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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.

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P-addict
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  • 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 '19 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 '19 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 '19 at 14:47

1 Answers1

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

dan_fulea
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