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All the extensions with this form $\frac{\mathbb Z_p[x]}{\langle p(x)\rangle}:\mathbb Z_p$ are Galois extensions?

where $p$ is prime, $p(x)$ is irreducible and the extension is finite.

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    $\begingroup$ If $p$ is a prime (like $p=2,3,5,7,11...$) then what does your notation $p(x)$ mean? Maybe change it to $f(x) \in \mathbb{Z}_p[x]$ etc. $\endgroup$ – coffeemath Aug 1 '17 at 3:57
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A finite extension of $\mathbb{Z}_p$ is a finite vector space of dimension $n$ over $\mathbb{Z}_p$. It has $p^n$ elements and is the splitting field of $X^{p^n}-X$, so it is a normal extension and separable extension since $X^{p^n}-X$ is separable so it is a Galois extension.

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  • $\begingroup$ what if $p$ is not prime? $\endgroup$ – user441848 Aug 1 '17 at 4:17
  • $\begingroup$ Does the result still holds? $\endgroup$ – user441848 Aug 1 '17 at 4:20
  • $\begingroup$ or what if we write $\mathbb Q$ instead of $\mathbb Z?$ with all the conditions as above. $\endgroup$ – user441848 Aug 1 '17 at 4:34
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    $\begingroup$ @Anneliset. If $p$ is not a prime, then this is not a field because factors of $p$ are zero divisors. Over $\Bbb{Q}$ you do get an extension field, but not necessarily Galois. Over $\Bbb{Z}_p$ all finite extensions are Galois (with a cyclic Galois group). $\endgroup$ – Jyrki Lahtonen Aug 5 '17 at 17:45
  • $\begingroup$ @JyrkiLahtonen 'Over $\mathbb Z_p$ all finite $K/F$ are Galois' Knowing that, I don't see any problem with the solution of an exercise that I solved in my exam math.stackexchange.com/questions/2381892/… .There were 2 similar problems and my professor didn't gave any points on them, so I failed my exam. $\endgroup$ – user441848 Aug 5 '17 at 19:20
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To add on Tsemo's answer.

$q = p^k$ where $p$ is prime.

  • If $f \in \mathbb{F}_q[x], deg(f) = n$ is irreducible then $K = \mathbb{F}_q[x]/(f(x))$ is a finite integral domain and hence a field, with $q^n$ elements.

  • Since $p = 0$ in $K$, $(a+b)^p = a^p+b^p$ and $\phi(a) = a^p$ is an automorphism of $K$, as well as $\varphi(a) = a^q = \phi^k(a)$.

  • Since $\mathbb{F}_q^\times $ has $q-1$ elements, those satisfy $a^{q-1} = 1$ and hence $\varphi \in \text{Gal}(k/\mathbb{F}_q)$.

    So all the elements of $\mathbb{F}_q$ are roots of $x^q-x$. But this polynomial of degree $q$ has at most $q$ roots in the field $k$. Thus $\mathbb{F}_q$ is exactly the splitting field of $x^q-x$.

  • Let $G = \langle \varphi \rangle$ the cyclic group generated by $\varphi$. We obtained $$\mathbb{F}_q = K^G = \{ a \in K, \forall g \in G, g(a) = a\}= \{ a \in K, a^q-a=0\}$$ and hence $K/\mathbb{F}_q$ is a Galois extension with Galois group $G$.

  • If $\alpha \in K$ then its minimal polynomial over $\mathbb{F}_q$ is $$g(X) = \prod_{\beta \in G (\alpha)} (X-\beta)=\prod_{m=0}^{l-1} (X-\varphi^m(\alpha))=\prod_{m=0}^{l-1} (X-\alpha^{q^m}) \qquad \in \mathbb{F}_q[X]$$ where $l$ is the least positive integer such that $\alpha^{q^l} = \alpha$

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