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PostPosted: 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.


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PostPosted: Thu Jun 09, 2016 7:35 am 
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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.


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PostPosted: Thu Jun 09, 2016 7:36 am 

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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.


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PostPosted: Thu Jun 09, 2016 7:36 am 
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Thank you very much for your comments!


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