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Following the question What is the size of the smallest rigid extension field of the complex numbers?, where it was noted that the least cardinality of a rigid field containing $\mathbb{C}$ is $(2^{\aleph_0})^+$ I have the following question:

Is the rigid field of that cardinality (I.e. $(2^{\aleph_0})^+$) containing $\mathbb{C}$ unique (as a field)?

Or, Is there a canonical one?

PS: By "rigid field" I mean a field with non nontrivial field automorphisms; references are provided in the answer to the question I mentioned.

Edit: The constraint on the cardinality mentioned above may not be true. It was stated in the comments that the least cardinality of a rigid field containing $\mathbb{C}$ is rather $2^\omega$.

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user38200
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  • You should have reminded that "rigid field" means "with no nontrivial field automorphism", and have provided a link: (http://mathoverflow.net/questions/61058/what-is-the-size-of-the-smallest-rigid-extension-field-of-the-complex-numbers). Btw I'm puzzled by this question because by rigid extension of $K$ they mean "with trivial automorphism group as $K$-algebra", which then includes $K$. So it would be useful to clarify, in particular I'd like a precise statement and reference for the fact that there is no rigid extension (in which sense?) of $\mathbb{C}$ of cardinal $2^{\aleph_0}$. – YCor Mar 22 '15 at 11:39
  • I edited to add the link. Note that "unique" still has two possible meanings (as a field, or as a $\mathbf{C}$-field). Moreover the question you link at defines "rigid" with another meaning than yours (considering only automorphisms as $\mathbf{C}$-algebra). – YCor Mar 22 '15 at 11:45
  • Thank you; Maybe "unique" should mean as a $\mathbb{C}$-field. – user38200 Mar 22 '15 at 11:49
  • And you still define "rigid" as an abstract field? Besides I still can't find in the linked question any reference for the result that an extension $K$ of $\mathbf{C}$ of cardinal $2^{\aleph_0}$ has nontrivial automorphisms (of field? of $\mathbf{C}$-field assuming $K\neq\mathbf{C}$)? – YCor Mar 22 '15 at 11:51
  • Ok you are right, Unique should mean as a field. I don't know about a reference that there is no field of cardinality $2^{\aleph_0}$ which contains $\mathbb{C}$ and which has trivial automorphism group (over $\mathbb{Q}$). – user38200 Mar 22 '15 at 11:54
  • So what makes you believe that it is true? – YCor Mar 22 '15 at 13:20
  • In the comments to the answer to the linked question, they said the methods applied so far cannot avoid a jump in cardinality. – user38200 Mar 22 '15 at 13:24
  • OK, so this is definitely no claim that it's true! – YCor Mar 22 '15 at 13:25
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    I finally managed to get a copy of Dugas&Göbel (1997), Automorphism groups of fields II. They prove exactly what is stated in the MR review: for every group $G$ and field $F$, there is an extension $K$ of $F$ whose automorphism group is $G$, such that $|K|=\aleph_0\lvert G\rvert\lvert F\rvert$, with no further assumptions. So, $\mathbb C$ has a rigid extension of cardinality $2^\omega$. – Emil Jeřábek Mar 24 '15 at 18:40
  • Dear Emil: could you send me a copy please on hollowdead1@gmail.com? – user38200 Mar 25 '15 at 09:20
  • Also, is the extension you talked about rigid as a (abstract) field or as a $\mathbb{C}$-algebra? – user38200 Mar 25 '15 at 09:21
  • Rigid as a field (and therefore rigid as a $\mathbb C$-algebra, though that’s not a big feat: $\mathbb C$ itself is also a rigid $\mathbb C$-algebra). – Emil Jeřábek Mar 25 '15 at 17:15

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