Does there exist an $\alpha$ in an algebraic closure $\mathbb{Q}_p^{\rm alg}$ of $\mathbb{Q}_p$ such that $\frac{p}{p-1} \geq v(\alpha)>0$ and $1+\alpha$ is a $p$th power in $\mathbb{Q}_p(\alpha)$?, where $v$ is the extension of the $p$ adic valuation on $\mathbb{Q}_p$ to $\mathbb{Q}_p^{\rm alg}$, normalized so that $v(p)=1$. I know $1+\alpha$ is always a $p$th power in $\mathbb{Q}_p(\alpha)$ for any $\alpha$ such that $v(\alpha)>\frac{p}{p-1}$, because of the existence of the exponential map. But I would like to know if this is the only way this can happen. This might be a basic question, but I have not been able to decide one way or the other.
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Yes, that can happen. Not if you want $\alpha$ to be a uniformizer of $\mathbb{Q}_p(\alpha)$, of course: If $$ 1 + \alpha = (1+\beta)^p = \sum \binom{p}{i} \beta^p, $$ we must have $v_p(\beta) = \frac{1}{p} v_p(\alpha)$, since all other terms besides $1+\beta^p$ must have larger valuation than $\beta^p$. But for arbitrary $\alpha$, it can happen: Let $p=2$, $\beta=2^{1/2}$, $\alpha = (1+\beta)^2-1 = 2^{3/2} + 2$. Then $\mathbb{Q}_2(\alpha)=\mathbb{Q}_2(\beta)$ since we can write $\beta = \frac{\alpha-2}{2}$, and by construction $1+\alpha$ is a square. Also $v_2(\alpha)=1<\frac{p}{p-1}$.
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$\begingroup$ Here is a more "generic" argument: Let $K$ be a finite extension of $\mathbb{Q}_p$, and $\alpha$ an arbitrary element for which $(1+\alpha)$ is a $p$-th power. For $\gamma$ a primitive element of $K$, all but finitely many of the $\alpha + p^N \gamma$ are a primitive element of $K$ (either by the same argument as used in the proof of the primitive element theorem, or using that there are only finitely many proper subfields of $K$ by separability, and if two $\alpha+p^N\gamma, \alpha+p^M\gamma$ lie in the same, it follows that $\gamma$ does, too.) $\endgroup$ Commented Sep 4 at 8:25
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$\begingroup$ It follows that there are primitive elements $\alpha' = \alpha + p^N\gamma$ arbitrarily close to $\alpha$. For large enough $N$, all of these are also going to have $1+\alpha'$ a $p$-th power, so $K=\mathbb{Q}_p(\alpha')$ and $1+\alpha'$ is a $p$-th power with $v_p(\alpha')=v_p(\alpha)$. So any $K$ which contains elements of positive valuation $\leq \frac{1}{p-1}$ can be used to construct a counterexample. $\endgroup$ Commented Sep 4 at 8:27
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$\begingroup$ Sorry, I meant to take $p$ odd to have to right radius of convergence of the exponential. I am not following the generic argument. It seems to only work if $K$ contains elements $\beta \in K$ such that $\frac{p}{p-1} \geq v(\beta^p-1)>0$, so that one could take $\alpha:=\beta^p-1$. But how do we know such $\beta$ exists? $\endgroup$– Math FoxCommented Sep 4 at 14:51
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$\begingroup$ I see. We can just start with an arbitrary $\alpha$ with $\frac{p}{p-1} \geq v(\alpha) >0$, and define $\beta$ to be $(1+\alpha)^{\frac{1}{p}} \in \mathbb{Q}_p^{\rm alg}$, $K:=\mathbb{Q}_p(\beta)$. $\endgroup$– Math FoxCommented Sep 4 at 15:25