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Here's another proof of the fact that any root of a (non-zero) $p$-adic power series (convergent on the open unit disc) is algebraic. The argument doesn't rely on the Weierstrass preparation theorem, but instead proof is a little silly in that it uses Tate's theorem on Galois invariants of ${\mathbb C}_p$ (which is quite deep) instead of the Weierstrass preparation theorem (which is fairly elementary). But here it is in any case.

The key fact I need is the following which I'll prove after explaining how it completes the proof.

Claim: Any transcendental element of ${\mathbb C}_p$ has uncountably many conjugates (under $\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p))$.

Accepting this claim for the moment, let $\alpha$ be a transcendental zero of $f(x)$, our convergent power series. Then all conjugates of $\alpha$ are also zeroes of $f(x)$ in the open unit disc. Thus, by the above claim, $f(x)$ has uncountably many zeroes in the open unit disc of ${\mathbb C}_p$. However, any uncountable set in a separable metric space (e.g. ${\mathbb C}_p$) has an accumulation point. But then the zeroes of $f(x)$ have an accumulation point, necessarily in the open unit disc since that is closed in ${\mathbb C}_p$, which forces $f(x)$ to be identically zero.

Returning now to the claim, let $G=\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p)$ and let $H$ denote the subgroup of $G$ which stabilizes $\alpha$. Since the conjugates of $\alpha$ are in one-to-one correspondence with $G/H$, we will show that $H$ has uncountable index.

First note that $G$ acts continuously on ${\mathbb C}_p$ where we give ${\mathbb C}_p$ the $p$-adic topology (not the discrete topology). Thus $H$ is a closed subgroup and hence of the form $\text{Gal}(\overline{\mathbb Q}_p/M)$ for some algebraic extension$M/{\mathbb Q}_p$.

Assume $G/H$ is finite, and thus that $M$ is a finite extension of ${\mathbb Q}_p$. But then by Tate's theorem, $\alpha$ must be in $M$ as it is fixed by $\text{Gal}(\overline{\mathbb Q}_p/M)$. This is impossible as $\alpha$ is transcendental.

Thus, $G/H$ is infinite, and we must now show that it is uncountable. We use the following (standard) fact:

Fact: Any infinite compact Hausdorff space with no isolated points is uncountable.

Since $G$ is compact, $G/H$ is compact. The coset space $G/H$ is Hausdorff since $H$ is closed. To see that $G/H$ has no isolated points, note that if it has one isolated point, then all of its points are isolated as $G$ acts transitively (by left multiplication) on $G/H$ by homeomorphisms. But then $G/H$ is discrete which is impossible as it is infinite and compact.

Thus, $G/H$ is uncountable, and $\alpha$ has uncountably many conjugates.

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Here's another proof of the fact that any root of a (non-zero) $p$-adic power series (convergent on the open unit disc) is algebraic. The argument doesn't rely on the Weierstrass preparation theorem, but instead uses Tate's theorem on Galois invariants of ${\mathbb C}_p$.

The key fact I need is the following which I'll prove after explaining how it completes the proof.

Claim: Any transcendental element of ${\mathbb C}_p$ has uncountably many conjugates (under $\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p))$.

Accepting this claim for the moment, let $\alpha$ be a transcendental zero of $f(x)$, our convergent power series. Then all conjugates of $\alpha$ are also zeroes of $f(x)$ in the open unit disc. Thus, by the above claim, $f(x)$ has uncountably many zeroes in the open unit disc of ${\mathbb C}_p$. However, any uncountable set in a separable metric space (e.g. ${\mathbb C}_p$) has an accumulation point. But then the zeroes of $f(x)$ have an accumulation point, necessarily in the open unit disc since that is closed in ${\mathbb C}_p$, which forces $f(x)$ to be identically zero.

Returning now to the claim, let $G=\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p)$ and let $H$ denote the subgroup of $G$ which stabilizes $\alpha$. Since the conjugates of $\alpha$ are in one-to-one correspondence with $G/H$, we will show that $H$ has uncountable index.

