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 Post subject: Galois Group and Solvability by Radicals Posted: Thu Jun 09, 2016 7:34 am
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Let $\displaystyle K = \mathbb{Z}_{p}(t)$ be the field of rational functions of t over $\displaystyle \mathbb{Z}_{p}$, where $\displaystyle p$ is a prime. Let $\displaystyle f(x) = x^p - x - t \in K[x],$ and let $\displaystyle E$ be the splitting field of $\displaystyle f(x)$ over $\displaystyle K$. Show that the Galois group of $\displaystyle f(x), \, G=Gal(E/K),$ is isomorphic to $\displaystyle \mathbb{Z}_{p},$ but $\displaystyle f(x)$ is not solvable by radicals.

Top   Post subject: Re: Galois Group and Solvability by Radicals Posted: Thu Jun 09, 2016 7:35 am
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Joined: Tue Nov 10, 2015 8:25 pm
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Let me post a solution to this exercise. Because i'm not sure whether it's correct or not, any comments or corrections would be appreciated.

Let $\displaystyle f(x) = x^p - x - t \in K[x],$ where $\displaystyle K = \mathbb{Z}_{p}(t),$ and let $\displaystyle E/K$ be the splitting field of $\displaystyle f(x).$ Now, let $\displaystyle \rho \in E$ be a root of $\displaystyle f(x).$ Then $\displaystyle f(\rho) = 0 \implies \rho^p -\rho -t = 0$Notice that
\begin{align*}
f(\rho+1) &= (\rho+1)^p - (\rho+1) - t \\
&= \rho^p +1^p -\rho - 1 - t \\
&= (\rho^p -\rho -t) + (1-1) \\
&= 0
\end{align*}
since $\displaystyle char\left(E\right) = 0$, which means that $\rho + 1$ is also a root of $\displaystyle f(x)$. Because $\displaystyle f(x)$ has at most p roots in E, $\rho, \, \rho + 1, \, \dots , \, \rho + (p-1)$ are all roots of $\displaystyle f(x).$ It follows that $\displaystyle f(x)$ is seperable and hence $\displaystyle E/K$ is a Galois extension. So $\displaystyle E/K$ is simple and obviously $\displaystyle E = K(\rho).$

We, now, observe that the mappings
$\sigma_{k} : E \longrightarrow E \; , \; \sigma_{k}(\rho) = \rho + k$ are exactly the elements of Galois group of the extension $\displaystyle E/K.$ Since $\sigma_{1}^{i}(\rho) = \sigma_{i}(\rho) \, , \, \forall i \in \{ 1, \dots ,p \},$ the element $\sigma_{1}$ is a generator of $\displaystyle Gal\left( E/K \right) ,$ which means that $\displaystyle Gal\left( E/K \right)$ is a cyclic group of order p and thus isomorphic to $\displaystyle \mathbb{Z}_{p}.$

Note that $\displaystyle Gal\left( E/K \right)$ is abelian, since it is cyclic, and hence solvable. We will, however, show that $\displaystyle f(x)$ is not solvable by radicals.

Let us suppose the opposite, that is $\displaystyle f(x)$ is solvable by radicals. Then there exists a radical extension $\displaystyle B_{t} / K$ which contains the splitting field $\displaystyle E/K$ of $\displaystyle f(x)$. (It can be easily proved that) we can suppose that $\displaystyle B_{t}=E$. Since the extension $\displaystyle E/K$ is radical, there exists a radical tower
$\displaystyle B_{0} = K \, \subset \, B_{1} \, \subset \, \dots \, \subset \, B_{r-1} \, \subset \, B_{r} = E,$that is every extension $\displaystyle B_{i}/B_{i-1}$ is pure of type $\displaystyle m_{i} \in \mathbb{N}.$ Since
$\displaystyle \left[ E : K \right] = \left[ Gal(E/K) \right] : 1 ] = p,$from the degree formula we have that
$\displaystyle p = \left[ B_{r} : B_{r-1} \right] \cdot \left[ B_{r-1} : B_{r-2} \right] \cdot \dots \cdot \left[ B_{1} : B_{0} \right] \in \mathbb{P}$
Thus, there is a $\displaystyle j \in \{ 1, \dots , p \}$ such that $\displaystyle \left[ B_{j} : B_{j-1} \right] = p,$ and the following hold: $\displaystyle \left[ B_{i} : B_{i-1} \right] = 1 , \; \forall i \neq j.$ Since $\displaystyle E$ is the splitting field of $\displaystyle f(x)$, it follows that $\displaystyle j=r=1$, so the tower becomes $K \subset E=K(\rho)$ and, additionally, we have that $\displaystyle \rho^{p} \in K.$ Then
$\displaystyle irr( \rho, K ) = x^p - \rho^p = (x-\rho)^p \in K[x]$ However, $\displaystyle f(x)$ is irreducible over $\displaystyle K$, monic of degree p and has $\displaystyle \rho$ as a root. Due to the fact that $\displaystyle irr(\rho,K)$ is unique, we conclude that $\displaystyle f(x) = irr(\rho,K)$, which cannot happen. Thus, we have reached a contratiction. Therefore, $\displaystyle f(x)$ cannot be solved by radicals.

$\color{red}{Comment:}$ It is known that if the galois group of a polynomial with coefficients over a field of characteristic 0 is solvable, then this polynomial is solvable by radicals over this field. The above exercise shows that this isn't true for polynomials with coefficients over a field of characteristic p, where p is a prime number.

Top   Post subject: Re: Galois Group and Solvability by Radicals Posted: Thu Jun 09, 2016 7:36 am

Joined: Mon Nov 09, 2015 11:52 am
Posts: 77
Location: Limassol/Pyla Cyprus
I believe there is a small problem towards the end of the solution. It is not necessary that $\rho^p \in K$. What we can say is that there is an element $\alpha \in K(\rho)$ with $\alpha \notin K$ such that $\alpha^p \in K$. From this we can derive a contradiction as follows:

Since $\alpha \notin K$ then $[K(\alpha):K] > 1$ and so $[K(\alpha):K] = p$. Therefore $K(\alpha) = K(\rho)$ and so there is a polynomial $g(x) \in K[x]$ such that $\rho = g(\alpha)$. But then $\rho^p = g(\alpha)^p = g(\alpha^p) \in K$ a contradiction as $\rho^p = \rho+t \notin K$

I now realise that there is something else that we do need to check: We need to show that $f(x)$ has no root in $K$. [Simple by assuming that it has a root of the form $a(t)/b(t)$. Then we get $a(t^p) = a(t)b(t)^{p-1} + tb(t^p)$ and we can derive a contradiction by comparing the degrees of the polynomials.] Once this is shown we also need to show that $f(x)$ is irreducible but this is rather simple.

Top   Post subject: Re: Galois Group and Solvability by Radicals Posted: Thu Jun 09, 2016 7:36 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314

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