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Consider a field $K$. Now, consider the class of all algebraic extensions of $K$. Is this a set? Since I think it isn't, how to prove it isn't?

If the class were of all extensions of $K$, I think I could argue with Zorn's Lemma to get a maximal element and enlarge it by taking the field of fractions of the polynomials with coefficients in that maximal element (proving, thus, that it is a not a set), but I can't repeat this argument for algebraic extensions. So, what to do ?

Aloizio Macedo
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  • Any ZF object (set) can be a member of an algebraic extension. – André Nicolas May 13 '14 at 04:40
  • I am more than almost sure that this is a duplicate. – Asaf Karagila May 13 '14 at 04:41
  • Maybe this one: http://math.stackexchange.com/questions/621944/proof-of-existence-of-algebraic-closure-too-simple-to-be-true/ – Asaf Karagila May 13 '14 at 04:47
  • "For instance, the field of all algebraic numbers is an infinite algebraic extension of the rational numbers." - Wikipedia. So there's a maximal element of any chain $\Bbb{Q} \subset \dots$. Let $\Bbb{Q}'$ be the maximal element. It's possible that $\Bbb{Q}' \subsetneq $ your field of fractions yet is isomorphic to it. – Daniel Donnelly May 13 '14 at 04:54

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