How to show that $\sqrt[4]{13}$ is not in $\mathbb{Q}_{13}(\sqrt[4]{26})$ Approach


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*I was able to show that $[\mathbb{Q}_{13}(\sqrt[4]{26}):\mathbb{Q}_{13}] = 4$ since $x^4 - 26$ in an irreducible polynomial in $\mathbb{Q}_{13}[x]$ (this can be shown by using Eisenstein's Criterion and the Gaussian Lemma). Therefore, a basis of $\mathbb{Q}_{13}(\sqrt[4]{26})$ as a $\mathbb{Q}$-vector space is $$ (1, \sqrt[4]{26}, \sqrt{26}, \sqrt[4]{26^3}).$$

*If $\sqrt[4]{13}$ would be in $\mathbb{Q}_{13}(\sqrt[4]{26})$, then there would be coefficients $c_0,c_1,c_2,c_3 \in \mathbb{Q}_{13}$ such that $$(\sum_{k=0}^4 c_k \sqrt[4]{26}^k)^4 = 13. $$
However, getting a contradiction out of this equation seems to be super complicated.


Could you please help me with this problem? Thank you!
 A: Let $\pi_K$ an uniformizer, $v(p) = 1,v(\pi_K) = 1/e$ and the residue field $O_K/(\pi_K) = F_{p^f} = \sum_{j=0}^{f-1} \zeta_{p^f-1}^j F_p$ then $O_K = \sum_{m=0}^{e-1} \sum_{j=0}^{f-1} \zeta_{p^f-1}^j \pi_K^m Z_p$ so $[O_K:Z_p] = e f$. 
Here $K = Q_p[a], a= 2^{1/4} 13^{1/4}, v(a)=1/4, e \ge 4$, since $a$ is the root of a polynomial of degree $4$ it means $ef \le 4,\pi_K = a, e=4,f = 1$ and $13^{1/4} \in O_K$ iff $2^{1/4} \in O_K$ iff (as $x^4-2\bmod p$ is separable, Hensel lemma) $2^{1/4} \in O_K/(\pi_K)=F_p$.
A: Note that $2^{(13^2-1)/4}=(2^6)^7$ is congruent to $-1$ mod $13$, thus $2$ is not a fourth root in a quadratic extension of $\mathbb{F}_{13}$. 
Therefore, $X^4-2$ is irreducible in $\mathbb{F}_{13}$, thus in $\mathbb{Z}_{13}$ thus in $\mathbb{Q}_{13}$. 
So $K=\mathbb{Q}_{13}[X]/(X^4-2)=\mathbb{Q}_{13}(2^{1/4})$ is a field with dimension $4$ over $\mathbb{Q}_{13}$. 
So it is enough to prove that ${\sqrt{13}} \notin K$. 
Now, if $\sqrt{13} \in K$, there are $a,b,c,d \in \mathbb{Q}_{13}$ such that $(a+b2^{1/4}+c2^{2/4}+d2^{3/4})^2=13$.
The equations are rewritten as $a^2+4bd+c^2=13$, $2ac+b^2+2d^2=0$, $4dc+2ab=0$, $2ad+2bc=0$. 
Thus $ab=-2cd$, $ad=-bc$, $2ac+b^2+d^2=0$, $a^2+4bd+c^2=13$. 
Assume $a=0$, then $c=0$ or $d=0$ and $b=0$ or $c=0$. If $c=0$, then $b^2+d^2=0$ and $4bd=13$, which is impossible because the $13$-adic valuations of $b$ and $d$ must both be the same and have an odd sum. 
So if $a=0$, then $c \neq 0$, thus $b=d=0$, therefore $2c^2=13$, impossible. 
So $a \neq 0$, thus $b=-2cd/a$, thus $d=-bc/a=2c^2d/a^2$, hence $d=0$ or $(c/a)^2=1/2$. The latter is impossible, because $2$ is not a square root mod $13$ (and $\mathbb{Z}_{13}$ is integrally closed and surjects to $\mathbb{F}_{13}$), thus $b=d=0$, thus $2ac=0$, thus $c=0$, thus $a^2=13$, impossible. 
So $\sqrt{13} \notin K$. 
Thus there is no field extension of $\mathbb{Q}_{13}$ with dimension $4$ containing $2^{1/4}$ and $13^{1/2}$, let alone $26^{1/4}$ and $13^{1/4}$. 
