# Extension of degree 5 not obtained by adjoining a 5th root and whose normal closure contains a primitive 5th root of unity.

The question is mainly what the title says, but here is the setup in more details.

Let $$K$$ be a field not containing a primitive 5th root of unity (for this question, the case $$K = \mathbb{Q}$$ seems already interesting). Let $$L$$ be a degree $$5$$ extension such that $$L\setminus K$$ contains no element $$x$$ with $$x^5 \in K$$. Is it possible for the normal closure of $$L$$ to contain a primitive 5th root of unity?

EDIT: I am trying to find an example using local fields, and there are some good candidates in $$\mathbb{Q}_5$$. Indeed, using the table of local fields of Jones and Roberts, I see that $$\mathbb{Q}_5$$ has $$4$$ field extensions of degree $$5$$ with discriminant having $$5$$-adic valuation $$7$$, Galois group $$F_5\cong C_5\rtimes C_4$$ and which are totally ramified (because the inertia group is the same as the Galois group). Those fields are not of the form $$\mathbb{Q}_5[X]/(X^5-a)$$ (because the discriminant would be of $$5$$-adic valuation either $$5$$ or $$9$$ depending on the valuation of $$a$$). The only thing I'm unsure of is how to check whether their normal closure contain a primitive $$5$$th root of unity. Any idea on how to do that?

I'm actually interested in this question when one replaces all occurrences of $$5$$ by an arbitrary prime number. Note that for $$p = 3$$, the answer is no, because the normal closure of L contains a primitive 3rd root of unity if and only if the fundamental discriminant is $$-3$$, which is equivalent to $$L$$ being a pure cubic field. But already for $$p=5$$ I have no idea of how to approach the question, and any suggestion is welcome.

Let us try to approach the problem starting with the minimal assumptions: $$p$$ is an odd prime, and all the fields we consider have characteristic $$\neq p$$, and if necessary, adding hypotheses as we go along.

1. For any finite extension $$E/F$$, we want to express in a "functorial way" your cumbersome condition that $$E\setminus F$$ contains no $$x\in E^*$$ s.t. $$x^p \in {F^*}$$. Introduce the natural homomorphism $$\delta_{E/F} : F^*/{F^*}^p \to E^*/{E^*}^p$$, with kernel $$\Delta_{E/F}$$. Then your condition is obviously equivalent to the injectivity of $$\delta_{E/F}$$, i.e. the triviality of $$\Delta_{E/F}$$.

2. Coming back to your question, let $$L$$ be a finite extension of $$K$$ with trivial kernel $$\Delta_{L/K}$$. Suppose that $$N$$ contains the group $$\mu_p$$ of $$p$$-th roots of unity. Then $$N$$ contains $$\mathcal K=K(\mu_p)$$ and $$\mathcal L=L(\mu_p)$$, and $$N/\mathcal K$$ is galois, say with group $$G$$. We can apply Kummer theory to describe $$\Delta_{N/\mathcal K}$$, taking the $$G$$-cohomology of the exact sequence $$1\to\mu_p\to N^* \to {N^*}^p \to 1$$. This yields an exact sequence of groups $$1\to \mu_p\to {\mathcal K}^*\to {\mathcal K}^* \cap {N^*}^p \to H^1(G,\mu_p)=Hom(G,\mu_p)$$ [since $$G$$ acts trivially on $$\mu_p$$]$$\to H^1(G,N^*)=0$$ [by Hilbert's thm. 90]. Summarizing, we have a canonical isomorphism $$\Delta_{N/\mathcal K}\cong Hom(G,\mu_p)$$, and thus this kernel is completely known through $$G$$. If $$p$$ divides the order of $$G$$ (as is the case in your question), certainly $$\Delta_{N/\mathcal K}$$ is not trivial. To go further we need information on $$N$$, but in general we don't even know whether $$\mathcal L/\mathcal K$$ is galois.

3. From now on, we add your hypothesis that $$[L:K]=p$$, but even this is not sufficient to get hold of $$N$$ (see "hint" at the end). Extra simplification : suppose $$N=\mathcal L$$, as seems to happen in the example given by @user293657, and consequently $$\Delta_{\mathcal L/\mathcal K}$$ is cyclic of order $$p$$. But $$[\mathcal K:K]$$ divides $$p-1$$, so that $$\mathcal K/K$$ and $$L/K$$ are linearly disjoint, and $$H:=Gal(\mathcal L/L)\cong Gal(\mathcal K/K)$$ has order prime to $$p$$. Here we are in the so called semi-simple case, where any $$\mathbf Z_p [H]$$-module $$M$$ can be decomposed into the direct sum of its isotypical components, using the idempotents of the group algebra. If $$M$$ is killed by $$p$$, it will be convenient to vizualize it as an $$\mathbf F_p$$-vector space on which a generator $$h$$ of $$H$$ acts linearly, and then the isotypical decomposition is just the decomposition of $$M$$ into eigenspaces corresponding to the eigenvalues of $$h$$.

