2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $k(u)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.