A question related to this one Can a field be isomorphic to its subfield?: are there field extensions K/E and E/F such that K and F are isomorphic but E is not isomorphic to them?
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We construct an example. There are many.
Let $t_0,t_1,t_2,t_3,\dots$ be a countably infinite collection of algebraically independent transcendentals.
Let $F$ be the algebraic closure (in $\mathbb{C}$) of $\mathbb{Q}(t_1,t_2,t_3,\dots)$ and let $K$ be the algebraic closure of $\mathbb{Q}(t_0,t_1,t_2, t_3,\dots)$. Let $E=F(t_0)$.
André Nicolas
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Dear André, thank you for the answer. It follows that there exist two fields (E and F, above) which can be embedded to each other BUT not isomorphic. This may be a NEW equivalence relation in the category of fields. If it is indeed new, we may call it the Nicolas-equivalence. – Oleg Smirnov Nov 16 '14 at 20:38
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Why are $F$ and $E$ not isomorphic though? Isomorphism over what field? $F$-isomorphism ? $\mathbb{Q}$-isomorphism? Are they not isomoprhic as rings (fields) or are they not isomorphic as $F$-fields – Werner Germán Busch Apr 29 '16 at 01:17
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1It has been a while, and I am post dinner and wine. They are not isomorphic as fields, since $F$ is algebraically closed but $E$ is not. – André Nicolas Apr 29 '16 at 02:19
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why one being algebraically closed but the other not imply they are not isomorphic though? – Werner Germán Busch May 06 '16 at 21:30
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Why is $E$ not algebraically closed? – Werner Germán Busch May 06 '16 at 21:38
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(i) $\sqrt{t_0}\not\in E$ (the $t_i$ are algebraically independent; (ii) Any isomorphism preserves any algebraic property, for example the property of having a root. – André Nicolas May 06 '16 at 21:48
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thanks a lotttt! – Werner Germán Busch May 06 '16 at 22:06
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I credited you for your answer ;) – Werner Germán Busch May 06 '16 at 22:08
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You are welcome. I had left some gaps to fill in, and you correctly focused on these. – André Nicolas May 06 '16 at 22:10