Let $k$ be a field and let $K=k(x)$ be the rational function field in one variable over $k$. Let $\DeclareMathOperator{\aut}{Aut}\sigma, \tau \in \aut(K)$ s.t. $$\sigma\left(\frac {f(x)}{ g(x)}\right)=\frac {f(1/x)}{g(1/x)}\qquad \&\qquad \tau\left(\frac {f(x)}{ g(x)}\right)=\frac {f(1-x)}{g(1-x)}.$$ Find the fixed field $F$ of $<\sigma, \tau>$. Determine $\DeclareMathOperator{\gal}{Gal}\gal(k/F)$. Find an $h \in F$ so that $F=k(h)$.

Now for the first automorphism $\sigma$, I think if I concentrate on $k[x]$ it will give me some idea. Now, if $f(x)=a_0+a_1x+a_2x^2+ \cdots + a_nx^n \in k[x]$ then if $x^n \sigma(f(x))=f(x)$ the case will be $a_i=a_{n-i}$ $\forall i=0(1)[n/2]$. So all the funtions in $K$ which will be fixed by $\sigma$ are of the form $\frac {f(x)}{ g(x)}$ where $f(x)=a_0+a_1x+a_2x^2+ \cdots + a_nx^n$ and $g(x)=b_0+b_1x+b_2x^2+ \cdots + b_nx^n$ such that $a_i=a_{n-i}$ $\forall i=0(1)[n/2]$ and $b_i=b_{n-i}$ $\forall i=0(1)[n/2]$.

Please check whether am I right or not?

And any ideas for the rest of the part?

  • $\begingroup$ Take a look at this older question. It is relatively well known (but not immediately obvious!) that $\langle \sigma,\tau\rangle\cong S_3$. $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 20:29
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    $\begingroup$ A comment on your attempt: $\sigma$ doesn't fix any polynomials. You are considering so called palindromic polynomials (of degree $n$), but those are fixed under the operation $$f(x)=x^nf(\frac1x),\qquad(*)$$ i.e. $\sigma$ followed by multiplication by $x^n$. And you are right in that the quotient of any two polynomials satisfying $(*)$ for the same $n$ is fixed under $\sigma$. It turns out that the fixed field of $\sigma$ is generated by $x+1/x=(x^2+1)/x$. $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 20:41
  • $\begingroup$ Yeah correct. I am sorry. $\endgroup$ – Ri-Li Aug 28 '15 at 20:43
  • $\begingroup$ No need to be sorry! After all, you (intuitively?) described the correct set of rational functions! $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 20:45
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    $\begingroup$ Yeah okay from the relations $\sigma^2=1=(\sigma \tau)^3$ and $(\sigma \tau)\sigma=\sigma(\sigma \tau)^{-1}$ we get that the group will be $S_3$ $\endgroup$ – Ri-Li Aug 28 '15 at 21:10

You can get a complete answer by checking out the related answer that @JyrkiLahtonen mentions in his comment. But I see this as not so much an algebraic question, as one in complex variable theory and geometry.

Do you know the relationship between $2\times2$ complex matrices and fractional-linear transformations of the extended complex plane, also known as the Riemann sphere? You get a homomorphism from nonsingular matrices $\pmatrix{a&b\\c&d}$ to transformations $z\mapsto\frac{az+b}{cz+d}$ of the Riemann sphere. The kernel is the group of scalar matrices $aI$, where $I$ is the identity matrix. Then your $\sigma$ is $z\mapsto1/z$, coming from $\pmatrix{0&1\\1&0}$ and your $\tau$ comes from $\pmatrix{-1&1\\0&1}$. The product of these in either order gives you something of order three.

You check that your six transformations permute the subset $\{0,1,\infty\}$ of the Riemann sphere, and so it makes sense \, if you want something fixed by all six, to try a rational function $h(z)$ with poles at $0$, $1$,and infinity. In fact, it’s a theorem in algebraic curves that the field extension degree $[\mathbb C(z)\colon\mathbb C(h(z))]$ is the total number of poles of $h$, counting multiplicity. So you should expect that if my strategy works, $h$ will have double poles at the three points I mentioned.

If we put $z^2(z-1)^2$ in the denominator of $h$, and look for a sextic numerator, then there will also be a double pole at infinity. What about special points of your two given transformations? If you look at $1/2$, you see that it’s fixed under $\tau$, but sent to $2$ by $\sigma$. Then you check that the set $\{-1,1/2,2\}$ is invariant under both $\sigma$ and $\tau$. So why not choose $$ h=\frac{(z+1)^2(2z-1)^2(z-2)^2}{z^2(z-1)^2}\,? $$ There it is, sextic above, obviously invariant under $\tau$, and almost as obviously invariant under $\sigma$.

EDIT: Here’s my response to your request for a description of how to find an irreducible sextic, showing that you readyy do have a generator of the subfield $F$. You start with your rational function known to be fixed under $\sigma$ and $\tau$. It will be a suitable $f(x)/g(x)$, and the max of the degrees of $f$ and $g$ will be six. All our computed examples have $g$ of smaller degree, and for my explanation, I will assume that this is the case, but the general situation is only a little more complicated.

We have $u=f(x)/g(x)$, fixed under the order-six group $G$ generated by $\sigma$ and $\tau$. We know, from general Galois-theoretical considerations, that the fixed field $F$ has $[k(x)\colon F]=6$. Since $u\in F$, we know that $[k(x)\colon k(u)]\ge6$. On the other hand consider the sextic polynomial $f(T)-g(T)u\in k(u)[T]$. I have used $\deg(g)<\deg(f)=6$ to conclude that this polynomial in $T$ is of degree six. Now, $x$ is clearly a root of our polynomial, so $[(k(u))(x)\colon k(u)]\le 6$; but of course $(k(u))(x)=k(x)$, so that $[k(x)\colon k(u)]=6$, and hence $F=k(u)$. You get as a bonus the conclusion that our polynomial in $T$ is irreducible over $k(u)$.

  • $\begingroup$ you are completely right but the thing is I have to solve it algebraically. So please anyone give me any algebraic proof. $\endgroup$ – Ri-Li Aug 29 '15 at 18:44
  • $\begingroup$ Awright, without applying any geometric intuition or knowledge, take s plain linear, such as $x-3$, and apply to it all six elements of the group. Now take the product of these six. You’re taking the norm of $x-3$, from $k(x)$ down to the fixed field. There you are, end of story. $\endgroup$ – Lubin Aug 29 '15 at 20:21
  • $\begingroup$ Oh okay thanks. I think I am getting it. But please explain your last comment... $\endgroup$ – Ri-Li Aug 29 '15 at 21:50
  • $\begingroup$ You have a group, call it $G$, generated by $\sigma$ and $\tau$. It’s necessary to check that this group has order $6$. It follows that the fixed field $F$ had $[K\colon F]=6$. I told you how to get an element $u$ of $F$. You only need to show that this element generates all of $F$, not a proper subfield. One way to do this is find a sextic polynomial over $k(u)$ that has $x$ as a root. This shows that $[k(x)\colon k(u)]\le 6$. But what you already know shows that $[k(x)\colon k(u)]\ge 6$. Thus $F=k(u)$. If you want to see the sextic polynomial of which $x$ is a root, let me know. $\endgroup$ – Lubin Aug 30 '15 at 4:48
  • $\begingroup$ Yes I want to know how to find the sextic polynomial in an algebraic way.. $\endgroup$ – Ri-Li Aug 30 '15 at 9:28

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