1). A nice way of concluding! – Jyrki Lahtonen Oct 31 '17 at 10:22
  • @dragonballz I think that in the real case you run into problems before you have that $\hat\alpha$. $\Bbb{R}$ is not algebraically closed, so you are not guaranteed to be able extend the endomorphism from $\Bbb{Q}(S)$, $S$ a transcendence basis, to all of $\Bbb{R}$. For example, some elements of a given transcendence basis may be negative and some positive, but only the latter have square roots in $\Bbb{R}$. More complicated obstructions may easily block your way. – Jyrki Lahtonen Oct 31 '17 at 11:46
  • @dragonballz Actually if we continue with that transcendence basis $S$ of $\Bbb{R}$ over $\Bbb{Q}$, you see immediately that we can permute them only monotonically. For if $s<s'$ are two elements of $S$, then $s'-s$ has a square root in $\Bbb{R}$. Therefore so must $\alpha(s')-\alpha(s)$. Meaning that we must have $\alpha(s')>\alpha(s)$. – Jyrki Lahtonen Oct 31 '17 at 11:49