Blog 1

Random Talk on Random Thoughts

Read a Proof of Existence of Algebraic Closure

| Comments | ∞

Background

In this Chinese New Year holiday, I read E. Artin’s proof of the existence of algebraic closure, which could be found online.1

Problem

Two days ago, I was stuck at the point where the ideal $I$ of $F[\dots,X_f,\dots]$ was defined. I mistakenly thought that one needed to fix an arbitrary polynomial $f$ in the polynomial ring $F[X]$ first because the goal of this theorem is to show the existence of an algebraic field extension $E$ of $F$ so that $E$ is algebraically closed, and for any nonconstant polynomial $f$ in $F[X]$, $f$ has a root (a.k.a. zero) in $E$.

Resolution

This noon, I finally saw “indexed by $f$”, and I found I should interpret “the ideal generated by $f(X_f)$” in this way:

\[ I := \left\{\sum_{m = 1}^n g_m (\dots,X_f,\dots) f_m (X_{f_m}) : n \in \Z_{>0}, f_i \in F[X], g_i (\dots,X_f,\dots) \forall i \in \{1,\dots,n\}\right\}, \]

instead of viewing $f(X)$ as a fixed polynomial in $F[X]$.

  1. Milne, J.S. (2014). Fields and Galois Theory (v4.50) (pp. 87). Available at http://www.jmilne.org/math/ ↩

Comments