First note that $G$ acts continuously on ${\mathbb C}_p$ where we give ${\mathbb C}_p$ the $p$-adic topology (not the discrete topology). Thus $H$ is a closed subgroup and hence of the form $\text{Gal}(\overline{\mathbb Q}_p/M)$ for some algebraic extension$M/{\mathbb Q}_p$.

Assume $G/H$ is finite, and thus that $M$ is a finite extension of ${\mathbb Q}_p$. But then by Tate's theorem, $\alpha$ must be in $M$ as it is fixed by $\text{Gal}(\overline{\mathbb Q}_p/M)$. This is impossible as $\alpha$ is transcendental.

Thus, $G/H$ is infinite, and we must now show that it is uncountable. We use the following (standard) fact:

Fact: Any infinite compact Hausdorff space with no isolated points is uncountable.

Since $G$ is compact, $G/H$ is compact. The coset space $G/H$ is Hausdorf Hausdorff since $H$ is closed. To see that $G/H$ has no isolated points, note that if it has one isolated point, then all of its points are isolated as $G$ acts transitively (by left multiplication) on $G/H$ by homeomorphisms. But then $G/H$ is discrete which is impossible as it is infinite and compact.

Thus, $G/H$ is uncountable, and $\alpha$ has uncountably many conjugates.

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Here's another proof of the fact that any root of a (non-zero) $p$-adic power series (convergent on the open unit disc) is algebraic. The argument doesn't rely on the Weierstrass preparation theorem, but instead uses Tate's theorem on Galois invariants of ${\mathbb C}_p$.

The key fact I need is the following which I'll prove at after explaining how it completes the endproof.

Claim: Any transcendental element of ${\mathbb C}_p$ has uncountably many conjugates (under $\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p))$.

Accepting this claim for the moment, let $\alpha$ be a transcendental zero of $f(x)$, our convergent power series. Then all conjugates of $\alpha$ are also zeroes of $f(x)$ in the open unit disc. Thus, by the above claim, $f(x)$ has uncountably many zeroes in the open unit disc of ${\mathbb C}_p$. However, any uncountable set in a separable metric space (e.g. ${\mathbb C}_p$) has an accumulation point. But then the zeroes of $f(x)$ have an accumulation point, necessarily in the open unit disc since that is closed in ${\mathbb C}_p$, which forces $f(x)$ to be identically zero.

Returning now to the claim, let $G=\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p)$ and let $H$ denote the subgroup of $G$ which stabilizes $\alpha$. Since the conjugates of $\alpha$ are in one-to-one correspondence with $G/H$, we will show that $H$ has uncountable index.

First note that $G$ acts continuously on ${\mathbb C}_p$ where we give ${\mathbb C}_p$ the $p$-adic topology (not the discrete topology). Thus $H$ is a closed subgroup and hence of the form $\text{Gal}(\overline{\mathbb Q}_p/M)$ for some algebraic extension$M/{\mathbb Q}_p$.

Assume $G/H$ is finite, and thus that $M$ is a finite extension of ${\mathbb Q}_p$. But then by Tate's theorem, $\alpha$ must be in $M$ as it is fixed by $\text{Gal}(\overline{\mathbb Q}_p/M)$. This is impossible as $\alpha$ is transcendental.

Thus, $G/H$ is infinite, and we must now show that it is uncountable. We use the following (standard) fact:

Fact: Any infinite compact Hausdorf Hausdorff space with no isolated points is uncountable.

Since $G$ is compact, $G/H$ is compact. The coset space $G/H$ is Hausdorf since $H$ is closed. To see that $G/H$ has no isolated points, note that if it has one isolated point, then all of its points are isolated as $G$ acts transitively (by left multiplication) on $G/H$ by homeomorphisms. But then $G/H$ is discrete which is impossible as it is infinite and compact.

Thus, $G/H$ is uncountable, and $\alpha$ has uncountably many conjugates.

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