Let $\mathbb{Q}_p$ be the $p$-adic rational field.

I want to verify that the field extension $\mathbb{Q}_p/ \mathbb{Q}$ is infinite therefore $\dim_{\mathbb{Q}}(\mathbb{Q}_p) = \infty$.

My considerations: Assume the extension is finite. Since $char(\mathbb{Q})=0$ the extension is separable and we can apply the theorem of primitive element and obtain $\mathbb{Q}_p= \mathbb{Q}[T]/f$ for an irreducible $f \in \mathbb{Q}[T]$.

How to obtain the contradiction?

Or is there another way to prove the claim?

Update: Preferably I'm keen curiuos how the agument mentioned by @hunter below using Hensel's lemma (which version) works to solve this problem.

  • $\begingroup$ two possible approaches: 1) show that $\mathbb{Q}_p$ is uncountable, or 2) similar to yours -- use Hensel's lemma to produce roots of infinitely many polynomials whose extensions are linearly disjoint over the rationals (e.g. quadratic extensions), $\endgroup$ – hunter Apr 16 at 0:19
  • $\begingroup$ @hunter:could you explain the secound argument using Hensel lemma an bit more? Hensel allows to lift pairs (or more general finite products) of pairwise coprime polynomials from $\mathbb{F}_p[T]$ to $\mathbb{Q}_p[T]$. Do you mean that in the sense that for arbitrary $n \in \mathbb{N}$ it's easier to find $n$ coprim polynomials and then lift them? Or did I misunderstood your Hensel argument? $\endgroup$ – KarlPeter Apr 16 at 0:36
  • $\begingroup$ If $p \ne 2$ a special case of Hensel lemma is for $q\equiv 1\bmod p$, $q^{1/2}= \sum_{k=0}^\infty {1/2 \choose k} (q-1)^k \in \Bbb{Z}_p$. Next $\Bbb{Q}(q_1^{1/2},\ldots,q_n^{1/2})/\Bbb{Q}$ is an extension of degree $2^n$ for coprime non-square integers $q_j$. $\endgroup$ – reuns Apr 16 at 1:36
  • $\begingroup$ @reuns:So essentially the contribute of Hensel here is to lift a root $\bar{r}$ of polynomial $X^2 -\bar{q} \in \mathbb{F}_p[X]$ to a root $r$ of $X^2 -q \in \mathbb{Z}_p[X]$ given by your explicit formula,right? $\endgroup$ – KarlPeter Apr 16 at 2:08
  • $\begingroup$ Have you looked at the proof of Hensel lemma ? For a polynomial with derivative $\in \Bbb{Z}_p^\times$, from the root $\bmod p^m$ it constructs a sequence of roots $\bmod p^k$ converging to the lift. In the special case of $x^n-(1+pa)$ the root $\bmod p$ is $1$ and the lift is $\sum_{k=0}^\infty {1/n \choose k} (pa)^k$. $\endgroup$ – reuns Apr 16 at 2:24

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