18. p-adic numbers¶

Add up \(1 + 2 + 4 + 8 + 16 + \cdots\) and the sum runs off to infinity. But stop after \(n\) terms and you have \(2^n - 1\), which is \(-1\) plus a number divisible by a huge power of 2. If "divisible by a huge power of 2" counted as "tiny", the sum would close in on \(-1\). This chapter takes that idea seriously. For each prime \(p\) it builds a new way to measure the size of a number, finds that the rational numbers have holes in it, fills the holes, and gets a new number system, the \(p\)-adic numbers \(\mathbb{Q}_p\). There you can solve equations one digit at a time (Hensel's lemma), you meet the only possible ways of measuring size on \(\mathbb{Q}\) (Ostrowski's theorem), and you learn to ask the question that drives much of modern number theory: if an equation can be solved modulo every power of every prime, and in the real numbers, can it be solved in the rationals?

In this chapter

You will learn:

the \(p\)-adic valuation and absolute value, the strong triangle inequality, and why every \(p\)-adic triangle is isosceles
the product formula that ties all the absolute values of a rational number together
how \(\mathbb{Q}_p\) is built from \(\mathbb{Q}\) as the reals are, and how to write \(p\)-adic numbers as digit strings that run to the left
Hensel's lemma, with a full proof, and how to use it to find square roots and roots of unity in \(\mathbb{Q}_p\)
Ostrowski's theorem: the real absolute value and the \(p\)-adic ones are the only ones on \(\mathbb{Q}\)
the local-global principle, the Hasse-Minkowski theorem for quadratic forms, and equations where it fails

You need: congruences and inverses mod \(n\) (Chapter 4), quadratic residues (Chapter 8) and some comfort with limits of sequences. Rings of integers (Chapter 17) appear only in remarks. For limits and Cauchy sequences, the U of T guide MAT159 Unit 3: Sequences and Cauchy sequences is a good refresher.

Levels: 18.1 is intermediate (only divisibility); 18.2 and 18.3 are advanced, with full proofs; 18.4 and 18.5 are expert, with honest statements and the proofs that fit.

18.1 A new size for numbers¶

Level: intermediate. When are two whole numbers "close"? On a ruler, 1000 and 1001 are close and 1 and 1001 are far apart. But for a number theorist, 1 and 1001 have something in common: \(1001 - 1 = 1000 = 2^3 \cdot 5^3\), so they agree modulo 8 and modulo 125. And \(1\) and \(1 + 5^{20}\) agree modulo \(5^{20}\): as far as any question about remainders mod 5, 25, 125, ..., \(5^{20}\) is concerned, they are the same number. The idea of this chapter is to turn "agree modulo a high power of \(p\)" into "are close together", and see what follows.

The p-adic valuation¶

Fix a prime \(p\). Every nonzero rational number \(x\) can be written as

\[ x = p^k \cdot \frac{a}{b}, \]

where \(k\) is an integer (possibly negative) and \(a\), \(b\) are integers not divisible by \(p\). This is just unique factorization (Chapter 2): pull every factor \(p\) out of the top and the bottom. The exponent \(k\) is the \(p\)-adic valuation of \(x\), written \(v_p(x)\). It counts how many times \(p\) divides \(x\), with negative counts for \(p\) in the denominator. By convention \(v_p(0) = +\infty\) (zero is divisible by every power of \(p\)).

Examples with \(p = 2\): \(v_2(12) = 2\) because \(12 = 2^2 \cdot 3\); \(v_2(3/8) = -3\); \(v_2(5) = 0\). With \(p = 3\): \(v_3(12) = 1\), \(v_3(1/9) = -2\), \(v_3(-81/4) = 4\).

Two rules make \(v_p\) useful. For all rational \(x\) and \(y\):

\(v_p(xy) = v_p(x) + v_p(y)\);
\(v_p(x + y) \ge \min(v_p(x), v_p(y))\).

Proof. Write \(x = p^k a/b\) and \(y = p^m c/d\) with \(a, b, c, d\) prime to \(p\). Then \(xy = p^{k+m} (ac)/(bd)\), and \(ac\) and \(bd\) are prime to \(p\) because \(p\) is prime (Euclid's lemma), so rule 1 holds. For rule 2, say \(k \le m\). Then

\[ x + y = p^k \left( \frac{a}{b} + p^{m-k} \frac{c}{d} \right) = p^k \cdot \frac{ad + p^{m-k} bc}{bd}. \]

The denominator \(bd\) is prime to \(p\), and the numerator is an integer, so \(v_p(x+y) \ge k\). If one of \(x\), \(y\) is 0 the rules are clear. \(\square\)

Notice that rule 2 can be an inequality: \(v_3(1) = v_3(2) = 0\) but \(v_3(1 + 2) = 1\). But if \(k < m\) strictly, the numerator \(ad + p^{m-k}bc\) is \(ad\) plus a multiple of \(p\), so it is prime to \(p\), and then \(v_p(x + y) = k\) exactly.

The p-adic absolute value¶

Now turn the valuation into a size. The \(p\)-adic absolute value of \(x\) is

\[ |x|_p = p^{-v_p(x)}, \qquad |0|_p = 0. \]

The more often \(p\) divides \(x\), the smaller \(|x|_p\) is. So \(|12|_2 = 2^{-2} = 1/4\), \(|1/9|_3 = 3^2 = 9\), \(|5|_2 = 1\), and \(|2^{100}|_2 = 2^{-100}\), a very small number. In this world the powers \(p, p^2, p^3, \ldots\) shrink to 0, and the distance between \(x\) and \(y\) is \(|x - y|_p\): so \(1\) and \(1 + 5^{20}\) are at 5-adic distance \(5^{-20}\).

The two rules for \(v_p\) become, for \(|\cdot|_p\):

\(|xy|_p = |x|_p \, |y|_p\) (multiplicative);
\(|x + y|_p \le \max(|x|_p, |y|_p)\), the strong triangle inequality, also called the ultrametric inequality.

The second is stronger than the usual triangle inequality \(|x + y| \le |x| + |y|\), because the larger of two non-negative numbers is at most their sum. An absolute value obeying it is called non-archimedean. The name comes from the archimedean property of the reals: however small \(\varepsilon > 0\) is, adding it to itself enough times exceeds any bound. In \(\mathbb{Q}\) with \(|\cdot|_p\), adding 1 to itself never gives anything bigger than 1: \(|n|_p \le 1\) for every integer \(n\).

Worked example. Take \(x = -12/35\). Then \(-12/35 = -2^2 \cdot 3 / (5 \cdot 7)\), so

\[ |x|_2 = \tfrac14,\quad |x|_3 = \tfrac13,\quad |x|_5 = 5,\quad |x|_7 = 7,\quad |x|_p = 1 \text{ for every other } p, \]

and the ordinary absolute value is \(|x|_\infty = 12/35\). Multiply them all: \(\tfrac{12}{35} \cdot \tfrac14 \cdot \tfrac13 \cdot 5 \cdot 7 = 1\).

That is no accident. Writing \(|\cdot|_\infty\) for the ordinary absolute value, we have the

Product formula. For every nonzero rational \(x\),

\[ |x|_\infty \prod_{p} |x|_p = 1, \]

where the product is over all primes, and only finitely many factors differ from 1.