We need first to know $$\Delta_{\mathcal K/K}$$ and $$\Delta_{\mathcal L/L}$$. It suffices to repeat the demonstration of 2), knowing that $$\mathcal K$$ (resp. $$\mathcal K$$) does not contain $$\mu_p$$. This implies that $${\mathcal K}^* \cong {{\mathcal K}^*}^p$$, hence straightforwardly $$\Delta_{\mathcal K/K}\cong {({\mathcal K}^* / {{\mathcal K}^*}^p)}^H$$ $$= N_H ({\mathcal K}^* / {{\mathcal K}^*}^p)$$ , where $$N_H (.)$$ denotes the image of the norm map of the group algebra, and $${(.)}^H$$ the fixed points under $$H$$ ( = the eigenspace corresponding to the eigenvalue 1). Second, notice that $$\mathcal L$$ is of the form $$\mathcal K (\sqrt [p] a)$$ for a certain $$a\in {\mathcal K}^* / {{\mathcal K}^*}^p$$ (slightly abusig language). Fix a generator $$h$$ of $$H$$ and a generator $$\zeta$$ of $$\mu_p$$. It's a classial exercise in Galois theory to show that $$\mathcal L/ K$$ is galois iff $$a$$ belongs to the eigenspace which corresponds to the eigenvalue $${\omega (h)}^{-1}$$, with $$h(\sqrt [p] a)={(\sqrt [p] a)}^{\omega (h)}$$, and this is of course distinct from the previous eigenspace corresponding to 1. Finally, put $$b=N_{\mathcal K/K} (a)$$. From what we have just seen, $$b$$ is not trivial, and by your injectivity hypothesis, $$\delta_{L/K} (b)$$ is a non trival class of $$L^*/{L^*}^{p}$$. However $$a$$ becomes trivial in $${\mathcal L}^* / {{\mathcal L}^*}^p$$, hence $$N_{\mathcal L/L} (\delta_{\mathcal L/\mathcal K} (a))$$ is trivial (draw a commutative Galois diagram): contradiction . This gives an affirmative answer to your question (and seems to contradict @user293657).

Hint. To build more examples, perhaps you could use miscellaneous results on radical extensions in Kaplanski's book "Fields and Rings", I, 12.

• @user293657 I checked your example using a direct (hence much simpler) application of my general approach. So $K=\mathbf Q_5$, L is the explicit extension of degree 5 which you gave, $\mathcal L$ its normal closure. Putting $\mathcal K=K(\mu_5)$ (my notation), you assert that $\mathcal L=L(\mu_5)$. I accept your proof that $\delta_{L/K}$ (in my language) is injective. Let me show a contradiction. Because of linear disjointness of $L$ and $\mathcal K$ over $K$, we see that $\mathcal L/\mathcal K$ is cyclic of degree 5, hence, by Kummer theory, $\mathcal L = \mathcal K (\alpha)$, with... Commented Mar 14, 2019 at 21:05
• ... ${\alpha}^5=a \in {{\mathcal K}^*}^5$. Since $G(\mathcal L/L)\cong G(\mathcal K/K)$ , use the same notation $N$ for the corresponding norm maps. Recall that the normality of $\mathcal L/K$ implies that $b=N(a)$ is not a 5-power in $K^*$, as I showed, hence $b$ is not a 5-power in $L^*$ either because of the injectivity of $\delta_{L/K}$. But, on the other side, $b=N(\alpha)^5$ : contradiction. If I'm correct, the conclusion is that the assertion above, that $\mathcal L$ is the normal closure of $L$, is incorrect. Commented Mar 14, 2019 at 21:25
• Dear @nguyen quang do, after some thinking about your answer, I finally have some doubts about it. The key problem is when you claim that $b=N_{\mathcal{K}/K}(a)$ is non-trivial (I guess you mean "as an element of $K^*/(K^*)^p$"). Setting $k=\omega (h)$ (where $\omega (h)$ is as in your answer), we can actually compute that this norm is $a^{1+k+\dots +k^{p-2}}\lambda^p$ for some $\lambda \in \mathcal{K}^*$. Hence one can show that $b=N_{\mathcal{K}/K}(a)$ is non-trivial in $K^*/(K^*)^p$ if and only if $k=1$. How do you prove exactly that $k=\omega (h)$ has to be $1$? Commented Mar 19, 2019 at 10:54
CORRECTION: The following example is wrong, because the discriminant is not an invariant of a field extensions, it's only an invariant up to a square! Actually, using the continuity of automorphisms, we have that $$\omega (h) = 1$$ (using the notation of nguyen quang do's answer). Hence the argument in nguyen quang do's answer applies and proves that $$L$$ is of the form $$K[X]/(X^5-a)$$ for some $$a\in \mathbb{Q}_5$$.
So it turns out an example can be constructed using $$K=\mathbb{Q}_5$$. Taking $$L = K[X]/(X^5+15X^3+5)$$, we get a field extension of discriminant having 5-adic valuation $$7$$, hence $$L$$ is not isomorphic to $$K[X]/(X^5-a)$$ for any $$a\in \mathbb{Q}_5$$ because this would have discriminant $$5^5.a^4$$ (hopefully the discriminant is an invariant of the extension). Finally, according to what is written in the proof of Proposition 2.3.1 of the paper explaining the table of local fields, the cyclic 4 extension of $$\mathbb{Q}_5$$ that one has to adjoin to obtain the normal closure of $$L$$ is $$\mathbb{Q}_5[X]/(X^4-1125)$$, which is indeed isomorphic to $$\mathbb{Q}_5[X]/(X^4+X^3+X^2+X+1)$$ (again using the table of local fields of Jones and Roberts).