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Proposition 31 of Chapter 13 in Dummit and Foote's Abstract Algebra states as follows:

Let $K$ be an algebraically closed field and let $F$ be a subfield of $K$. Then the collection of elements of $\bar{F}$ of $K$ that are algebraic over $F$ is an algebraic closure of $F$. An algebraic closure of $F$ is unique up to isomorphism.

The first sentence of the proof reads: By definition, $\bar{F}$ is an algebraic extension of F.

My question is how do we know that $\bar{F}$ is necessarily a field in its own right? We are given that $K$ and $F$ are fields, and as far as I can tell $\bar{F}$ is just some 'set' that lies between $F$ and $K$. If we assume that $\bar{F}$ is a field then the remaining parts of the proof seem to follow but I can't seem to figure out why we exactly know that $\bar{F}$ will be a field.

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  • $\begingroup$ To prove this you can check that every element is invertible in the algebraic closure. Just pick one and think what you can say. $\endgroup$
    – Fabrizio
    Apr 18, 2021 at 7:53

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Consider $x,y \in \widetilde{F}$. and $E = F(x,y)$. Both $x$ and $y$ are algebraic and thus $E/F$ is a finite extension. In particular, the subextensions $F(x+y), F(xy), F(x^{-1})$ are also finite, hence algebraic, and so $x+y, xy, x^{-1} \in \widetilde{F}$.

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