Proof. Both sides are multiplicative in \(x\), and both are 1 for \(x = -1\) (since \(|-1|_p = 1\) for every \(p\)). So it is enough to check \(x = q\) a prime, since every nonzero rational is \(\pm\) a product of primes and their inverses. For a prime \(q\): \(|q|_\infty = q\), \(|q|_q = 1/q\), and \(|q|_p = 1\) for \(p \ne q\). The product is \(q \cdot (1/q) = 1\). \(\square\)

The product formula says that the real size and all the \(p\)-adic sizes of a rational number are not independent: if a rational number is small in some of them, it must be large in others. It is one of the reasons number theorists speak of the reals as "the prime at infinity", one place among the others.

Strange geometry: every triangle is isosceles¶

The strong triangle inequality makes \(p\)-adic geometry feel odd at first.

Fact 1. If \(|x|_p \ne |y|_p\), then \(|x + y|_p = \max(|x|_p, |y|_p)\).

Proof. Say \(|x|_p > |y|_p\). Then \(|x + y|_p \le |x|_p\). Also \(x = (x + y) - y\), so \(|x|_p \le \max(|x+y|_p, |y|_p)\). The maximum cannot be \(|y|_p\), which is smaller than \(|x|_p\), so it is \(|x + y|_p\), and \(|x|_p \le |x + y|_p\). Both inequalities give equality. \(\square\)

(This is the absolute value form of what we saw for \(v_p\): when \(k < m\), \(v_p(x + y) = k\).)

Fact 2: every triangle is isosceles. Take three points \(x\), \(y\), \(z\). The sides are \(a = x - y\), \(b = y - z\), and \(a + b = x - z\). If \(|a|_p \ne |b|_p\), Fact 1 says \(|a + b|_p\) equals the larger of them. So the two longest sides of the triangle are always equal.

Fact 3: every point of a disc is its centre. The closed disc (or ball) of radius \(r\) around \(a\) is \(B(a, r) = \{x : |x - a|_p \le r\}\). If \(b\) is in \(B(a, r)\) then \(B(b, r) = B(a, r)\).

Proof. If \(|x - a|_p \le r\), then \(|x - b|_p = |(x - a) + (a - b)|_p \le \max(|x-a|_p, |a-b|_p) \le r\), so \(B(a,r) \subseteq B(b,r)\). Swap \(a\) and \(b\) for the other inclusion. \(\square\)

A consequence: two \(p\)-adic discs are either disjoint or one is inside the other. They never partly overlap. That is exactly how the residue classes behave: the integers that are \(\equiv 4 \pmod 9\) all lie in the class \(\equiv 1 \pmod 3\), and the classes \(1 \bmod 3\) and \(2 \bmod 3\) do not meet. The disc \(B(4, 1/9)\) in the integers is the class \(4 + 9\mathbb{Z}\), and it sits inside \(B(1, 1/3) = 1 + 3\mathbb{Z}\).

Check. \(|10 - 1|_3 = |9|_3 = 1/9\), \(|10 - 4|_3 = |6|_3 = 1/3\), \(|4 - 1|_3 = |3|_3 = 1/3\). Two sides of length \(1/3\), one shorter: isosceles, as promised.

18.2 The p-adic numbers \(\mathbb{Q}_p\)¶

Level: advanced. The rational numbers have holes for the ordinary absolute value: the decimals \(1, 1.4, 1.41, 1.414, \ldots\) get closer and closer together but have no rational limit, because \(\sqrt 2\) is irrational. Filling those holes gives the real numbers \(\mathbb{R}\). Exactly the same thing happens for \(|\cdot|_p\), and filling the holes gives \(\mathbb{Q}_p\).

A sequence with no rational limit¶

Look for integers \(x\) with \(x^2 \equiv -1\) modulo higher and higher powers of 5. Modulo 5, \(x = 2\) works: \(2^2 + 1 = 5\). Modulo 25, try \(x = 2 + 5t\): we need \((2 + 5t)^2 + 1 = 5 + 20t + 25t^2 \equiv 0 \pmod{25}\), that is \(5 + 20t \equiv 0 \pmod{25}\), that is \(1 + 4t \equiv 0 \pmod 5\), so \(t = 1\) and \(x = 7\). Check: \(49 + 1 = 50\). Going on the same way gives

\[ x_1 = 2,\quad x_2 = 7,\quad x_3 = 57,\quad x_4 = 182,\quad x_5 = 2057, \ldots \]

with \(x_n^2 + 1\) divisible by \(5^n\) and \(x_{n+1} \equiv x_n \pmod{5^n}\). (Check: \(57^2 + 1 = 3250 = 2 \cdot 5^3 \cdot 13\) and \(182^2 + 1 = 33125 = 5^4 \cdot 53\).) So \(|x_{n+1} - x_n|_5 \le 5^{-n}\): the terms get 5-adically close together. In the language of analysis, the sequence is Cauchy: for every \(\varepsilon > 0\) there is an \(N\) with \(|x_m - x_n|_5 < \varepsilon\) for all \(m, n > N\). (For \(m > n\), the strong triangle inequality gives \(|x_m - x_n|_5 \le \max_{n \le k < m} |x_{k+1} - x_k|_5 \le 5^{-n}\).)

If the sequence had a rational limit \(L\), then \(|x_n^2 - L^2|_5 \to 0\), and since \(|x_n^2 + 1|_5 \le 5^{-n} \to 0\), we would get \(L^2 = -1\). No rational number squares to \(-1\). So \(\mathbb{Q}\) has a hole for \(|\cdot|_5\) exactly where "\(\sqrt{-1}\)" should be.

Completing \(\mathbb{Q}\)¶

The construction is the one that builds \(\mathbb{R}\) from Cauchy sequences, with \(|\cdot|_p\) in place of \(|\cdot|\):

A point of \(\mathbb{Q}_p\) is a Cauchy sequence of rationals for \(|\cdot|_p\), where two sequences count as the same point if their difference tends to 0.
Add and multiply sequences term by term. The rationals sit inside as constant sequences.
The absolute value extends: \(|(x_n)|_p = \lim |x_n|_p\). (This limit exists, and in fact \(|x_n|_p\) is eventually constant for a sequence not tending to 0, by Fact 1 of 18.1.)

One checks, as for the reals, that this gives a field \(\mathbb{Q}_p\) with an absolute value that still obeys the strong triangle inequality, in which \(\mathbb{Q}\) is dense (every point is a limit of rationals) and which is complete (every Cauchy sequence converges). It is the completion of \(\mathbb{Q}\) for \(|\cdot|_p\). The proofs are word for word those for \(\mathbb{R}\); Gouvêa's book (see Further reading) writes them out.

One new feature: the values of \(|\cdot|_p\) on \(\mathbb{Q}_p\) are still just the powers \(p^k\) and 0, since the limit of the eventually constant sequence \(|x_n|_p\) is a power of \(p\). So the valuation \(v_p\) extends to \(\mathbb{Q}_p\) too.

The p-adic integers¶

The \(p\)-adic integers are

\[ \mathbb{Z}_p = \{ x \in \mathbb{Q}_p : |x|_p \le 1 \}. \]

By the strong triangle inequality, sums and products of elements of size at most 1 still have size at most 1, so \(\mathbb{Z}_p\) is a ring. It contains \(\mathbb{Z}\), and also every rational \(a/b\) with \(p \nmid b\), such as \(1/3\) in \(\mathbb{Z}_2\). Its units (elements with an inverse in \(\mathbb{Z}_p\)) are the \(x\) with \(|x|_p = 1\): if \(|x|_p = 1\) then \(|1/x|_p = 1\) too. Every nonzero \(x \in \mathbb{Q}_p\) is uniquely \(x = p^k u\) with \(k = v_p(x)\) and \(u\) a unit.

