# What is the difference between Algebraic extension and Algebraic closure?

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.

• The algebraic closure can be thought of as "the biggest possible algebraic extension". – Baby Dragon Mar 16 '13 at 21:26
• What did you not understand in en.wikipedia.org/wiki/Algebraic_closure ? – lhf Mar 16 '13 at 21:33
• 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. – anon Mar 16 '13 at 21:53

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 $F$.