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Find $ Aut (\mathbb{C})$ , the group of automorphisms of field of complex numbers $\mathbb{C}$ .

My solution :

Take any $\phi\in Aut(\mathbb{C})$ . I think its enough if we know possible values of $\phi(i)$ because if we know $\phi(i)= a$ then we know $1=\phi(1)=\phi(i^4)=a^4$ . So $a=$ roots of equation $z^4=1$ in complex numbers .Lets name them as $a_1,a_2,a_3,a_4$

Now i can show that $\phi(x)=x a_1$ $\forall x\in \mathbb{Q}$ . I'm not sure what do about $\phi(x)$ when $x\in\mathbb{R}$ . If get possible values for $\phi(x)$ when $x\in\mathbb{R}$ i can find $\phi(z)$ for every $z\in \mathbb{C}$ ( if $z=a+ib $ $ a,b\in \mathbb{R}$ we have $\phi(z=\phi(a)+\phi(i)\phi(b)$ )

So i guess finding possible $\phi(x)$ for all $x\in \mathbb{R}$ will solve the problem . Alos i would like to share the link if its sheds some light on how to go about solving this problem

Is an automorphism of the field of real numbers the identity map?

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Suman Kundu
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    If your automorphism is continuous then you can easily find them. Else I believe this is pretty complicate : $\mathbb C \cong \overline{\mathbb Q(X_1, \dots, X_n, \dots )}$ if I am not mistaken, so they are lot of automorphisms as you can see if you don't ask for continuity. –  Mar 27 '17 at 22:18
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    I think that group is very, very large....something like $;2^{\aleph_0};$ , at least ...If you mean "continuous" automorphisms there are much less: two, in fact. – DonAntonio Mar 27 '17 at 22:19
  • Nothing about continuity is mentioned in the book . Ironically the problem is from a analysis book . – Suman Kundu Mar 27 '17 at 22:22
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    I guess it's understood that by automorphism the book mean sbiholomorphism $\Bbb C\to\Bbb C$; in that case, all and only such functions are $z\mapsto az+b$ where $a,b\in\Bbb C,,a\neq0$. – Joe Mar 27 '17 at 22:22
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    Assuming the axiom of choice, a transcendence basis for $\mathbb C$ over $\mathbb Q$ must have $2^{\aleph_0}$ elements, and a field automorphism can independently map each of the elements of this basis to, say, either itself or its negative. That gives us at least $2^{2^{\aleph_0}}$ automorphisms. On the other hand $2^{2^{\aleph_0}}$ is also the number of functions $\mathbb C\to\mathbb C$, no matter what their properties are. So there are $2^{2^{\aleph_0}}$ automorphisms. – hmakholm left over Monica Mar 27 '17 at 22:27
  • If we were to assume continuity of automorphism then it seems $\phi(x)=x a_i , i=1,2,3,4 $ would give complete list of continuous automorphisms of $\mathbb{C}$ , ami i right ? – Suman Kundu Mar 27 '17 at 22:34
  • No, because $\phi$ must map $1$ to itself. – D_S Mar 27 '17 at 22:53

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It depends on what you mean by automorphism. Do you mean an automorphism group under composition over holomorphic functions on $\mathbb{C}$? Then your solution is $\phi(z) = \lambda z + \beta$ for all $\lambda,\beta \in \mathbb{C}$ $\lambda \neq 0$.

Do you mean a continuous automorphism of the field $\mathbb{C}$? Then your choices are $\phi(z) = z, \overline{z}$.

Do you mean an automorphism of the field $\mathbb{C}$ that is not continuous? Well then you need the Absolute Galois Group.

As you can see, your question can be interpreted many ways, it isn't clear what you are asking.

Since this is from an analysis workbook, the answer is probably the first.