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.

Could someone please enlighten me?

  • $\begingroup$ $a_{i+1}\equiv a_i \; (\text{mod} \: p_i)$, 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.) $\qquad$ $\endgroup$ – Michael Hardy Apr 6 '17 at 17:17
  • $\begingroup$ Oh, OK, good to know $\endgroup$ – 35T41 Apr 6 '17 at 17:19
  • 6
    $\begingroup$ Hint: there are quite a lot $p$-adic numbers, and only few of them are algebraic. Count! $\endgroup$ – franz lemmermeyer Apr 6 '17 at 17:20
  • 1
    $\begingroup$ 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... $\endgroup$ – 35T41 Apr 6 '17 at 17:24
  • 1
    $\begingroup$ @JulianRosen: Do you mind recapping that argument as an answer? $\endgroup$ – 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 $d+|C|$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.