4
$\begingroup$

I'm trying to determine some explicit invariant subfields of $L=\mathbb{Q}(X)$. We have defined $\sigma_i\in$ Aut$(L)$ with $$\sigma_1(X)=1/X,\ \sigma_2(X)=1-X.$$ We are now asked to determine the invariant subfields $L^{\langle\sigma_i\rangle}$ for i=1,2. Also we must show that $\rho=\sigma_1\sigma_2$ has order 3 in Aut$(L)$ and we have to determine $L^{\langle\rho\rangle}$. For the first part I think it's almost complete.$$\text{Claim:}\ L^{\langle\sigma_1\rangle}=\mathbb{Q}(X+1/X).$$ We know $\langle\sigma_1\rangle=\{id,\sigma_2\}$ and by our definition of $L$ that $\mathbb{Q}(X+1/X)\subseteq L^{\langle\sigma_1\rangle}\subseteq \mathbb{Q}(X)$. We now say the minimum polynomial of $1/X$ is $f_{\mathbb{Q}(X+1/X)}^{1/X}=T^2-(X+1/X)T+1\in \mathbb{Q}(X+1/X)[T]$. Because the degree of $f_{\mathbb{Q}(X+1/X)}^{1/X}$ is 2, $[\mathbb{Q}:\mathbb{Q}(X+1/X)]=2$. We also know that $X\notin L^{\langle\sigma_1\rangle}$ so $L^{\langle\sigma_1\rangle}\neq\mathbb{Q}(X).$ Hence the claim is true. $$\text{Claim:}\ L^{\langle\sigma_2\rangle}=\mathbb{Q}(X(1-X)).$$ The claim is a guess because I suppose by $\sigma_2$ we have that $X\mapsto1-X$ and conversely $1-X\mapsto X$ (so together they make $X$?). Lastly, because in Aut$(L)$ we have that $X\mapsto1$ and therefore $$\rho^3=(\sigma_1\sigma_2)^3=(1/X-1)^3=1/X^3-3/X^2+3/X-1=1-3+3-1=0$$ and the order of $\rho$ is 3. Am I thinking on the right path or is this not even close? All hints and/or answers are highly appreciated!

$\endgroup$
9
  • 1
    $\begingroup$ The $\sigma_{1}$ part is well done. The case of $\sigma_{2}$ is similar, as $X \notin L^{\langle\sigma_2\rangle}$, and $X$ is a root of $T^{2} - T + X(1 - X)$. Finally we have, writing permutations on the right, $X \rho = X \sigma_{1} \sigma_{2} = (1/X) \sigma_{2} = 1/(1-X)$. Then $X \rho^{2} = 1/(1 - 1/(1-X)) = 1 - 1/X$ and $X^{\rho^{3}} = 1 - 1/(1/(1-X)) = X$, so $\rho$ has order $3$. $\endgroup$ Apr 23, 2018 at 12:39
  • $\begingroup$ Andreas Caranti Nice, thanks for your answer! When being given the polynomial it seems very obvious, but is there a general approach for finding them? $\endgroup$
    – Algebear
    Apr 24, 2018 at 14:26
  • $\begingroup$ In this case it is easy, because the group $\langle\sigma_2\rangle$ has order $2$, so if $X \notin L^{\langle\sigma_2\rangle}$, its minimal polynomial over it is $(T - X) (T - X \sigma_{2}) = (T - X) (T - (1 - X)) = T^{2} - T + X (1 - X)$, the coefficients $X + X \sigma_{2}$ and $X \cdot X \sigma_{2}$ being obviously fixed by $\sigma_{2}$. $\endgroup$ Apr 24, 2018 at 14:48
  • $\begingroup$ See my answer here $\endgroup$ Apr 25, 2018 at 12:01
  • $\begingroup$ A prime candidate for a duplicate. I answered that one also, so won't cast the first vote to close. Particularly as Jens has worked independently (+1), and as a new user hardly had a reason to suspect that this has been covered earlier. $\endgroup$ Apr 25, 2018 at 12:03

1 Answer 1

1
$\begingroup$

Let $L = \mathbb{Q} (X)$, let $\sigma_1, \sigma_2 \in \text{Aut}(L)$, with $\sigma_1 (X)= 1/X$ and $\sigma_2 (X) = 1 - X$. Let $\rho = \sigma_1 \sigma_2$.

Part 1) Show that $|\rho| = 3$ in $\text{Aut}(L)$.

We have $ \rho(X) = (\sigma_1 \sigma_2)(X)=\sigma_1(\sigma_2 (X)) = \sigma_1(1-X)=1-\frac{1}{X} = \frac{X-1}{X} $

$ \rho^2(X) =\rho(\frac{X-1}{X}) = \frac{\frac{X-1}{X}-1}{\frac{X-1}{X}} = \frac{-1}{X} \frac{X}{X-1} = \frac{1}{1-X}$

$ \rho^3(X) = \rho(\frac{1}{1-X}) = \frac{1}{1-\frac{X-1}{X}} = \frac{1}{(\frac{X-X+1}{X})} = X.$ Hence $|\rho| = 3$.

Part 2) Determine $L^{<\rho>}$. If we let $T1 = X, T2 = \rho(X), T3 = \rho^2(X)$, then every symmetric polynomial in $\mathbb{Q}[T1,T2,T3]$ is invariant under $\rho$, because $\rho$ has order $3$. We consider

$s_1 = T1 + T2 + T3 = X + \frac{X-1}{X} + \frac{1}{1-X} = \frac{1-3X+X^3}{(-1+X)X}.$

Let $K = \mathbb{Q}(X + \frac{X-1}{X} + \frac{1}{1-X})$, then we get the tower of extensions $K \subset L^{<\rho>} \subset L$. We want to determine $ [K : L] = \text{deg}f_K^{X}.$

Let $f \in K[T]$ be defined by $f=(X + \frac{X-1}{X} + \frac{1}{1-X})(T(T-1))-T^3+4T-1$. Then $f(X) = 0$. As the degree of $f$ is 3, we have $\text{deg}f_K^X \le 3$. Since $\rho(X) \not = X$, we have that $L^{<\rho>} \subsetneq L$, and by the tower of extensions also $K \subsetneq L$, hence $1 < \text{deg}f_K^X$. Hence $\text{deg}f_K^X$ is either $2$ or $3$, which are both primes. And since we already had that $L^{<\rho>} \subsetneq L$, we find that it must hold that $L^{<\rho>} = K$. $\Box$

$\endgroup$
3
  • $\begingroup$ Why can you define T2 and T3 like so? $\endgroup$
    – Algebear
    Apr 25, 2018 at 18:03
  • $\begingroup$ I fixed a mistake: the symmetric polynomials had to be in $\mathbb{Q}[T1,T2,T3]$. We can define the T2, and T3 because they are elements of $\mathbb{Q}(X)$ $\endgroup$ Apr 25, 2018 at 18:28
  • $\begingroup$ I think $\sigma_1(1-X)=\frac{1}{1-X}$ in the first line of computation. Then we get $$\rho^2(X)=\frac{1}{1-\frac{1}{1-X}}=\frac{1-X}{1-X-1}=\frac{X-1}{X}.$$ But what happens then? $\endgroup$
    – Algebear
    Dec 18, 2018 at 18:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .