Proof of an existence of an algebraic closure of a given field

Question

I am studying the field theory with Abstract Algebra by Dummit & Foote.

A proof of an existence of an algebraic closure by constructing such extension using the Zorn's lemma is given in this textbook.

But I can't not understand it fully.

Basically, this proof consists of 2 parts. First, it constructs the extension $E$ of a given field $F$ to have all the roots of every element of $F[x]$. Second, obtain the union using this extension countable infinitely to get algebraically closed field.

I can understand the second part of this proof which is not too hard.

This proof begins with the following sentence:

For every nonconstant monic polynomial $f(x)$, let $x_f$ denote an indeterminate and consider the polynomial ring $F[\cdots,x_f,\cdots]$ generated over $F$ by the variables $x_f$.

I can not understand even this first sentence because I don't get what $F[\cdots,x_f,\cdots]$ does mean. Is this a polynomial ring? What was in "$\cdots$" originally?

And the proof shows that $I=(f(x_f))$ should be proper by assuming not proper so that there are $g_i$ with $$\sum_{i=1}^ng_if_i(x_{f_i})=1$$ and leads the left side of above equation to $0$, a contradiction, $0=1$.

Where did $n$, $f_i$ come from?

Please help me to understand this proof. I'll edit to add some more information if you need. Thanks.

Answer 1

For every nonconstant monic polynomial $f(x)$, let $x_f$ denote an indeterminate and consider the polynomial ring $F[\cdots,x_f,\cdots]$ generated over $F$ by the variables $x_f$.

This is a polynomial ring in infinitely many variables - it has one variable $x_f$ for each nonconstant monic polynomial $f \in F[x]$.

And the proof shows that $I=(f(x_f))$ should be proper by assuming not proper so that there are $g_i$ with $$\sum_{i=1}^ng_if_i(x_{f_i})=1$$ and leads the left side of above equation to $0$, a contradiction, $0=1$.

Suppose the ideal $I = (f(x_f): f \in F[x])$ is not a proper ideal, i.e. it is equal to all of the huge polynomial ring (let's call it $R$) mentioned above. Then $1 \in I$, so we can write $1$ as a finite $R$-linear sum of elements of $\{f(x_f): f \in F[x]\}$, i.e. $$\sum_{i=1}^ng_if_i(x_{f_i})=1$$ where the $g_i \in R, f_i \in F[x]$

I hope this helps, let me know if you're still confused.

Answer 2

For what it's worth, the second part of the proof you mention is technically redundant: the extension $E$ is already an algebraic closure of $F$ but this is not easy to show. It's a theorem of Robert Gilmer, see here or Theorem 2 here. For a similar construction where it's easier to see you get an algebraic closure in just one step, see here.