# clarification of algebraic closure and algebraically closed field

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.

• It might be preferable to discuss what needs to be proved first so then you can try yourself. $L$ is the set of all elements in $K$ that are roots of nonzero polynomials with coefficients in $F$. $L$ being algebraically closed means that if you take any non-constant polynomial with coefficients in $L$, it has a root in $L$.
– Dave
Sep 10, 2014 at 0:09
• Your definition of algebraic closure is not mine. I would say $K$ algebraically closed, and no proper subfield of $K$ that contains $F$ is algebraically closed. Sep 10, 2014 at 0:20
• @AndréNicolas could you be a little clearer..Thanx Sep 10, 2014 at 0:21
• Doesn't matter, yours is equivalent. Sep 10, 2014 at 0:22

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.

• @prof Lubin.........thanx a lot....your definitions were very clear and very helpful... Sep 10, 2014 at 18:51

I think the problem you have is the definition of algebraic closure of a field within an extension.

Most likely the definition is as follows: the algebraic closure of $F$ in $L\supseteq F$ is the set of roots of polynomials with coefficients in $F$. The point of the exercise, it seems to me, is that this "algebraic closure in $L$" operation is indeed a closure operation, i.e. that it is idempotent (so a polynomial whose coefficients are roots in $L$ of polynomials over $F$ is divided by a polynomial whose coefficients are in $F$).

Take a non zero polynomial $\;f(x)=a_0+a_1x+\ldots+a_nx^n\in L[x]\;$ , then $\;F(a_0,...,a_n)/F\;$ is algebraic (why?) .

But since clearly $\;f(x)\in K[x]\;$ and $\;K\;$ is alg. closed, there exists $\;k\in K\;\;s.t.\;\;f(k)=0\;$, so $\;F(a_0,...,a_n,k)/F\;$ algebraic (why?), whick means $\;k\;$ is algebraic over $\;F\;$ and thus $\;k\in L\;$ and we're done.