7
$\begingroup$

I'm looking at automorphisms $t \to 1-t$ and $t \to \frac{1}{t}$ of the field $\mathbb{Q}(t)$. By looking at the relations between these I think I've found the group generated by them to be $S_3$.

Now I'm trying to find the fixed fields, call these $F$. For the first one I noticed that the polynomial $f = t(t-1)$ is invariant and as we know that $[\mathbb{Q}(t) : F] \ge 2$ and then the polynomial $g = t(t-1)-f$ is a two degree polynomial in $\mathbb{Q}(f)[t]$ then I deduced $F = \mathbb{Q}(f)$.

However I'm not having trouble applying this method to find the fixed field of the second automorphism. I can't actually find any rational functions that are fixed by it.

Thanks for any help

$\endgroup$
  • $\begingroup$ How about all of the constants? Or $t+\frac1t$? $\endgroup$ – Henning Makholm Nov 27 '14 at 17:20
  • $\begingroup$ Thank you for your helpful comments. @HenningMakholm I don't know how I didn't spot that! so if $f = t + \frac{1}{t}$ then we can take $g = t^2 + 1 - tf$ and that is a 2 degree polynomial and hence $F = \mathbb{Q}(f)$? $\endgroup$ – Wooster Nov 27 '14 at 17:46
4
$\begingroup$

Let $F$ be the fixed field.

Since the symmetry group is finite, there is a trace map $\mathbf{Q}(t) \to F$:

$$ \operatorname{Tr}_{\mathbf{Q}(t) / F}(f(t)) = \sum_{\sigma \in S_3} f(\sigma(t)) $$

In fact, we can see that $f(t) \in F$ if and only if $f(t) = \frac{1}{6} \operatorname{Tr}_{\mathbf{Q}(t) / F}(f(t))$.

Usually, for problems like these, I just compute the trace of a few simple elements, and stop as soon as I've used them to generate a field where it's easy to see that $[\mathbf{Q}(t) : F] = 6$.

The norm map works too (as do any of the other elementary symmetric polynomials in the conjugates), but I usually use the trace map for this purpose.


Although now that I've thought about it some more, the norm map might be the better choice when working with a rational function field. Every automorphism of $K(t)$ is a permutation on the set of rational functions whose numerator and denominator are both degree 1 or less. Thus, the norm of such a rational function down to $K(t)^G$ will be a rational function whose numerator and denominator are of degree $d$ or less, where $d$ is the size of $G$.

Thus, if I pick some $f(t)$ whose numerator and denominator are degree 1 or less and set

$$ u := \frac{a(t)}{b(t)} := \operatorname{Nm}_{K(t) / K(t)^G}(f(t)) $$

then $b(t) u - a(t)$ is a polynomial of degree $|G|$ or less over the field $K(u)$ that has $t$ as a root.

If this polynomial is nonzero (i.e. if $u$ is not constant), then $t$ is a root of the polynomial, and we have

$$ [ K(t) : K(u) ] \leq |G| $$ $$ [ K(t) : K(t)^G ] = |G| $$ $$ K(u) \subseteq K(t)^G $$ and thus $$ [K(t)^G : K(u)] \leq 1 $$

and thus we see that $K(u) = K(t)^G$.

As an example, using $f(t) = t$, the algorithm above reproduces the fact that the fixed field of $t \mapsto 1-t$ is indeed $K(t(1-t))$ .

For your problem, $f(t) = t$ doesn't work, but $f(t) = 1+t$ does, and gives:

$$ u = -\left(\frac{(t+1)(t-2)(2t-1)}{t(t-1)} \right)^2 $$

$\endgroup$
3
$\begingroup$

Consider the following rational function $$ w=\frac{(t^3-3t^2+1)(t^3-3t+1)}{t^2(t-1)^2}. $$ It is relatively easy to check that $w$ is invariant under both $\sigma:t\mapsto 1-t$ and $\tau:t\mapsto 1/t$. The automorphism $\sigma$ maps the factors in the numerator to the negatives of each other (check this), and clearly does the same to the factors of the denominator. Invariance under $\tau$ follows essentially from the observation that the factors in the numerator are reciprocal polynomialss of each other. The degrees match nicely to cancel the effect of $\tau$ to the denominator.

Clearly $t$ is a zero of the sextic polynomial $$ p(x)=(x^3-3x^2+1)(x^3-3x+1)-x^2(x-1)^2w\in\Bbb{Q}(w)[x], $$ so $[\Bbb{Q}(t):\Bbb{Q}(w)]\le6$. OTOH, as pointed out by Hurkyl, the fixed field $K$ of $G=\langle\sigma,\tau\rangle$ ($\subset Aut(\Bbb{Q}(t))$) satisfies $[\Bbb{Q}(t):K]=6$. We just saw that $\Bbb{Q}(w)\subseteq K$, so we can conclude that $$ \operatorname{Inv}(G)=\Bbb{Q}(w). $$


A `funny' thing happend in that $\operatorname{Inv}(G)\simeq\Bbb{Q}(t)$ (send $w$ to $t$). This is not a coincidence. Lüroth's theorem says that all the intermediate fields properly between $\Bbb{Q}$ and $\Bbb{Q}(t)$ are isomorphic to $\Bbb{Q}(t)$.


How did I find $w$? In this earlier question I used the trace to find a non-trivial element $$u=t+\frac1{1-t}+\frac{t-1}t=\frac{t^3-3t+1}{t(t-1)}$$ of the fixed field for the index two subgroup $\langle \sigma\tau\rangle\le G$. Here $w=-u\sigma(u)$ and that is clear the invariant of both $\sigma$ and $\sigma\tau$, hence all of $G$.

$\endgroup$
  • $\begingroup$ I suspect that there may be an easier way for finding a generator. I just wanted to reuse the result of that earlier question. May be you can see something simpler in the intersection of the fixed fields of $\tau$ and the fixed field of $\sigma$? $\endgroup$ – Jyrki Lahtonen Nov 27 '14 at 18:05
  • $\begingroup$ Thanks for your answer. So I think I've managed to find that the fixed fields are $\mathbb{Q}(t(t-1))$ and $\mathbb{Q}(t + 1/t)$ but I'm not quite sure how we would then go about finding the intersection of those two? $\endgroup$ – Wooster Nov 27 '14 at 18:09
  • $\begingroup$ Here in $w$ numerator has degree $6$ and denominator has degree $4$. So can it be invariant under $\tau$? $\endgroup$ – Ri-Li Aug 28 '15 at 21:15
  • $\begingroup$ @user152715: Yes it can. If you expand the denominator as $a_0+a_1t+\cdots +a_6t^6$ you will find that also $a_6=a_0=0$, which is the condition you came up with in your own attempt. Sorry about not making that clear. $\endgroup$ – Jyrki Lahtonen Aug 28 '15 at 21:19
  • $\begingroup$ Yeah, sorry. It is right. Now please come up with my question in the morning. $\endgroup$ – Ri-Li Aug 28 '15 at 21:33

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.