Definition of Algebraic closure: An extension $K$ of $F$ is called an algebraic closure of $F$ if
(a) $F \subset K$ is algebraic;
(b) $K$ is algebraically closed.
Given the above definition, I have been trying to solve the following question:
Q. Let $F \subset K$ where $K$ is algebraically closed. Let $L$ be the algebraic closure of $F$ in $K$. Then prove that $L$ is also algebraically closed.
I feel that the answer is evident from the definition, but It involves a proof.
Please make some clear distinctions between 'algebraic closure' and 'algebraically closed' and explain these concepts by helping me solve the above question.
Best Answer
I’d like to expand on @tomasz”s answer in several ways.
There are really three concepts here: (1) an algebraically closed field; (2) an algebraic closure of a field; and (3) the algebraic closure of one field in another.
I’m going to sweep some delicate points under the rug by choosing a very handy definition of an algebraically closed field: Def. A field $F$ is algebraically closed if it has no proper algebraic extensions. Now, given a field $k$, an algebraic closure of $k$ is an algebraically closed field that is algebraic over $k$. Finally, if $L\supset k$ is an extension of fields, the algebraic closure of $k$ in $L$ is the set of elements of $L$ that are algebraic over $k$. For instance, the algebraic closure of $\mathbb Q$ in $\mathbb R$ is the set (field, really) of all real algebraic numbers. The algebraic closure of $k$ in $L$ is not generally algebraically closed, nor is it an algebraic closure of $k$.
Notice that there is no unique algebraic closure of a field $k$, and that it’s necessary to prove that any two such are isomorphic (as fields containing $k$).
Now to your question. It’s to take $F\subset L$, where $L$ is algebraically closed, and to define $K$ to be the algebraic closure of $F$ in $L$, and show that $K$ is also algebraically closed. Well: $K$ is certainly an algebraic extension of $F$ (all elements are algebraic over $F$). So, let $K'$ be an algebraic extension of $K$. We have $F\subset K\subset K'$, both inclusions being algebraic. But algebraic over algebraic is still algebraic, so $K'$ is algebraic over $F$, i.e. every element of $K'$ is algebraic over $F$, so every element of $K'$ is in $K$, and you’ve shown that $K$ has no proper algebraic extensions.