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.

  • 1
    $\begingroup$ 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 $\displaystyle\varprojlim_{n\to \infty} R/p^n R$ $\endgroup$
    – reuns
    Dec 21, 2019 at 4:49
  • 1
    $\begingroup$ 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. $\endgroup$
    – reuns
    Dec 21, 2019 at 4:53
  • 1
    $\begingroup$ $(1+3)^{1/2} = \sum_{k \ge 0} {1/2 \choose k} 3^k \equiv 1+{1/2 \choose 1} 3\bmod 9$, this is a square root of $4\bmod 9$ $\endgroup$
    – reuns
    Dec 21, 2019 at 4:59
  • 1
    $\begingroup$ $f(x)=(x+1)^2-x^2\in R[x]$ then $\frac12 f(x)\equiv x+1/2\bmod p$ $\endgroup$
    – reuns
    Dec 21, 2019 at 5:04
  • 1
    $\begingroup$ 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. $\endgroup$
    – Lubin
    Dec 21, 2019 at 5:46

1 Answer 1


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.

  • 1
    $\begingroup$ Thanks a lot for the detailed answer :) $\endgroup$ Dec 22, 2019 at 10:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .