SKW

3 The valuation function \(s\)

3.1 General properties

Definition 32
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For \(a \in \mathbb {Z}\) and a maximal ideal \(\mathfrak {Q}\) of \(\mathcal{O}_L\), let \(s_{\mathfrak {Q}}(a) \in \mathbb {N}\) be the multiplicity of \(\mathfrak {Q}\) in the ideal \((g_a)\). We write \(s(a) = s_{\mathfrak {P}}(a) = v_{\mathfrak {P}}(g_a)\).

Lemma 33
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If \(a \equiv b \pmod{q-1}\), then \(s(a) = s(b)\).

Proof ▶

Indeed \(g_a = g_b\) by 18.

Lemma 34
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If \((q - 1) \mid a\), then \(s(a) = 0\).

Proof ▶

By 19, \(g_a = -1\) is a unit, so \((g_a) = \mathcal{O}_L\).

Lemma 35
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If \((q-1) \nmid (a+b)\), then \(s(a) + s(b) = s(a+b) + v_{\mathfrak {P}}(J(a, b))\).

Proof ▶

Take valuations in 21. All the elements involved are nonzero by 27, so the valuations are finite.

Lemma 36
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For all \(a, b \in \mathbb {Z}\), \(s(a + b) \leq s(a) + s(b)\).

Proof ▶

If \((q-1) \mid (a+b)\), then \(s(a+b) = 0\) by 34. Otherwise use 35.

Lemma 37
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If \((q-1) \nmid a\), then \(s(a) + s(-a) = f \cdot e(\mathfrak {P}\mid p)\).

Proof ▶

Take valuations in 22: the factor \(\omega ^{-a}(-1)\) is a root of unity, hence a unit, and \(v_{\mathfrak {P}}(p^f) = f \cdot e(\mathfrak {P}\mid p)\).

Lemma 38
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Assume \((q-1) \mid k\). Then

\[ s(a) + s(k - a) = \begin{cases} 0 & \text{if } (q-1) \mid a, \\ f \cdot e(\mathfrak {P}\mid p) & \text{otherwise.} \end{cases} \]
Proof ▶

We have \(s(k - a) = s(-a)\) by 33. Conclude with 34 and 37.

Lemma 39
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We have

\[ 2 \sum _{a=0}^{q-2} s(a) = (q-2) \cdot f \cdot e(\mathfrak {P}\mid p). \]
Proof ▶

Pair the term \(a\) with \(q - 1 - a\). By 38 with \(k = q - 1\), each of the \(q - 2\) values \(1 \leq a \leq q-2\) gives \(s(a) + s(q-1-a) = f \cdot e(\mathfrak {P}\mid p)\), and \(s(0) = s(q-1) = 0\) by 34.

Let \(\mathfrak {Q}\) be a maximal ideal of \(\mathcal{O}_L\) not lying above \(p\). Then \(s_{\mathfrak {Q}}(a) = 0\) for all \(a \in \mathbb {Z}\).

Proof ▶

If \((q-1) \mid a\), use 34. Otherwise, if \(g_a \in \mathfrak {Q}\), then the norm of \(\mathfrak {Q}\) divides \(N_{L/\mathbb {Q}}(g_a)\), which is a power of \(p\) by 26, so \(\mathfrak {Q}\) lies above \(p\).

Lemma 41
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For all \(\sigma \in \mathrm{Gal}(L/F)\) and \(a \in \mathbb {Z}\), \(s_{\sigma (\mathfrak {P})}(a) = s_{\mathfrak {P}}(a \cdot \mathfrak {n}(\sigma ^{-1}))\).

Proof ▶

The multiplicity of \(\sigma (\mathfrak {P})\) in \((g_a)\) is the multiplicity of \(\mathfrak {P}\) in \((\sigma ^{-1}(g_a))\), and \(\sigma ^{-1}(g_a) = g_{a \mathfrak {n}(\sigma ^{-1})}\) by 30.

3.2 Upper bound by the sum of digits

Lemma 42
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We have \(s(1) = 1\).

Proof ▶

By 24, \(g_1 \in \mathfrak {P}\), so \(s(1) \geq 1\). By 28, \(g_1 \equiv -(\zeta _p - 1) \pmod{\mathfrak {P}^2}\), and \(\zeta _p - 1 \notin \mathfrak {P}^2\) by 106, so \(s(1) \leq 1\).

Lemma 43
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For all \(a \in \mathbb {N}\), \(s(a) \leq a\).

Proof ▶

By induction on \(a\), using \(s(0) = 0\) (34), \(s(a+1) \leq s(a) + s(1)\) (36) and \(s(1) = 1\) (42).

Lemma 44
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Let \(a \in \mathbb {N}\) and let \(a = a_0 + a_1 p + \cdots + a_{t-1} p^{t-1}\) be its \(p\)-adic expansion. Then

\[ s(a) \leq a_0 + a_1 + \cdots + a_{t-1}. \]
Proof ▶

By induction on the list of digits. Write \(a = a_0 + p a'\). Then \(s(a) \leq s(a_0) + s(pa') \leq a_0 + s(a')\) by 36, 43, and \(g_{pa'} = g_{a'}\) (20).

3.3 The digit formula

Proposition 45
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Let \(0 \leq a \leq q - 2\) and let \(a = a_0 + a_1 p + \cdots + a_{f-1} p^{f-1}\), \(0 \leq a_i \leq p - 1\), be the \(p\)-adic expansion of \(a\). Then

\[ s(a) = a_0 + a_1 + \cdots + a_{f-1}. \]
Proof ▶

By 105, \(e(\mathfrak {P}\mid p) = p - 1\), so 39 gives \(\sum _{a=0}^{q-2} s(a) = (q-2)f(p-1)/2\). A direct count shows that the sum of the digit sums of \(0, \ldots , q-2\) has the same value. Since each term satisfies \(s(a) \leq a_0 + \cdots + a_{f-1}\) by 44, all these inequalities are equalities.

Lemma 46
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Let \(b {\gt} 1\), \(l \geq 0\) and \(0 \leq x {\lt} b^l - 1\). Then

\[ (b - 1) \sum _{i=0}^{l-1} (x b^i \bmod (b^l - 1)) = (b^l - 1) \cdot S_b(x), \]

where \(S_b(x)\) is the sum of the base \(b\) digits of \(x\).

Proof ▶

Modulo \(b^l - 1\), multiplication by \(b\) cyclically rotates the \(l\) base \(b\) digits of \(x\). Hence each digit of \(x\) occurs exactly once in each position in the sum, which is therefore \(S_b(x)(1 + b + \cdots + b^{l-1})\).

Lemma 47
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Let \(m, d\) be positive integers with \(q - 1 = md\), and let \(a \in \mathbb {N}\) with \(da \leq q - 2\). Then

\[ m \cdot s(da) = (p-1) \sum _{i=0}^{f-1} (ap^i \bmod m). \]
Proof ▶

Multiply both sides by \(d\). Since \(d (ap^i \bmod m) = dap^i \bmod (q-1)\), this follows from 45 and 46 applied to \(b = p\), \(l = f\) and \(x = da\).