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Is it true that all extension of field $k$ are subfields of $\bar k$ (algebraic closure of $k$)? The same question for all algebraic extension.


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(a) Let $k$ be the rationals. Are the complex numbers a subfield of the algebraic numbers? Indeed for every $k$ there are extension fields of $k$ that are not isomorphic to $\overline{k}$. (b) Up to isomorphism. –  André Nicolas Sep 4 '11 at 19:11
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up vote 7 down vote accepted

All algebraic extensions of $k$ will embed in $\bar k$ over $k$. It's easy to do this for something like $k(\alpha)$ with $\alpha$ algebraic over $k$ [One way to think about this: $k(\alpha)$ is isomorphic to $k[X]/(f)$, where $f$ is the irreducible polynomial of $\alpha$ over $k$; what does it mean to give a homomorphism of $k$-algebras $k[X]/(f) \to \bar{k}$?], and you can prove it for arbitrary algebraic extensions using Zorn's lemma. You can read the proof in, for example, Milne's notes, Theorem 6.6.

On the other hand, there will always be extensions of $k$ that are not algebraic — these are the transcendental extensions. These fields have elements, such as $X$ in the field of rational functions $k(X)$, that do not satisfy any polynomial with coefficients in $k$. An embedding of such a field into $\bar k$ over $k$ would contradict this last property.

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For a concrete example, $\mathbb R$ is an extension of $\mathbb Q$ but is not an algebraic extension of $\mathbb Q$. –  lhf Sep 4 '11 at 19:11
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