# Question Regarding Chapter 13, Proposition 31's Proof (Dummit and Foote)

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.

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

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 $$x+y, xy, x^{-1} \in \widetilde{F}$$.