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Let $K = \mathbb{Q}_{13}(\sqrt[4]{13})$ and $L = K(\sqrt[4]{26})$.

I know that $L/K$ is an unramified quartic extension by the following line of arguments:

  • It is $L = K(\sqrt[4]{2})$ since $\sqrt[4]{2} = \frac{\sqrt[4]{26}}{\sqrt[4]{13}}$.
  • Next, consider the polynomial $x^4-2$ over $K$ whose root will define $\sqrt[4]{2}$ (up to conjugacy, of course).
    By reducing the equation $x^4-2=0$ over $\mathbb{F}_{13}$, the residue field of $K$, we can find out that this equation has no solutions over $\mathbb{F}_{13}$. This means that we must extend the residue field.
  • It is $\operatorname{ord}_{13}(2) = 12$. Suppose $\alpha$ is an element satisfying $\alpha^4=2$ over some extension of $\mathbb{F}_{13}$. Then the order of $\alpha$ is $4 \cdot 12 = 48$. The smallest natural number $k$ such that $13^k-1$ is divisible by $48$ is $k=4$, i.e. $\mathbb{F}_{13}[\alpha] = \mathbb{F}_{13^4}$ and therefore the inertial degree of $L/K$ must be $4$.
  • Since we already found a polynomial of degree $4$ where $\sqrt[4]{2}$ vanishes over $K$, it must mean that $L/K$ is unramified of degree $4$.

Now my actual question: If I would have skipped the first step and considered the polynomial $x^4-26$ over $K$, the argument would not work anymore since $x^4 - 26 \equiv x^4$ modulo $\sqrt[4]{13}$ (the uniformizer of $K$).
Why exactly does this not work, considering the element $\sqrt[4]{2}$ works fine?

Thank you!

Jyrki Lahtonen
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Diglett
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  • One reason: integrally, $\mathcal O_K[\sqrt[4]{26}]\subsetneq \mathcal O_L$, and the index $[\mathcal O_L:\mathcal O_K[\sqrt[4]{26}]]$ is divisible by $13$. – Mathmo123 Mar 06 '20 at 13:36
  • It is unramified because $x^4-2\in \Bbb{F}_{13}[x]$ is separable – reuns Mar 07 '20 at 05:27

1 Answers1

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Well your second step states the relation between a local field extension and its residue field extension in a very vague way.

Concretely, it is true that the assertion "there is a minimal polynomial of a primitive element of $L\vert K$ which (is in $\mathcal{O}_L[x]$ and) reduces to a polynomial without solutions in the residue field of $K$" implies that "the residue field of $L$ is a proper extension of the one of $K$".

But that does not mean that the assertion "there is a minimal polynomial of a primitive element of $L\vert K$ which (is in $\mathcal{O}_L[x]$ and) reduces to a polynomial with solutions in the residue field of $K$" would imply that the residue field of $L$ is not a proper extension of the one of $K$; for which your element $\sqrt[4]{26}$ is an example.

Plainly, $(\exists x: A(x) )\implies B$ does not imply $(\exists x: \neg A(x)) \implies \neg B$.

To see it in an easier example, set $K := \mathbb Q_3$ and $L:=K(\sqrt2)$. Note that $L=K(\sqrt{18})$ as well. Now the polynomial $x^2-2 \in \mathbb{Z}_3[x]$ reduces to $x^2-2 \in \mathbb{F}_3[x]$, which has no zeroes in $\mathbb F_3$, which here allows us to quickly conclude that $L\vert K$ must be unramified. On the other hand, the polynomial $x^2-18 \in \mathbb{Z}_3[x]$ modulo $3$ reduces to just $x^2 \in \mathbb F_3[x]$, which of course has the double zero $0$, but this just does not tell us anything, and in particular is no contradiction to the other statement.

Torsten Schoeneberg
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