How is $E = S(x)$, where $E = k(x), k$ a field, and S= k(I) of all rationals functions of $I = I(x) = \frac{(x^2 -x+1)^3}{x^2(x-1)^2}$? In Section G, Part II of Emil Artin's Galois Theory, in the first example.of the section, the author says :
Here, $k$ is a field, $E=k(x)$ is the space of all rational functions of variable $x$. We say an element $a$ of $E$ is a fixed point under isomorphisms $\sigma_{1}, \sigma_{2},....,\sigma_{n}$ of $E$ into a field $E'$, if $\sigma_{1}(a) = \sigma_{2}(a) =...... =\sigma_{n}(a)$. For the purpose of this question, we consider fixed points of the following six automorphisms of $E$ : $f(x)$ mapped to $f(x), f(1-x), f(1/x), f(1/(1-x)), f(1-1/x)$ and $f(x/(x-1))$.$F$ denotes the fixed point field, i.e., the subfiled of $E$ consisting of all fixed points of $E$. $S= k(I)$ is the field of all rationals functions of $I = I(x) = \frac{(x^2 -x+1)^3}{x^2(x-1)^2}$

We contend : $F= S$ and $(E/F) = 6$.
Indeed, from Theorem 13, we obtain $(E/F) \geq 6$.Since $ S \subset F$, it suffices to show that $(E/S) \leq 6$. Now, $E= S(x)$. It is thus sufficient to find some $6$-th degree equation with coefficients in $S$ that is satisfied by $x$.
The following one is obviously satisfied;
$(x^2 -x+1)^3 - x^2(x-1)^2 = 0.$

I have three doubts in the above passage:

*

*How is $E = S(x)$?

*Is $x$ an element or just a variable? I got confused because it says "It is thus sufficient to find some $6$-th degree equation with coefficients in $S$ that is satisfied by $x$."

*How is the last equation satisfied?

Please help me with these doubts.
 A: Let $k$ be some field.
Let us consider the diagram of fields over $k$:
$\require{AMScd}$
\begin{CD}
    E @= \text{The field $k(x)$}
    \\
    @AAA\\
    F @= \text{Subfield of $E$ fixed by $S_3$ acting by $x\to 1-x$ and $x\to1/x$}
    \\
    @AAA\\
    S=k(I) @= 
    \text{Subfield of $E$ generated by $I=(x^2-x+1)^3)/(x^2(x-1)^2)$}
\end{CD}


*

*We have $E=k(x)$, the smallest field over $k$ in the transcendental variable $x$, i.e.$E$ is the fraction field of the polynomial ring $k[x]$. Now $k\subseteq S$, so adjoining $x$ we get the inclusion of subfields of $E$, $k(x)\subseteq S(x)$. The notation "something$(x)$" is here considered as a construction in the category of subfields of $E$, we add $x$ to the "something" object in the category of sets, then take the smallest field containing both. Thus
$$
E=k(x)\subseteq S(x)\subseteq E\ .
$$

*$x$ is an element in $E$. When using polynomials and polynomial equations satisfied by $X$ i will use in the sequel polynomials in a new variable $X$.

*The degree of $E:F$ is insured to be $\ge 6$. To show it is $\le 6$ we show that even $E:S$ has degree $\le 6$. The extension $E:S$ is simple, $E$ being $S(x)$. We write now a polynomial in $f=f(X)\in S[X]$, which has the root $x$ and it is of degree $6$. The polynomial is:
$$
f=f(X)
=(X^2-X+1)^3 -X^2(X-1)^2\cdot I
\in S[X]=k(I)[X]
\ .
$$
Then 
$$f(x)
=
(x^2-x+1)^3 -x^2(x-1)^2\cdot 
\underbrace{
\frac{(x^2-x+1)^3}{x^2(x-1)^2}
}_{=I}
=0\ .
$$
