# Confused about definition of algebraic closure

1. Jul 27, 2011

### nonequilibrium

Hello,

After a theorem stating that the product, sum, etc of two elements of a field extension that are algebraic over the original field are also algebraic, my course states the following result (translated into english):

but later in my course it defines "the algebraic closure of F" as a field extension of F that is
(i) algebraically closed (in the sense that every polynomial has a root)
(ii) algebraic across F

These seem to be different concepts, am I right? Because the former doesn't need to be algebraically closed (despite its name...), because for example "the algebraic closure of $\mathbb Q$ in $\mathbb R$" still has no solution for X²+1=0, yet "the algebraic closure of $\mathbb Q$" (full stop) does.

So is the only difference seperating these two concepts the suffix/appendix "in E"?

2. Jul 27, 2011

### micromass

Staff Emeritus
Yes, that is correct. These are two quite different concepts with the same name. So you'll need to be quite careful!

3. Jul 27, 2011

### disregardthat

If I'm not mistaken:
For every field F, we can construct a field E containing F such that E is algebraically closed. Taking the subset of E containing all elements algebraic over F will yield an algebraically closed field K containing F such that every algebraically closed field containing F will contain an algebraically closed subfield containing F isomorphic to K. In this sense such a construction is unique up to isomorphism, and informally it is this isomorphism class we refer to when we are talking about the algebraic closure of F. The important thing is that such a minimal algebraic closure exists, and any two of them are isomorphic.

The construction is not simple, but can be found in most books on abstract algebra I believe.

4. Jul 27, 2011

### nonequilibrium

Thank you both!