I am studying field theory with Algebra: From the Viewpoint of Galois Theory by Siegfried Bosch.
In the proof of the existence of algebraic closure on Page 103-104, I wonder how we can derive the algebraic condition.
Here is the outline of the proof in this book.
-
Construct a polynomial ring $ K[\textbf{X}] $ where $\textbf{X}$ is an infinite set of indeterminants indexed by all polynomials with positive degree in $ K[x] $, i.e. $\textbf{X} = \{X_f|f\in K[x], \deg f>0\}$.
-
Consider the proper ideal $I$ in $K[\textbf{X}]$, $I = \left< \{f(X_f)|f\in K[x], \deg f>0\}\right> $. (I already know why it is proper.)
-
Since every proper ideal is contained in some maximal ideal, let $M$ be the maximal ideal admitting $I$. Hence $K[\textbf{X}]/M$ is a field which we denote as $L_1$.
-
Repeat the same procedure for $L_1$ (for countably many times) and take the union of $L_1, L_2, \cdots$. Prove that the union is algebraically closed and algebraic over $K$.
To achieve 4, the book says that $L_{n+1}$ is an extension of $L_n$ generated by a set of algebraic elements. I know this is true when the ideal $I$ is generated by a finite set. Could anyone help me with the infinite case?
Best Answer
Basically what you are asking is why $L:=K[\textbf{X}]/M$ (where $\textbf{X},M$ are defined in item 1 and 3 in your question) is an extension of $K$ generated by a set of algebraic elements over $K$.
(by taking $(L_n,L_{n+1})=(K,L)$).
Here is an outline :
To elaborate a little more on this :
is to use the fact that $L$ is generated by $x_f$ over $K$ ($L$ being a quotient of $K[\textbf{X}]$).
is by the definition of $M$ : for each $f$, one has $f(x_f)=0$ in $L$.
Notice that we don't need $M$ (or the ideal $I$ constructed in your question) to be finitely generated in $K[\textbf{X}]$ in these lines.