Algebraic extension says every element of extension field E of field F is algebraic over F. But that's also the definition of Algebraic closure. I'm confused. Please explain the difference.

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    $\begingroup$ The algebraic closure can be thought of as "the biggest possible algebraic extension". $\endgroup$ – Baby Dragon Mar 16 '13 at 21:26
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    $\begingroup$ What did you not understand in en.wikipedia.org/wiki/Algebraic_closure ? $\endgroup$ – lhf Mar 16 '13 at 21:33
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    $\begingroup$ It's not also the definition of algebraic closure. This is the same as the difference between a bag of apples, and a bag of all of the apples in existence. $\endgroup$ – anon Mar 16 '13 at 21:53

Let's look at an example to see the difference.

Let $F=\mathbb{Q}$ and consider the extension field $E=\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}.$ Then any element $a+b\sqrt{2}\in E$ satisfies the polynomial $f(x)=x^2-2ax+a^2-2b^2\in\mathbb{Q}[x]$, and hence we see that $E$ is algebraic over $F$.

An algebraic closure of $\mathbb{Q}$ is a field $E$ such that every polynomial in $E[x]$ has a root in $E$. Consider the polynomial $x^2-3$. This polynomial has no root in $\mathbb{Q}(\sqrt{2})$ (you can check this using the standard form of an element in $\mathbb{Q}(\sqrt{2})$ as above). It follows that $\mathbb{Q}(\sqrt{2})$ cannot be an algebraic closure of $\mathbb{Q}$.

Note that up to isomorphism, algebraic closures are unique, so we often refer to the algebraic closure of $F$.

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    $\begingroup$ "An algebraic closure of Q is a field E such that every polynomial in Q[x] has a root in E. " I don't think that would be sufficient. Every polynomial in E[x] should have a root in E! Thus the field E admits not algebraic extensions. $\endgroup$ – Myself Mar 16 '13 at 21:48
  • $\begingroup$ You're absolutely right. I've edited my response. $\endgroup$ – Jared Mar 16 '13 at 22:30

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