Digits that run to the left¶

Every real number has a decimal expansion, with finitely many digits to the left of the point and possibly infinitely many to the right. \(p\)-adic numbers are the other way round.

Theorem (p-adic expansion). Every \(x \in \mathbb{Z}_p\) can be written in exactly one way as a convergent series

\[ x = a_0 + a_1 p + a_2 p^2 + a_3 p^3 + \cdots, \qquad a_i \in \{0, 1, \ldots, p-1\}. \]

Every \(x \in \mathbb{Q}_p\) is \(p^{-k}\) times such a series, so it has finitely many digits after the "point".

Proof. First a key fact: every \(x \in \mathbb{Z}_p\) is congruent modulo \(p^n\) to exactly one integer in \(\{0, 1, \ldots, p^n - 1\}\), meaning \(|x - m|_p \le p^{-n}\). Existence: \(x\) is a limit of rationals, so some rational \(a/b\) has \(|x - a/b|_p \le p^{-n}\); then \(|a/b|_p \le 1\), so \(p \nmid b\), and \(b\) has an inverse \(b'\) modulo \(p^n\); take \(m \equiv a b' \pmod {p^n}\). Then \(a/b - m = (a - bm)/b\) and \(a - bm \equiv a - b a b' \equiv 0 \pmod{p^n}\), so \(|a/b - m|_p \le p^{-n}\), and by the strong triangle inequality \(|x - m|_p \le p^{-n}\). Uniqueness: two such integers differ by a multiple of \(p^n\) less than \(p^n\) in size, so they are equal.

Now write \(m_n\) for the integer in \([0, p^n)\) congruent to \(x\) modulo \(p^n\). Then \(m_{n+1} \equiv m_n \pmod{p^n}\) (both are congruent to \(x\)), so \(m_{n+1} = m_n + a_n p^n\) for a unique digit \(a_n \in \{0, \ldots, p-1\}\). So \(m_n = a_0 + a_1 p + \cdots + a_{n-1}p^{n-1}\) and \(|x - m_n|_p \le p^{-n} \to 0\): the series converges to \(x\). Different digit strings give different \(m_n\) for some \(n\), so different limits. For \(x \in \mathbb{Q}_p\) with \(v_p(x) = -k < 0\), apply this to \(p^k x \in \mathbb{Z}_p\). \(\square\)

In \(\mathbb{Q}_p\) a series converges as soon as its terms tend to 0. (Proof: the partial sums \(s_n\) have \(|s_m - s_n|_p \le \max_{n < k \le m} |a_k|_p\) by the strong triangle inequality, so they form a Cauchy sequence, and \(\mathbb{Q}_p\) is complete.) Nothing like this is true in \(\mathbb{R}\), where \(\sum 1/n\) diverges although \(1/n \to 0\). This is why \(p\)-adic series are so pleasant.

We write \(p\)-adic numbers like ordinary numerals, digits read from right to left, with the dots on the left: \(x = \ldots a_3 a_2 a_1 a_0\) in base \(p\).

Example: \(-1\). In \(\mathbb{Q}_5\), \(4 + 4\cdot 5 + 4 \cdot 25 + \cdots = 4(1 + 5 + 25 + \cdots) = 4 \cdot \frac{1}{1 - 5} = -1\). (The geometric series \(\sum p^n\) converges in \(\mathbb{Q}_p\) because \(|p^n|_p \to 0\), and its sum is \(1/(1-p)\) by the usual telescoping: \((1-p)(1 + p + \cdots + p^{n-1}) = 1 - p^n \to 1\).) So \(-1 = \ldots 44444\) in base 5. Add 1 and every digit carries: \(\ldots 44444 + 1 = \ldots 00000\). This is the same as the odometer of a car rolling over.

And the opening question: in \(\mathbb{Q}_2\), \(1 + 2 + 4 + 8 + \cdots = 1/(1 - 2) = -1\). The partial sums \(2^n - 1\) really do converge to \(-1\), because \(|(2^n - 1) - (-1)|_2 = |2^n|_2 = 2^{-n}\).

Example: \(1/3\) in \(\mathbb{Q}_5\). We need the digits of \(m_n \equiv 3^{-1} \pmod{5^n}\). Modulo 5, \(3^{-1} = 2\) (since \(3 \cdot 2 = 6 \equiv 1\)). Modulo 25 it is 17 (\(3 \cdot 17 = 51\)), and \(17 = 2 + 3 \cdot 5\). Modulo 125 it is 42 (\(3 \cdot 42 = 126\)), and \(42 = 2 + 3\cdot 5 + 1 \cdot 25\). Continuing,

\[ \tfrac13 = \ldots 1313132 \text{ (base 5)}. \]

Check. Multiply \(\ldots 1313132\) by 3 by hand, from the right: \(3 \cdot 2 = 6 = 1 + 1 \cdot 5\), write 1 carry 1; \(3 \cdot 3 + 1 = 10 = 0 + 2 \cdot 5\), write 0 carry 2; \(3 \cdot 1 + 2 = 5\), write 0 carry 1; \(3 \cdot 3 + 1 = 10\), write 0 carry 2; and the pattern repeats forever. Result: \(\ldots 0001 = 1\).

Example: \(1/3\) and \(-1/3\) in \(\mathbb{Q}_2\). Similarly \(1/3 = \ldots 0101011\) and \(-1/3 = \ldots 0101010101\) in base 2. Check the second: \(\ldots 010101 = 1 \cdot 1 + 0 \cdot 2 + 1 \cdot 4 + \cdots = 1 + 4 + 16 + \cdots = 1/(1-4) = -1/3\).

As with decimals, a \(p\)-adic number is rational exactly when its digits are eventually periodic. One direction is the geometric series trick we just used: a periodic tail is a rational multiple of \(1/(1 - p^r)\). The other direction comes from the long division that produces the digits: for \(a/b\) with \(p \nmid b\), the digits are produced by a process whose state is a remainder that, after the first few steps, can only take finitely many values, so it must repeat.

Arithmetic with these digit strings is ordinary school arithmetic with carries, done right to left, never stopping. That is how a computer handles \(\mathbb{Z}_p\) in practice: it keeps the last \(n\) digits, which means it computes in \(\mathbb{Z}/p^n\mathbb{Z}\). The map \(\mathbb{Z}_p \to \mathbb{Z}/p^n\mathbb{Z}\) that keeps the last \(n\) digits is a ring homomorphism, and a \(p\)-adic integer is the same thing as a list of residues \(m_1, m_2, m_3, \ldots\) modulo \(p, p^2, p^3, \ldots\) that are compatible (\(m_{n+1} \equiv m_n \bmod p^n\)). Algebraists write this as \(\mathbb{Z}_p = \varprojlim \mathbb{Z}/p^n\mathbb{Z}\), the inverse limit.

