# Profit of defining evaluation of a rational at a p-adic?

To recall some definitions, given a prime $$p$$ and a rational $$a/b$$, we can define the evaluation of the prime $$p$$ at the rational $$a/b$$.

If $$b \equiv 0 \bmod p$$, then we say that the rational $$a/b$$ has a pole at $$p$$. Otherwise, $$b$$ must be invertible mod p. Let the inverse of $$b$$ be $$b'$$ modulo $$p$$: $$bb' \equiv 1 \bmod p$$.

We define the evaluation of $$a/b$$ at $$p$$ as $$ab' \bmod p$$.

How is this definition profitable? I would like some theorem, proof technique, or insight we gain from the perspective of evaluating a prime at a rational.

I feel that there should be some topology (perhaps related to the Zariski topology of $$\mathbb Z$$) that governs what happens, but I don't know enough to justify what the conjecture is supposed to be.

• The natural is setting is the ring $R=(\Bbb{Z}-(p))^{-1}\Bbb{Z}$ of rational numbers with denominator coprime with $p$ then $pR$ is (the unique) maximal ideal of $R$ and there is a ring homomorphism $R\to R/pR\cong \Bbb{Z/pZ}$. The $p$-adic integers are $\displaystyle\varprojlim_{n\to \infty} R/p^n R$ Dec 21, 2019 at 4:49
• Reducing modulo $p$ a rational number (with denominator not divisible by $p$) is useful in general. Knowing how to construct the $p$-adic numbers too. Dec 21, 2019 at 4:53
• $(1+3)^{1/2} = \sum_{k \ge 0} {1/2 \choose k} 3^k \equiv 1+{1/2 \choose 1} 3\bmod 9$, this is a square root of $4\bmod 9$ Dec 21, 2019 at 4:59
• $f(x)=(x+1)^2-x^2\in R[x]$ then $\frac12 f(x)\equiv x+1/2\bmod p$ Dec 21, 2019 at 5:04
• Oh, @reuns, you missed a teachable moment, in not pointing out that the $3$-adic binomial expansion for $\sqrt{1+3}$ gives you not $2$ but $-2$ as its value. Dec 21, 2019 at 5:46

You’ve asked me to go into depth on the validity in the $$p$$-adic context, of the binomial expansion $$(1+t)^z=\sum_{j=0}^\infty\binom zjz^i$$, when $$z$$ is a $$p$$-adic integer, i.e. $$z\in\Bbb Z_p$$.

Let me first set some notations, all of them fairly standard in this subject. Let us agree first that $$v_p:\Bbb Q\to\Bbb Z\cup\{\infty\}$$, the $$p$$-adic additive valuation measures divisibility of a number by $$p$$, so that $$v_p(p^m)=m$$, no matter the $$m\in\Bbb Z$$, and $$v_p(q)=v_p(1)=0$$ for other primes $$q$$ than $$p$$. Extend by multiplicativity, $$v(zw)=v(z)+v(w)$$, and set $$v(0)=\infty$$, so that now $$v$$ is defined throughout $$\Bbb Q$$. (Multiplicativity forces $$v(-1)=0$$ as well.) The definition forces $$v(z+w)\ge\min\bigl(v(z),v(w)\bigr)$$. This is the additively notated form of the triangle inequality, as you may check. As a result of this inequality, you see that by choosing $$\varepsilon$$ to be any real number with $$0<\varepsilon<1$$, when we set $$\mid z\mid=\varepsilon^{v(z)}$$, $$\mid\bullet\mid$$ becomes a good absolute value, and $$d(z,w)=\lvert z-w\rvert$$ gives a good metric $$d$$ on $$\Bbb Q$$. Likely you know all this. Since I prefer to work with the additive valuation, I will not use the absolute-value notation. But you must remember that $$p$$-adically, the number $$p$$ is considered to be small, its positive powers going to zero in limit, so that to say that $$\lim_nz_n=w$$ is exactly to say that $$\lim_nv(w-z_n)=\infty$$.

Since the $$p$$-adic metric gives us a good metric-space structure on $$\Bbb Q$$, we may complete by proclaiming that Cauchy sequences in $$\Bbb Q$$ have limits in the completion $$\Bbb Q_p$$, the field of $$p$$-adic numbers. The valuation extends in a natural way, to make $$v$$ continuous: if $$\{z_j\}$$ is a Cauchy sequence, then either $$\{v(z_j)\}$$ is eventually constant, to a value that you proclaim to be the valuation of the corresponding $$\Bbb Q_p$$-number; or else $$v(z_j)\uparrow\infty$$, when you say that the limit is the $$\Bbb Q_p$$-number $$0$$. Again, you may know all this already.

