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(a) Let $K$ be an algebraically closed field extension of $F$. Show that the algebraic closure of $F$ in $K$ is an algebraic closure of $F$.

What is algebraic closure of $F$ in $K$? The definition of algebraic closure is:

If $K$ is an algebraic extension of $F$ and is algebraically closed, then $K$ is said to be an algebraic closure of $F$

In this case, $K$ is more than algebraic extension so, what algebraically closed extension? I'm a little confused by this.

(b) If $\mathbb{A} = \lbrace a \in \mathbb{C}\,|\,a\,\text{is algebraic over}\,\mathbb{Q}\rbrace$, then, assuming that $\mathbb{C}$ is algebraically closed, show that $\mathbb{A}$ is an algebraic closure of $\mathbb{Q}$.

I imagine that $\mathbb{C}$ is an algebraically closed field extension of $\mathbb{Q}$ and $\mathbb{A}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$.

So, I would like some help to understand these definitions for can answer the two items. Thanks for the advance!

It may be useful to put some definitions:

Lema 1. If $K$ is a field, then the following statements are equivalente:

  1. There are no algebraic extensions of $K$ other than $K$ itself.
  2. There are no finite extensions of $K$ other than $K$ itself.
  3. If $L$ is a field extension of $K$, then $K = \lbrace a \in L\,|\,a\,\text{is algebraic over}\,K\rbrace$.
  4. Every $f(x) \in K[x]$ splits over $K$.
  5. Every $f(x) \in K[x]$ has a root in $K$.
  6. Every irreducible polynomial over $K$ has degree $1$.

Definition 1. If $K$ satisfies the equivalent conditions of Lema 1, then $K$ is said to be algebraically closed.

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1 Answer 1

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The algebraic closure of $F$ in $K$ denotes the set of elements in $K$ which are algebraic over $F$.

And, yes, $\mathbf C$ is an algebraically closed field (that's what D'Alembert-Gauß' theorem asserts) and it is an extension of $\mathbf Q$.

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  • $\begingroup$ So, and "$K$ algebraically closed extension of $F$" is just a extension of $F$ which is algebraically closed? $\endgroup$
    – Lucas
    May 3, 2018 at 23:04
  • $\begingroup$ ${}$Absolutely! $\endgroup$
    – Bernard
    May 3, 2018 at 23:10

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