# What Are the Indeterminates in the Polynomials in This Proof?

To prove that every field $F$ has an algebraic closure, the first step is to prove that there exists an extension $F_1$ such that each polynomial in $F[x]$ has a root in $F_1$. Whereas in the construction of a splitting field we build a quotient ring directly from $F[x]$, to build the algebraic closure, we build a quotient ring from a multi-variable polynomial ring.

To that end, for each $f \in F[x]$, define a variable $x_f$. Let $S = \{ x_f : f \in F[x] \}$, and $F[S]$ is our multivariable-polynomial ring. It can be shown that the following is a proper ideal in $F[S]$. $$U = \left\{ \sum g_if_i(x_{f_i}) : g_i \in F[S] \text{ and } f_i \in F[x] \right\}.$$

By Zorn, there exists a maximal ideal $M \subset F[S]$ containing $U$. It follows that $\frac{F[S]}{M}$ is a field. The element $x_f + M$ is a root of $f$ ...

My understanding is that the $x_{f_1}, x_{f_2} \ldots$ are "place-holders" for the roots of the $f_i \in F[x]$. Let $L$ be a field extension holding the roots, say $\alpha_1, \alpha_2, ... , \alpha_m$, of $f_1, f_2, ... , f_m$. Therefore, we obtain the following contradiction, which proves that $U$ is proper $$0 = \sum_{i=1}^{m} g_i(\alpha_1, \alpha_2, ... ,\alpha_m)f_i(\alpha_i) = 1 \in U.$$ So there is nothing wrong with expressions like $f_1(x_{f_2})$ or $f_1(x_{f_2} + M)$ except that they do not serve our proof. For example $f_1(\alpha_2)$ may not be zero. Can you kindly confirm or correct my understanding here?

• Viewed as multi-variable polynomials, $f_1(x_{f_1}) \ne f_1(x_{f_2})$, so $f_1(x_{f_2} + M)$ may not be zero either .... – Andy Tam Jul 13 '15 at 17:35