The tree of p-adic integers¶

Since a \(p\)-adic integer is a path of choices, one digit at a time, \(\mathbb{Z}_p\) looks like an infinite tree. The drawing shows the first two levels for \(p = 3\): the root is all of \(\mathbb{Z}_3\), which splits into three discs (the classes mod 3), each of which splits into three smaller discs (the classes mod 9), and so on forever. A 3-adic integer is an infinite path down the tree. The orange path is \(-1 = \ldots 222\), which goes to \(2 \bmod 3\), then \(8 \bmod 9\), then \(26 \bmod 27\), always taking the last branch.

The picture explains Fact 3 of 18.1: a disc is a whole subtree, and two subtrees are either nested or disjoint. It also suggests two facts we state without proof: \(\mathbb{Z}_p\) is compact (every sequence has a convergent subsequence: some branch at the first level contains infinitely many terms, then some branch below it, and so on), and it is totally disconnected (any two different points can be separated into different discs).

Try it in Python. The last \(n\) digits of a rational \(a/b\) with \(p \nmid b\):

def padic_digits(a, b, p, n):
    m = a * pow(b, -1, p**n) % p**n       # a/b modulo p^n
    return [(m // p**i) % p for i in range(n)]   # a_0, a_1, ..., a_(n-1)

print(padic_digits(1, 3, 5, 8))    # [2, 3, 1, 3, 1, 3, 1, 3]  so 1/3 = ...31313132 in base 5
print(padic_digits(-1, 3, 2, 8))   # [1, 0, 1, 0, 1, 0, 1, 0]  so -1/3 = ...01010101 in base 2

18.3 Hensel's lemma¶

Level: advanced. In 18.2 we built \(\sqrt{-1}\) in \(\mathbb{Q}_5\) by hand, one digit at a time. Every step was the same: we had a root modulo \(5^n\), wrote the next candidate as \(x + t \cdot 5^n\), and found that the condition on \(t\) was a linear congruence modulo 5. Hensel's lemma says this always works when the derivative does not vanish modulo \(p\).

For a polynomial \(f(x) = c_0 + c_1 x + \cdots + c_d x^d\), its derivative is the polynomial \(f'(x) = c_1 + 2c_2 x + \cdots + d c_d x^{d-1}\), defined by this formula with no limits needed.

Hensel's lemma. Let \(f\) be a polynomial with coefficients in \(\mathbb{Z}_p\), and let \(a \in \mathbb{Z}_p\) with

\[ f(a) \equiv 0 \pmod p \quad \text{and} \quad f'(a) \not\equiv 0 \pmod p. \]

Then there is exactly one \(\alpha \in \mathbb{Z}_p\) with \(f(\alpha) = 0\) and \(\alpha \equiv a \pmod p\).

(Here \(x \equiv y \pmod{p^n}\) for \(p\)-adic integers means \(|x - y|_p \le p^{-n}\), that is, \(x - y \in p^n\mathbb{Z}_p\).)

Proof. We build \(a_1 = a, a_2, a_3, \ldots\) with \(f(a_n) \equiv 0 \pmod{p^n}\) and \(a_{n+1} \equiv a_n \pmod{p^n}\), and show each \(a_{n+1}\) is determined modulo \(p^{n+1}\).

The tool is Taylor's formula for polynomials, which is pure algebra: expanding \(f(x + h)\) in powers of \(h\) gives

\[ f(x + h) = f(x) + f'(x)\, h + h^2 \, g(x, h) \]

for some polynomial \(g\) with coefficients in \(\mathbb{Z}_p\). (Each term \(c_k (x + h)^k\) expands by the binomial theorem to \(c_k x^k + k c_k x^{k-1} h + h^2 \cdot (\text{integer combination of } x \text{ and } h)\).)

Suppose \(f(a_n) \equiv 0 \pmod{p^n}\), so \(f(a_n) = p^n c\) with \(c \in \mathbb{Z}_p\). Try \(a_{n+1} = a_n + t p^n\) with \(t \in \mathbb{Z}_p\). Then

\[ f(a_n + tp^n) = p^n c + f'(a_n)\, t p^n + t^2 p^{2n} g = p^n \big( c + f'(a_n) t \big) + p^{2n}(\ldots). \]

Since \(2n \ge n + 1\), this is \(\equiv 0 \pmod{p^{n+1}}\) exactly when \(c + f'(a_n) t \equiv 0 \pmod p\). Now \(a_n \equiv a \pmod p\), so \(f'(a_n) \equiv f'(a) \not\equiv 0 \pmod p\) (a polynomial with \(p\)-adic integer coefficients respects congruences). So \(f'(a_n)\) is invertible modulo \(p\), and the condition has exactly one solution \(t \equiv -c \, f'(a_n)^{-1} \pmod p\). This gives \(a_{n+1}\), unique modulo \(p^{n+1}\) among numbers \(\equiv a_n \pmod{p^n}\).

The sequence \((a_n)\) is Cauchy (\(|a_{n+1} - a_n|_p \le p^{-n}\)), so it converges to some \(\alpha \in \mathbb{Z}_p\) (complete, and the disc \(\mathbb{Z}_p\) is closed). Since \(\alpha \equiv a_n \pmod{p^n}\), \(f(\alpha) \equiv f(a_n) \equiv 0 \pmod{p^n}\) for every \(n\), so \(|f(\alpha)|_p \le p^{-n}\) for all \(n\), and \(f(\alpha) = 0\).

Uniqueness: if \(\beta \equiv a \pmod p\) is another root, then by induction \(\beta \equiv a_n \pmod{p^n}\) for every \(n\) (at each stage, \(\beta\) reduced mod \(p^{n+1}\) is a root mod \(p^{n+1}\) lying over \(a_n\), and that lift is unique). So \(\beta = \alpha\). \(\square\)

The proof is an algorithm, and it is Newton's method in disguise: \(t p^n \equiv -f(a_n)/f'(a_n)\), so \(a_{n+1} \equiv a_n - f(a_n)/f'(a_n)\). If you use the full Newton step \(a \mapsto a - f(a)/f'(a)\), computed modulo a growing power of \(p\), the number of correct digits doubles at each step, just as Newton's method doubles the number of correct decimals in \(\mathbb{R}\).

Worked example: \(\sqrt{-1}\) in \(\mathbb{Q}_5\)¶

\(f(x) = x^2 + 1\), \(f'(x) = 2x\). Start at \(a = 2\): \(f(2) = 5 \equiv 0\) and \(f'(2) = 4 \not\equiv 0 \pmod 5\). Hensel's lemma gives a unique root \(\alpha \equiv 2 \pmod 5\) in \(\mathbb{Z}_5\). Digit by digit we found \(2, 7, 57, 182, 2057\) in 18.2, so

\[ \alpha = 2 + 1 \cdot 5 + 2 \cdot 5^2 + 1 \cdot 5^3 + 3 \cdot 5^4 + \cdots = \ldots 31212_5. \]

Newton's method from \(a = 2\), working modulo \(25\), then \(625\), then \(390625\): \(2 \mapsto 7 \mapsto 182 \mapsto 280182\), the precision doubling each time. (Check: \(182^2 + 1 = 33125 = 53 \cdot 625\).) Starting from \(a = 3\) instead gives the other root, which is \(-\alpha\), since \(-\alpha \equiv -2 \equiv 3 \pmod 5\). So \(\mathbb{Q}_5\) contains \(i\), and in fact two square roots of \(-1\), as it must.

