By definition, one says that an algebraic closure $K$ of a given field $F$ is an algebraic extension of $F$ who is algebraically closed. This makes me curious about whether there exists an extension $K'$ of $F$ who is algebraically closed, while $K'$ is not an algebraic extension?
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$\begingroup$ When you say $K'$ is not an algebraic extension, you need to qualify that. Specifically, $K'$ is not an algebraic extension of what field? $\endgroup$ – quasi Aug 18 '17 at 7:52
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$\begingroup$ @quasi, I am talking about extension of the field $F$. $\endgroup$ – Easy Aug 18 '17 at 7:54
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Sure, this occurs naturally. Consider $F = \mathbb Q$ and its algebraic closure $K = \bar{\mathbb Q}$. Now consider $\mathbb Q$ as a subfield of $K' = \mathbb C$.
$\mathbb C$ is algebraically closed, its an extension of $\mathbb Q$ but it is not algebraic over $\mathbb Q$.
edit: Here is another example. Let $F$ be a finite field and let $K$ be its algebraic closure. Let $F \subseteq F'$ be an extension of $F$ of sufficiently large cardinality (i.e. uncountable) and let $K'$ be the algebraic closure of $F'$. (*)
$K'$ is a field extension of $F$ that is algebraically closed but it cannot be algebraic over $F$ because its cardinality is too large. I.e. there are only countable many polynomial roots over $F[X]$ but $K'$ contains uncountably many elements - most of which cannot be algebraic over $F$. In this fashion you can get an arbitrarily large 'gap' between the cardinality of the algebraic closure of $F$ and an algebraically closed extension. (The fact that $F$ is finite is irrelevant here - I simply thought it might help to make this example more approachable.)
(*) Note that $F'$ exists by the existence of the infinite field $K$ and the upward Löwenheim-Skolem Theorem, but you can also just view it as a field extension $F' = F(x_i \mid i \in I)$ with indepenent $x_i$ and $I$ uncountable.
answered Aug 18 '17 at 7:55
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$\begingroup$ Sounds very easy, I should have thought about this by myself.. $\endgroup$ – Easy Aug 18 '17 at 7:58
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$\begingroup$ I liked your question - even though the answer may seem trivial in hindsight. Coming up with (counter-)examples while learning something new is always tough... $\endgroup$ – Stefan Mesken Aug 18 '17 at 8:08
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$\begingroup$ Maybe I can conclude that if I want to construct an algebraically closed extension $K$ of a field $F$ which is not algebraic then $K$ must be an extension of one of the algebraic closures of $F$? So in order to obtain such an extension we need to first construct an algebraic closure, then extend it by adjoining a transcendental element:) $\endgroup$ – Easy Aug 18 '17 at 11:51
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$\begingroup$ Yes, that's right. The algebraic closure of a given field is - up to isomorphism - the $\subseteq$-least algebraically closed field extension of said field. So any other algebraically closed extension must contain some such (or - depending on your point of view - the up to isomorphism unique) algebraic closure. $\endgroup$ – Stefan Mesken Aug 18 '17 at 12:15
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Let $K'$ be an algebraic closure of $F(x)$, where $F(x)$ is the field of rational functions over $F$. Then $K'$ is an extension of $F$, and is algebraically closed, but is not an algebraic extension of $F$.
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Sure. $\mathbb{C}$ contains $\pi$.
Let $F = \mathbb{Q}$. $\mathbb{C}$ is algebraicly closed (although it is not the algebraic closure of $\mathbb{Q}$, which is just the set of algebraic numbers) and is an extension of $\mathbb{Q}$. But $\mathbb{C}$ contains $\pi$, which is not algebraic over $\mathbb{Q}$, so $\mathbb{C}$ is an algebraically closed transcendental extension of $\mathbb{Q}$.
answered Aug 18 '17 at 7:59
