The extension $\mathbb{Q}_3 / \mathbb{Q}$ is not algebraic

I'll start from the beginning: the definition of $\mathbb{Z}_p$ (the ring of $p$-adic integers) we were taught is the set of infinite series $(a_1, a_2, \ldots)$ such that $a_{i+1}\equiv a_i \pmod{p^i}$ (I'm omitting some formal details, this is just a clarification). Then we defined $\mathbb{Q}_p$ as the field of fractions of $\mathbb{Z}_p$ (equivalence classes of pairs of $p$-adic integers...).

I'm saying all this because this is truly all I know about $p$-adic numbers. Oh, and also that there's a natural embedding of $\mathbb{Q}$ into $\mathbb{Q}_p$ (derived from viewing the $p$-adic integer $(k, k, k,\ldots)$ as the integer $k$).

So now I'm requested to show that the extension $\mathbb{Q}_3 / \mathbb{Q}$ is not algebraic, that is, there exists a $3$-adic rational that is not a root of any polynomial in $\mathbb{Q}[x].$

I can think of some $3$-adic rationals that are not in the embedded $\mathbb{Q}$, but showing one of them is not algebraic is quite harder.

• $a_{i+1}\equiv a_i \; (\text{mod} \: p_i)$, coded as a_{i+1}\equiv a_i \; (\text{mod} \: p_i) is rather complicated and non-standard. I changed it to $a_{i+1}\equiv a_i \pmod{p_i},$ coded as a_{i+1}\equiv a_i \pmod{p_i}. (And the last set of braces is not needed if only a single object follows \pmod, eg. $A\equiv B\pmod\infty$ is coded as A\equiv B\pmod\infty.) $\qquad$ – Michael Hardy Apr 6 '17 at 17:17
• Oh, OK, good to know – 35T41 Apr 6 '17 at 17:19
• Hint: there are quite a lot $p$-adic numbers, and only few of them are algebraic. Count! – franz lemmermeyer Apr 6 '17 at 17:20
• Oh, of course, cardinalities! Ok, that definitely solves the question, yet it doesn't really teach my anything. It would be very valuable to see a proof that a certain $p$-adic is non-algebraic, or even some sort of a characterisation for that... – 35T41 Apr 6 '17 at 17:24
• @JulianRosen: Do you mind recapping that argument as an answer? – Jyrki Lahtonen Apr 7 '17 at 6:46

If you just want to show that there exist transcendental elements of $\mathbb{Q}_p$, Franz Lemmermeyer's argument from the comments is a good way to go: $\mathbb{Q}_p$ is uncountable, but only countably many elements can be algebraic over $\mathbb{Q}$.

It is possible to write down an transcendental element $\mathbb{Q}_p$ using a $p$-adic version of Liouville's constant.

Claim: The $p$-adic number $$\alpha:=\sum_{n=0}^\infty p^{n!}$$ is transcendental over $\mathbb{Q}$.

First we need a lemma.

Lemma: Let $f(T)\in\mathbb{Z}[T]$ be a non-constant square-free polynomial of degree $d$, and suppose $x_0\in\mathbb{Z}_p\backslash\mathbb{Z}$ is a root of $f$. Then there is a constant $C$ such that for all $x\in\mathbb{Z}$ sufficiently close to $x_0$, $$v_p(x-x_0)\leq d\cdot\log_p(|x|) + C.$$

Proof: We can write $f(x)$ as a polynomial in $(x-x_0)$, say $$\tag{1}f(T)=\sum_{n=1}^d a_d(T-x_0)^n.$$ The constant term vanishes because $x_0$ is a root of $f$, and $a_1\neq 0$ because $f(T)$ is square-free. For $x$ sufficiently close to $x_0$, the biggest term in $(1)$ is the first, so $$v_p(f(x)) = v_p(a_1) + v_p(x-x_0).$$ Now $f(x)$ is a non-zero integer, so $v_p(f(x))\leq \log_p(|f(x)|)$. We also have $|f(x)|\leq A |x|^d$, where $A$ is the sum of the absolute values of the coefficieints of $f$. Thus we have shown $$v_p(x-x_0)=v_p(f(x))-v_p(a_1)\leq \log_p(A|x|^d)-v_p(a_1)=d\log_p(|x|)+\log_p(A)-v_p(a_1).$$ This completes the proof.

Proof of the claim: For the sake of contradiction, suppose $f(T)\in\mathbb{Z}[T]$ is irreducible of degree $d$ and has $\alpha$ as a root. For $N=0,1,2,\ldots,$ define $$x_N:=\sum_{n=0}^N p^{n!},$$ so that $x_N\to\alpha$ as $N\to\infty$. Let $C$ be be the constant guaranteed to exist by the lemma ($f(T)$ is square-free because it is irreducible). Then $$(N+1)!=v_p(\alpha-x_N)\leq d \log_p(x_N)+C\leq d (N!+1) + C.$$ This is a contradiction once $N$ is larger than $d+|C|$.

• Oh, good. I wondered, without being willing to do the necessary work, whether Liouville’s Theorem would apply $p$-adically. – Lubin Apr 10 '17 at 19:39