This is a striking difference between \(\mathbb{Q}_5\) and \(\mathbb{R}\): the field \(\mathbb{Q}_5\) contains a square root of \(-1\), so \(\mathbb{Q}_5\) cannot be ordered the way \(\mathbb{R}\) is. In \(\mathbb{Q}_3\), by contrast, \(x^2 + 1\) has no root even modulo 3, so there is no \(\sqrt{-1}\) in \(\mathbb{Q}_3\). Whether \(\sqrt{-1} \in \mathbb{Q}_p\) depends on whether \(-1\) is a quadratic residue mod \(p\), which by Chapter 8 is exactly when \(p \equiv 1 \pmod 4\).

Squares in \(\mathbb{Q}_p\)¶

Theorem. Let \(p\) be odd. A nonzero \(x = p^k u \in \mathbb{Q}_p\) (\(u\) a unit) is a square in \(\mathbb{Q}_p\) if and only if \(k\) is even and \(u \bmod p\) is a nonzero square modulo \(p\).

Proof. If \(x = y^2\) then \(v_p(x) = 2 v_p(y)\) is even, and the unit part of \(x\) is the square of the unit part of \(y\), so it is a square mod \(p\). Conversely, if \(k = 2j\) and \(u \equiv b^2 \pmod p\) with \(p \nmid b\), apply Hensel to \(f(x) = x^2 - u\) at \(a = b\): \(f(b) \equiv 0\) and \(f'(b) = 2b \not\equiv 0 \pmod p\) because \(p\) is odd. So \(u = \beta^2\) and \(x = (p^j \beta)^2\). \(\square\)

So for odd \(p\), a nonzero \(p\)-adic number is a square according to two yes/no tests (even valuation? residue a square?), and the group \(\mathbb{Q}_p^\times / (\mathbb{Q}_p^\times)^2\) of "numbers up to squares" has exactly four elements, represented by \(1, n, p, np\) where \(n\) is any non-residue mod \(p\).

Example. Is 2 a square in \(\mathbb{Q}_7\)? Yes: \(3^2 = 9 \equiv 2 \pmod 7\). Hensel lifts 3 to \(3, 10, 108, 2166, \ldots\) modulo \(7, 49, 343, 2401\), so \(\sqrt 2 = 3 + 1 \cdot 7 + 2 \cdot 49 + 6 \cdot 343 + \cdots\) in \(\mathbb{Q}_7\). (Check: \(10^2 = 100 = 2 + 2 \cdot 49\).) In \(\mathbb{Q}_5\), 2 is not a square, because the squares modulo 5 are 1 and 4.

When \(p = 2\), or when the derivative vanishes¶

For \(p = 2\) Hensel's lemma as stated never applies to \(x^2 - u\), since \(f'(x) = 2x \equiv 0 \pmod 2\). Something more is needed, and there is a stronger version.

Hensel's lemma, strong form. If \(f\) has coefficients in \(\mathbb{Z}_p\), \(a \in \mathbb{Z}_p\), and

\[ |f(a)|_p < |f'(a)|_p^2, \]

then there is a unique root \(\alpha\) of \(f\) with \(|\alpha - a|_p < |f'(a)|_p\).

When \(|f'(a)|_p = 1\) this is the version above. The proof is the same Newton iteration: with \(a_{n+1} = a_n - f(a_n)/f'(a_n)\), one shows from the Taylor formula that \(|f'(a_n)|_p\) stays equal to \(|f'(a)|_p\) and that \(|f(a_n)|_p / |f'(a)|_p^2\) is squared at each step, so it tends to 0 very fast. (Full details: Gouvêa's chapter on Hensel's lemma, or Koblitz, Chapter I.)

Squares in \(\mathbb{Q}_2\). An odd number squared is \(\equiv 1 \pmod 8\) (since \((2m+1)^2 = 4m(m+1) + 1\) and \(m(m+1)\) is even). So a unit \(u \in \mathbb{Z}_2\) can only be a square if \(u \equiv 1 \pmod 8\). Conversely, if \(u \equiv 1 \pmod 8\), take \(f(x) = x^2 - u\) and \(a = 1\): \(|f(1)|_2 = |1 - u|_2 \le 1/8\) and \(|f'(1)|_2^2 = |2|_2^2 = 1/4\). Since \(1/8 < 1/4\), the strong form gives a root. So:

A unit of \(\mathbb{Z}_2\) is a square exactly when it is \(\equiv 1 \pmod 8\). For example 17 is a square in \(\mathbb{Q}_2\); modulo 64 its square roots are \(9, 23, 41, 55\) (Check: \(9^2 = 81 = 17 + 64\)). And \(3, 5, 7\) are not squares in \(\mathbb{Q}_2\), so \(\mathbb{Q}_2^\times/(\mathbb{Q}_2^\times)^2\) has eight elements, represented by \(\pm 1, \pm 2, \pm 5, \pm 10\).

Roots of unity¶

Apply Hensel to \(f(x) = x^{p-1} - 1\). By Fermat's little theorem every \(a \in \{1, \ldots, p-1\}\) is a root mod \(p\), and \(f'(a) = (p-1)a^{p-2} \not\equiv 0 \pmod p\). So each lifts to a unique root in \(\mathbb{Z}_p\):

\(\mathbb{Q}_p\) contains exactly \(p - 1\) roots of \(x^{p-1} = 1\), one in each nonzero class mod \(p\). These are the Teichmüller representatives. In \(\mathbb{Q}_5\) they are \(1\), \(-1\) and the two square roots of \(-1\) found above (modulo 25: \(1, 7, 18, 24\)). In \(\mathbb{Q}_7\) there is a primitive cube root of unity, a root of \(x^2 + x + 1\); the one \(\equiv 2 \pmod 7\) is \(\equiv 30 \pmod{49}\). (Check: \(30^2 + 30 + 1 = 931 = 19 \cdot 49\).)

18.4 Ostrowski's theorem¶

Level: expert. We now have infinitely many ways to measure a rational number: \(|\cdot|_\infty\) and one \(|\cdot|_p\) for each prime. Are there more? Ostrowski's theorem (1916) says no.

Definition. An absolute value on a field \(K\) is a function \(|\cdot| : K \to [0, \infty)\) with

\(|x| = 0\) only for \(x = 0\);
\(|xy| = |x|\,|y|\);
\(|x + y| \le |x| + |y|\).

The trivial absolute value has \(|x| = 1\) for all \(x \ne 0\). Two absolute values are equivalent if they define the same notion of convergence; one can show this happens exactly when \(|x|_1 = |x|_2^{\,c}\) for a constant \(c > 0\). For example \(|\cdot|_\infty^{1/2}\) is an absolute value equivalent to \(|\cdot|_\infty\), and \(2^{-v_2(x)}\) and \(10^{-v_2(x)}\) are equivalent to each other (both give "2-adically close").

Ostrowski's theorem. Every nontrivial absolute value on \(\mathbb{Q}\) is equivalent to the ordinary absolute value \(|\cdot|_\infty\) or to \(|\cdot|_p\) for exactly one prime \(p\).

The theorem splits along a dividing line that we can prove.

Lemma. An absolute value on \(\mathbb{Q}\) is non-archimedean (obeys \(|x + y| \le \max(|x|, |y|)\)) if and only if \(|n| \le 1\) for every integer \(n\).

