# 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:

1. How is $$E = S(x)$$?
2. 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 $$x$$."
3. How is the last equation satisfied?

Please help me with these doubts.

• Please define all objects in the text (not only in the title) in the natural order. It is hard to digest a question starting almost immediately starting with $F=S$, where both $F$, $S$ are not defined yet. In the equation there is maybe some $I$ missing... – dan_fulea Jun 29 at 14:20
• Sure, I edited the question. In the book I have, there is a $1$ in front of $x^2(x-1)^2$ in the equation, but there's no $I$. – P-addict Jun 29 at 14:38
• But having an $I$ surely makes more sense, because $I \in S$ and the equation is also satisfied. – P-addict Jun 29 at 14:47

## 1 Answer

Let $$k$$ be some field.

Let us consider the diagram of fields over $$k$$:

$$\require{AMScd}$$ $$\begin{CD} E @= \text{The field k(x)} \\ @AAA\\ F @= \text{Subfield of E fixed by S_3 acting by x\to 1-x and x\to1/x} \\ @AAA\\ S=k(I) @= \text{Subfield of E generated by I=(x^2-x+1)^3)/(x^2(x-1)^2)} \end{CD}$$

1. 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 $$E=k(x)\subseteq S(x)\subseteq E\ .$$

2. $$x$$ 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 $$X$$.

3. 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 $$f(x) = (x^2-x+1)^3 -x^2(x-1)^2\cdot \underbrace{ \frac{(x^2-x+1)^3}{x^2(x-1)^2} }_{=I} =0\ .$$

• what is $S_3$ in your diagram? – P-addict Jun 29 at 21:56
• Thanks a lot for your answer. – P-addict Jun 29 at 21:58