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I'm studying abstract algebra and read a theorem on the uniqueness of the algebraic closure of a field. Namely, for any field $F$, the algebraic closure exists and is unique up to isomorphism.

But I have a question: for the reals $\mathbb{R}$ and the complex numbers $\mathbb{C}$, we may consider the field of rational functions $\mathbb{C}(x)$ on $x$. Then, we can extend $\mathbb{C}(x)$ to get the algebraic closure $\mathbb{D}$ of $\mathbb{C}(x)$, which obviously contains $\mathbb{C}$ itself. Then, $\mathbb{C}$ and $\mathbb{D}$ are isomorphic? It seems weird.. is there something I'm missing?

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    $\begingroup$ Your $\mathbb{D}$ is not an algebraic closure of $\mathbb{R}$, though. It contains plenty of elements which are not the solutions to polynomials over $\mathbb{R}$. $\endgroup$ – Hayden Mar 19 '15 at 11:20
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    $\begingroup$ To recap the definition: An algebraic closure of a given field is an algebraically closed field, which is algebraic over the given field. Maybe you want to relook at the proof of the uniqueness (up to isomorphism) of the algebraic closure and check where the fact, that the closure is algebraic over the given field, is needed. $\endgroup$ – MooS Mar 19 '15 at 11:22
  • $\begingroup$ Oh, thanks Hayden and MooS. I am an idiot.. hahaha.. $\endgroup$ – user224850 Mar 19 '15 at 11:23
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    $\begingroup$ Nevertheless the fields $\mathbb{C}$ and $\mathbb{D}$ are isomorphic. $\endgroup$ – Hagen Knaf Mar 19 '15 at 22:00

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