Proof. If it is non-archimedean then \(|n| = |1 + \cdots + 1| \le |1| = 1\). Conversely, suppose \(|n| \le 1\) for all integers. For \(x, y\) and any \(N \ge 1\), the binomial theorem gives

\[ |x + y|^N = \left| \sum_{k=0}^{N} \binom{N}{k} x^k y^{N-k} \right| \le \sum_{k=0}^N |x|^k |y|^{N-k} \le (N + 1) \max(|x|, |y|)^N. \]

Take \(N\)-th roots: \(|x + y| \le (N+1)^{1/N} \max(|x|, |y|)\). As \(N \to \infty\), \((N+1)^{1/N} \to 1\). \(\square\)

Proof of Ostrowski, non-archimedean case. Suppose \(|n| \le 1\) for all integers and \(|\cdot|\) is nontrivial. Then some integer has \(|n| < 1\) (if every nonzero integer had \(|n| = 1\), every rational would too, by multiplicativity). The set \(I = \{n \in \mathbb{Z} : |n| < 1\}\) is closed under addition (strong triangle inequality) and under multiplication by any integer (since \(|mn| = |m||n| \le |n|\)), so it is an ideal of \(\mathbb{Z}\), so \(I = p\mathbb{Z}\) for some \(p \ge 2\). It is a prime ideal: if \(|ab| < 1\) then \(|a| < 1\) or \(|b| < 1\). So \(p\) is prime. Let \(c = -\log |p| / \log p > 0\), so that \(|p| = p^{-c}\). Any integer \(n = p^k m\) with \(p \nmid m\) has \(|m| = 1\) (as \(m \notin I\)), so \(|n| = p^{-ck} = |n|_p^{\,c}\). By multiplicativity this holds for all rationals. \(\square\)

The archimedean case (idea). If some integer has \(|n| > 1\), one shows \(|n| = n^{c}\) for a fixed \(0 < c \le 1\), so \(|\cdot| = |\cdot|_\infty^{\,c}\). Ostrowski's trick: fix integers \(m, n > 1\) and write \(m^k\) in base \(n\). It has about \(k \log m / \log n\) digits, each less than \(n\), and the triangle inequality bounds \(|m^k|\) by (number of digits) \(\times\) (largest digit size) \(\times\, \max(1, |n|)^{\text{number of digits}}\). Taking \(k\)-th roots and letting \(k \to \infty\) kills the polynomial factors and leaves \(|m|^{1/\log m} \le |n|^{1/\log n}\), and by symmetry the two are equal. So \(\log |n| / \log n\) is a constant \(c\). The full proof, about two pages, is in Gouvêa's book and in Koblitz, Chapter I.

Why this matters. The completions of \(\mathbb{Q}\) are therefore exactly \(\mathbb{R}\) and the \(\mathbb{Q}_p\). Number theorists call them the places of \(\mathbb{Q}\), or its local fields, and they write \(\mathbb{Q}_\infty = \mathbb{R}\). The product formula of 18.1 is then a single relation linking all places at once. The same picture works for every number field \(K\) (Chapter 17): its non-archimedean places correspond to the prime ideals of its ring of integers, its archimedean places to its embeddings into \(\mathbb{R}\) and \(\mathbb{C}\), and a product formula holds again. This analogy, between primes and "points", and between number fields and curves, runs through the rest of the book.

18.5 Local-global: the Hasse-Minkowski theorem¶

Level: expert. Here is the main reason number theorists care about \(\mathbb{Q}_p\). Suppose you want to know whether an equation, say \(3x^2 + 5y^2 = 7z^2\), has a solution in rational numbers other than \((0,0,0)\). That is a global question (about \(\mathbb{Q}\)). Two kinds of easier local questions come first:

Does it have a nonzero solution in \(\mathbb{R}\)? (A question of signs and inequalities.)
Does it have a nonzero solution in \(\mathbb{Q}_p\), for each prime \(p\)?

A rational solution is also a real solution and a \(p\)-adic solution for every \(p\), since \(\mathbb{Q}\) sits inside all of these fields. So if any local test fails, there is no rational solution. The question is the converse.

What a p-adic solution means in practice¶

For a homogeneous equation \(F(x_1, \ldots, x_n) = 0\) with integer coefficients, a nonzero solution in \(\mathbb{Q}_p\) can be scaled to one in \(\mathbb{Z}_p\) with at least one coordinate a unit. Reducing modulo \(p^k\) gives a solution modulo \(p^k\) that is not all divisible by \(p\), for every \(k\). Conversely, if such "primitive" solutions exist modulo every \(p^k\), compactness of \(\mathbb{Z}_p\) (the tree argument of 18.2) produces a \(p\)-adic solution. So:

Solvable in \(\mathbb{Q}_p\) = solvable modulo every power of \(p\), not all coordinates divisible by \(p\).

That turns the local questions into finite computations, helped by Hensel's lemma, which says a good solution mod \(p\) (one where some partial derivative is nonzero mod \(p\)) already lifts.

A local obstruction¶

Example. \(x^2 + y^2 = 3z^2\) has no solution in rationals except \((0, 0, 0)\).

Proof (a 3-adic argument). Suppose there is a nonzero solution in \(\mathbb{Q}_3\); scale it so that \(x, y, z \in \mathbb{Z}_3\) and not all are divisible by 3. Modulo 3, squares are 0 or 1, so \(x^2 + y^2 \equiv 0 \pmod 3\) forces \(x \equiv y \equiv 0 \pmod 3\). Then \(9 \mid x^2 + y^2 = 3z^2\), so \(3 \mid z^2\) and \(3 \mid z\). All three are divisible by 3, a contradiction. So there is no nonzero solution in \(\mathbb{Q}_3\), hence none in \(\mathbb{Q}\). \(\square\)

The equation has real solutions, so the obstruction is purely 3-adic. This is Fermat's method of descent (Chapter 9) dressed in \(p\)-adic clothes.

The theorem¶

A quadratic form over \(\mathbb{Q}\) is a homogeneous polynomial of degree 2, \(Q(x_1, \ldots, x_n) = \sum_{i \le j} c_{ij} x_i x_j\) with rational coefficients. By completing squares (or by the diagonalisation of symmetric matrices, U of T guide MAT247 Unit 10) it can be brought, after a rational change of variables, to the form \(a_1 x_1^2 + \cdots + a_n x_n^2\). It represents zero (is isotropic) over a field \(K\) if \(Q(x) = 0\) for some nonzero \(x\) with coordinates in \(K\).

Hasse-Minkowski theorem. A quadratic form with rational coefficients represents zero over \(\mathbb{Q}\) if and only if it represents zero over \(\mathbb{R}\) and over \(\mathbb{Q}_p\) for every prime \(p\).

For forms in three variables this goes back to Legendre's theorem (1785): if \(a, b, c\) are nonzero, squarefree, pairwise coprime integers, not all of the same sign, then \(ax^2 + by^2 + cz^2 = 0\) has a nontrivial integer solution if and only if \(-bc\) is a square mod \(|a|\), \(-ca\) is a square mod \(|b|\) and \(-ab\) is a square mod \(|c|\). Minkowski proved the corresponding statements for forms with rational coefficients in the 1880s and 1890s in a different language, and Helmut Hasse, in his doctoral thesis of 1921, gave the statement in terms of \(\mathbb{Q}_p\) and proved it. The full proof fills a chapter: Serre, A Course in Arithmetic, Chapter IV, is the classic account, using the Hilbert symbol and quadratic reciprocity. Two consequences show its power.