The rationals $$z$$ for which $$v_p(z)\ge0$$ form a ring (you check this), which has no fully standard notation, but it’s sometimes called $$\Bbb Z_{(p)}$$. These are the rational numbers with no $$p$$ in the denominator, and they are the ones on which your “evaluation” is defined. (I had never heard this terminology!) In the same way, the set of all $$z\in\Bbb Q_p$$ with $$v(z)\ge0$$ is a ring, standardly denoted $$\Bbb Z_p$$, the ring of $$p$$-adic integers. It’s a local ring, i.e. has just one maximal ideal, namely $$p\Bbb Z_p$$, the $$p$$-adic numbers $$z$$ with $$v(z)>0$$, in this case $$v(z)\ge1$$.

Now I must make a digression, which you should feel free to ignore. If $$K$$ is a finite field extension of $$\Bbb Q_p$$, then thanks to the renowned Hensel’s Lemma, there is a unique extension of $$v_p$$ to $$K$$, under which $$K$$ also is complete. Again, the set of elements with $$v(z)\ge0$$ is a ring, called the (local) integers of $$K$$, often denoted $$\mathfrak o_K$$. Now $$v$$ can take values with a common denominator $$e$$, called the ramification index of $$K$$ over $$\Bbb Q_p$$. It’s always a divisor of $$[K:\Bbb Q_p]$$. Here, $$\mathfrak o_K$$ is still a local ring, and its maximal ideal $$\mathfrak m_K$$ is defined by the inequality $$v(z)>0$$. (End of digression)

Next, granting completeness of $$\Bbb Q_p$$, you see that a power series $$\sum_ja_jt^j$$ will be convergent to an element of $$\Bbb Q_p$$ if the $$a_j$$ are in $$\Bbb Q_p$$ with $$v(a_j)$$ bounded below (in particular if all $$a_j\in\Bbb Z_p$$) and if $$t$$ is evaluated to an element $$z$$ of $$\Bbb Q_p$$ with $$v(z)>0$$. (There are other convergent series than these, such as the logarithmic series $$-\sum_1^\infty(-t)^j/j$$.)

I need some Lemmas, easily proved:
Lemma. Let $$z,w\in\Bbb Z_p$$ with $$v(z-w)=m\ge1$$. Then $$v(z^p-w^p)\ge m+1$$.
Lemma. For each positive integer $$m$$, the polynomial $$\binom tm=\frac{t(t-1)\cdots(t-m+1)}{m!}$$ is continuous as a function on $$\Bbb Q_p$$ (more generally, as a function on the complete field extension $$K$$ of $$\Bbb Q_p$$).
Lemma. Let $$z\in\Bbb Z_p$$. If $$m$$ is a positive integer, then $$\binom zm\in\Bbb Z_p$$.

Maybe the third Lemma needs an argument to justify it. You know, I hope, that every $$p$$-adic integer $$z$$ can be written as limit of positive integers $$x_j$$. Just take your favorite infinite representation of $$z$$ and cut it off after $$j$$ terms to get your $$x_j$$. Now since $$z=\lim_jx_j$$, continuity of the binomial polynomial $$\binom tm$$ says that $$\bigl\lbrace\binom{x_j}k\bigr\rbrace$$ is $$p$$-adically convergent, necessarily to $$\binom zm$$. But the numbers $$\binom{x_j}m$$ are ordinary high-school-style binomial coefficients, that is, they’re integers. So their limit must be a $$p$$-adic integer. (It’s a fact, which I will not prove, that if $$z\in K$$, where $$K$$ is any complete extension of $$\Bbb Q_p$$, in particular if $$K$$ is a finite extension of $$\Bbb Q_p$$, and if for every $$m\ge0$$, $$v\binom zm\ge0$$, then $$z\in\Bbb Z_p$$.)

More “easy” Lemmas:
Lemma. If $$z\in\Bbb Z_p$$, then the series $$S_z(t)=\sum_{m=1}^\infty\binom zmt^m$$ has all coefficients in $$\Bbb Z_p$$, and consequently is convergent whenever $$t$$ is evaluated to an element $$w$$ with $$v(w)>0$$. That is, if $$w$$ is in the maximal ideal $$\mathfrak m$$ of the integers of the complete field $$K$$, then $$S_z(w)$$ is a well-defined element of $$\mathfrak o$$, the integers of $$K$$.
Lemma. In particular, if $$n$$ is a positive integer indivisible by $$p$$, then $$\sqrt[n]{1+t}=(1+t)^{1/n}=1+\sum_{j=1}^\infty\binom{1/n}jt^j$$ is convergent whenever $$t$$ is evaluated to an element $$w$$ with $$v(w)>0$$.

I’ve gone on at such length already that I don’t think I should give one of the arguments that the series for $$\sqrt[n]{1+t}$$ really does give an $$n$$-th root of $$1+t$$. But you did ask why the $$3$$-adic series for $$\sqrt{1+3}$$ yields $$-2$$ rather than $$2$$. The reason is easy to give: the series delivers a result that’s $$\equiv1\pmod3$$, that is, the value $$s$$ resulting has $$v_3(s-1)>0$$. The number $$s=-2$$ has this property, while $$2$$ does not.

• Thanks a lot for the detailed answer :) Dec 22, 2019 at 10:49