Only finitely many checks. For a diagonal form \(a_1 x_1^2 + \cdots + a_n x_n^2\) with integer coefficients, at every odd prime \(p\) dividing none of the \(a_i\), a form in \(n \ge 3\) variables always represents zero over \(\mathbb{Q}_p\) (by the Chevalley-Warning theorem there is a nonzero solution mod \(p\), and Hensel lifts it). So only the primes dividing \(2 a_1 \cdots a_n\), and the real place, need checking. The infinitely many local conditions are really finitely many.

Five variables. Every quadratic form in five or more variables represents zero over every \(\mathbb{Q}_p\) (Serre, Chapter IV). With Hasse-Minkowski this gives Meyer's theorem (1884): a quadratic form over \(\mathbb{Q}\) in at least five variables that takes both positive and negative values represents zero over \(\mathbb{Q}\).

Worked example. Does \(3x^2 + 5y^2 = 7z^2\) have a nonzero rational solution? It is real-solvable (mixed signs). In Legendre's form \(3x^2 + 5y^2 - 7z^2 = 0\) with \(a = 3\), \(b = 5\), \(c = -7\): we need \(-bc = 35 \equiv 2\) to be a square mod 3 (it is not: squares mod 3 are 0, 1). So there is no solution. A direct 3-adic check confirms it: modulo 3 the equation reads \(5y^2 \equiv 7z^2\), i.e. \(2y^2 \equiv z^2 \pmod 3\); since 2 is not a square mod 3, \(y \equiv z \equiv 0\), then \(9 \mid 3x^2\) so \(3 \mid x\), and descent finishes it.

When the principle fails¶

Hasse-Minkowski is about degree 2. For higher degree, local solvability everywhere does not imply global solvability. Two examples, one easy and one famous.

A one-variable example. The equation

\[ (x^2 - 2)(x^2 - 17)(x^2 - 34) = 0 \]

has no rational root, since 2, 17 and 34 are not squares. But it has a root in \(\mathbb{R}\) (namely \(\sqrt 2\)) and in every \(\mathbb{Q}_p\):

for \(p = 2\): \(17 \equiv 1 \pmod 8\), so 17 is a square in \(\mathbb{Q}_2\) (18.3);
for \(p = 17\): \(2 \equiv 6^2 \pmod{17}\) (\(36 = 2 + 34\)), so 2 is a square in \(\mathbb{Q}_{17}\);
for any other odd prime \(p\): 2 and 17 are units. If neither is a square mod \(p\), their product 34 is a square mod \(p\) (the Legendre symbol is multiplicative, Chapter 8). So one of 2, 17, 34 is a nonzero square mod \(p\), and Hensel lifts it.

So the equation is solvable everywhere locally and not globally. (It is not a quadratic form, so Hasse-Minkowski is not contradicted.)

Selmer's cubic. In 1951 Ernst Selmer showed that

\[ 3x^3 + 4y^3 + 5z^3 = 0 \]

has nonzero solutions in \(\mathbb{R}\) and in every \(\mathbb{Q}_p\), but none in \(\mathbb{Q}\). The local part is a finite check of the kind above (plus Hensel); the global part is genuinely hard and uses a descent on a related elliptic curve, the curve \(x^3 + y^3 + 60z^3 = 0\) (its Jacobian; \(60 = 3 \cdot 4 \cdot 5\)). A plane cubic curve with a rational point is an elliptic curve, the subject of Chapter 19, and Selmer's curve is a "twisted form" of one. Its failure to have a point is measured by a group attached to that elliptic curve, the Tate-Shafarevich group Ш, whose elements are, roughly, the curves of this kind that have points everywhere locally, with all the curves that have a rational point counted as the zero element. Ш appears again in the Birch and Swinnerton-Dyer conjecture (19.6).

Try it in Python. No small solutions, but local solutions exist; here is the search for small nonzero integer solutions, which finds nothing:

sols = [(x, y, z) for x in range(-30, 31) for y in range(-30, 31) for z in range(-30, 31)
        if (x, y, z) != (0, 0, 0) and 3*x**3 + 4*y**3 + 5*z**3 == 0]
print(sols)    # []  (Selmer: there are none at all, of any size)

A finite search proves nothing by itself; Selmer's theorem is what says the list is empty forever.

A short history¶

Kurt Hensel introduced \(p\)-adic numbers in a paper of 1897, thinking of them as analogues of power series expansions of functions around a point: an integer is like a polynomial, its base \(p\) digits like its Taylor coefficients at a point, and \(\mathbb{Z}_p\) like the ring of power series. That analogy between primes and points is the one we met at the end of 18.4. József Kürschák defined general absolute values (valuations) on fields in 1913, and Alexander Ostrowski classified those on \(\mathbb{Q}\) soon after. Hasse's 1921 thesis turned the \(p\)-adic numbers from a curiosity into a central tool by proving the local-global theorem for quadratic forms. Today \(p\)-adic methods are everywhere in number theory, from Wiles's proof of Fermat's last theorem (Chapter 20) to \(p\)-adic \(L\)-functions and \(p\)-adic Hodge theory.

Check yourself¶

1. Compute \(|600|_2\), \(|600|_3\), \(|600|_5\), \(|600|_7\), and check the product formula for \(600\).

\(600 = 2^3 \cdot 3 \cdot 5^2\). So \(|600|_2 = 1/8\), \(|600|_3 = 1/3\), \(|600|_5 = 1/25\), \(|600|_7 = 1\) (and \(|600|_p = 1\) for all \(p \ge 7\)). Product formula: \(|600|_\infty \cdot \frac18 \cdot \frac13 \cdot \frac1{25} = 600/600 = 1\).

2. Show that \(1 + 3 + 9 + 27 + \cdots = -\tfrac12\) in \(\mathbb{Q}_3\).

The partial sum is \(s_n = 1 + 3 + \cdots + 3^{n-1} = (3^n - 1)/2\). Then \(s_n - (-\tfrac12) = 3^n/2\), and \(|3^n/2|_3 = 3^{-n} \to 0\). So \(s_n \to -\tfrac12\) 3-adically. (Also the geometric series formula: \(1/(1-3) = -1/2\).)

3. Write \(-1\) and \(\tfrac12\) as 3-adic digit strings.

\(-1 = 2(1 + 3 + 9 + \cdots) = 2 \cdot (-\tfrac12)\), so \(-1 = \ldots 2222_3\). From question 2, \(-\tfrac12 = \ldots 1111_3\), so \(\tfrac12 = -\tfrac12 + 1 = \ldots 1112_3\). Check: \(2 \cdot \ldots 1112_3\): \(2 \cdot 2 = 4 = 1 + 1\cdot 3\), write 1 carry 1; \(2 \cdot 1 + 1 = 3\), write 0 carry 1; and so on forever, giving \(\ldots 0001 = 1\).

4. Is \(2\) a square in \(\mathbb{Q}_7\)? In \(\mathbb{Q}_5\)? In \(\mathbb{Q}_2\)?

\(\mathbb{Q}_7\): yes, \(2 \equiv 3^2 \pmod 7\) and Hensel lifts (18.3). \(\mathbb{Q}_5\): no, the nonzero squares mod 5 are 1 and 4, and 2 is a unit. \(\mathbb{Q}_2\): no, \(v_2(2) = 1\) is odd, and a square has even valuation.

5. Find the square root of 6 in \(\mathbb{Z}_5\) that is \(\equiv 1 \pmod 5\), modulo \(625\).

\(f(x) = x^2 - 6\), \(f'(1) = 2 \not\equiv 0 \pmod 5\). Lift to mod 25: \(x = 1 + 5t\), \(x^2 = 1 + 10t + 25t^2\), need \(1 + 10t \equiv 6 \pmod{25}\), i.e. \(2t \equiv 1 \pmod 5\), \(t = 3\), \(x = 16\). Mod 125: \(16^2 - 6 = 250 \equiv 0 \pmod{125}\), so \(x = 16\) already works (the next digit is 0). Mod 625: \(x = 16 + 125s\), \(x^2 - 6 = 250 + 4000s + \ldots \equiv 250 + 250 s \pmod{625}\), so need \(2 + 2s \equiv 0 \pmod 5\), \(s = 4\), \(x = 516\). Check: \(516^2 - 6 = 266250 = 426 \cdot 625\). So \(\sqrt 6 \equiv 516 \pmod{625}\); its digits are \(1, 3, 0, 4\), i.e. \(\sqrt 6 = \ldots 4031_5\).

6. Show that \(3\), \(5\) and \(7\) are not squares in \(\mathbb{Q}_2\), but \(-7\) is.

They are units of \(\mathbb{Z}_2\), and a unit is a square exactly when it is \(\equiv 1 \pmod 8\) (18.3). \(3, 5, 7\) are not \(\equiv 1 \pmod 8\), but \(-7 \equiv 1 \pmod 8\). So \(\sqrt{-7} \in \mathbb{Q}_2\).

7. Prove that \(\sum_{n \ge 0} n!\) converges in \(\mathbb{Q}_p\) for every prime \(p\).

In \(\mathbb{Q}_p\) a series converges when its terms tend to 0. For \(n \ge kp\), the product \(n!\) contains the \(k\) factors \(p, 2p, \ldots, kp\), so \(v_p(n!) \ge k\) and \(|n!|_p \le p^{-k}\). Hence \(|n!|_p \to 0\). (Whether this sum is rational is, as far as the author knows, an open question.)

8. Hensel needs \(f'(a) \not\equiv 0\). Show what goes wrong for \(f(x) = x^2 - 5\) in \(\mathbb{Z}_2\) with \(a = 1\).

\(f(1) = -4 \equiv 0 \pmod 2\), but \(f'(1) = 2 \equiv 0 \pmod 2\), so the simple form does not apply. The strong form needs \(|f(1)|_2 = 1/4 < |f'(1)|_2^2 = 1/4\), which fails too. And indeed there is no root: \(5 \not\equiv 1 \pmod 8\), so 5 is not a square in \(\mathbb{Q}_2\). A root mod 2 (even mod 4) need not lift.

9. Find the primitive cube root of unity in \(\mathbb{Z}_7\) that is \(\equiv 4 \pmod 7\), modulo \(49\).

\(f(x) = x^2 + x + 1\), \(f(4) = 21 \equiv 0 \pmod 7\), \(f'(4) = 9 \equiv 2 \not\equiv 0\). Put \(x = 4 + 7t\): \(f = 21 + 9 \cdot 7t + 49t^2 \equiv 7(3 + 9t) \pmod{49}\), need \(3 + 2t \equiv 0 \pmod 7\), \(t = 2\), \(x = 18\). Check: \(18^2 + 18 + 1 = 343 = 7 \cdot 49\). (The other root is \(30\), and \(18 + 30 = 48 \equiv -1\), as the sum of the roots of \(x^2 + x + 1\) must be.)

10. In \(\mathbb{Q}_p\), show that if \(|x - y|_p < |y|_p\) then \(|x|_p = |y|_p\).

Write \(x = y + (x - y)\). The two summands have different absolute values (\(|x - y|_p < |y|_p\)), so by Fact 1 of 18.1 \(|x|_p = \max(|y|_p, |x - y|_p) = |y|_p\). (This is why, in the completion, \(|x_n|_p\) is eventually constant for a Cauchy sequence not tending to 0.)

11. Show that \(x^2 + y^2 + z^2 = 7\) has no rational solution.

Suppose \(x = a/d\), \(y = b/d\), \(z = c/d\) with \(d > 0\) minimal. Then \(a^2 + b^2 + c^2 = 7d^2\). Squares mod 8 are \(0, 1, 4\). If \(d\) is odd, \(7d^2 \equiv 7 \pmod 8\), but no three squares sum to \(7\) mod 8 (the possible sums are \(0, 1, \ldots, 6\)). If \(d\) is even, \(7d^2 \equiv 0 \pmod 4\), which forces \(a, b, c\) all even (an odd square is \(1 \bmod 4\), and one or two or three of them would give a sum \(\equiv 1, 2, 3 \pmod 4\)). Then divide \(a, b, c, d\) by 2, contradicting minimality. In \(p\)-adic language: the form \(x^2 + y^2 + z^2 - 7w^2\) does not represent 0 over \(\mathbb{Q}_2\). (This is Legendre's three-square theorem for the number 7, Chapter 11.)

12. Use Legendre's criterion to decide whether \(x^2 + y^2 = 5z^2\) and \(x^2 + 2y^2 = 7z^2\) have nontrivial rational solutions; find one where it exists.

\(x^2 + y^2 - 5z^2\): \(a = 1\), \(b = 1\), \(c = -5\). Conditions mod \(|a| = 1\) and \(|b| = 1\) are empty; mod 5 we need \(-ab = -1\) to be a square mod 5: \(-1 \equiv 4 = 2^2\). Yes. Solution: \((1, 2, 1)\), since \(1 + 4 = 5\). \(x^2 + 2y^2 - 7z^2\): \(a = 1\), \(b = 2\), \(c = -7\). Mod 2: \(-ca = 7 \equiv 1\), a square. Mod 7: \(-ab = -2 \equiv 5\); squares mod 7 are \(1, 2, 4\), so 5 is not a square. No nontrivial solution.

What you learned¶

For each prime \(p\), \(|x|_p = p^{-v_p(x)}\) measures how divisible \(x\) is by \(p\); it obeys the strong triangle inequality, so every triangle is isosceles and discs are nested or disjoint.
The product formula \(|x|_\infty \prod_p |x|_p = 1\) links all the sizes of a rational number.
\(\mathbb{Q}_p\) is the completion of \(\mathbb{Q}\) for \(|\cdot|_p\); its elements are digit strings running to the left, \(\mathbb{Z}_p\) is a tree of residues mod \(p^n\), and series converge when their terms tend to 0.
Hensel's lemma lifts a simple root mod \(p\) to a unique \(p\)-adic root; it decides which numbers are squares in \(\mathbb{Q}_p\) and gives the \(p - 1\) roots of unity.
Ostrowski: \(|\cdot|_\infty\) and the \(|\cdot|_p\) are the only absolute values on \(\mathbb{Q}\), up to equivalence.
Hasse-Minkowski: a rational quadratic form represents 0 over \(\mathbb{Q}\) exactly when it does over \(\mathbb{R}\) and every \(\mathbb{Q}_p\). For higher degree the local-global principle can fail, as Selmer's cubic \(3x^3 + 4y^3 + 5z^3 = 0\) shows.

Next: Chapter 19, cubic curves whose rational points form a group, and the questions about them that lead to the frontier.