p-adic analysis 2

Shu Sasaki

10th February 2011

Abstract

Lecture notes for p-adic analysis, 24/01/2011 and a part of 31/01/2011.

1 p-adic numbers and non-archimedean fields

1.1 Completions

1.1.1 F = Q and the p-adic numbers

See “p-adic analysis 1”.

1.1.2 Completions and inverse limits on F = Q

PROPOSITION. Z_{p} is compact (any covering by open sets has a finite refinement) and Q_{p} is locally compact (every point has a neighbourhood which is compact). Proof. Suppose Z_{p} is compact. Since it is a neighbourhood of zero, for any x ∈ Q_{p}, x + Z_{p} defines a compact neighbourhood (note that “adding x” in Q_{p} defines a continuous map and the image x + Z_{p} of a compact set Z_{p} is compact) and therefore Q_{p} is locally compact. STP Z_{p} is compact. Since Z_{p} is complete with respect to the metric topology, it suffices to show that it is precompact (see Bourbaki’s “General topology”, II, 4,2, Corollary: compact if and only if Hausdorff, complete and precompact in the sense of Bourbaki), which follows, by Bourbaki’s “GT”, II, 4,2, Theorem 3, for every r > 0, there exists a finite set S such that {C(r)(x)}_{x∈S} cover Z_{p} (recall they are “open subsets” of Z_{p}; immediate if you have done an exercise!). It suffices to check r = p^{−n}, n ∈ Z_{≥0}, in which case {x + p^{n}Z_{p}}_{x∈{0,1,...,p^{n}−1}} = {C(p^{−n})(x)}_{x∈{0,...,p^{n}−1}} cover Z_{p} (Why!?(rhetorical)). □

Let lim_{←n} Z/p^{n}Z denote the closed (check!) subset of ∏_{n≥1} Z/p^{n}Z (with the discrete topology on each Z/p^{n}Z) of elements (..., x_{n}, ...) such that x_{n} ∈ Z/p^{n}Z maps to x_{n−1} ∈ Z/p^{n−1}Z by the natural surjection Z/p^{n}Z → Z/p^{n−1}Z.

PROPOSITION. Z_{p} ≃ lim_{←} Z/p^{n}Z.

For x ∈ Z_{p}, there exists a sequence (x_{n}), n ∈ N, x_{n} ∈ Z such that (1) 0 ≤ x_{n} ≤ p^{n} − 1, (2) x_{n} is congruent to x_{n−1} mod p^{n−1} (3) x mod p^{n} equals x_{n} mod p^{n}, in other words, ||x − x_{n}||_{p} ≤ p^{−n}. This follows since Z has dense image in Z_{p}. The (x_{n}) is a Cauchy sequence converging to x. Conversely, given a Cauchy sequence with these properties, we get a element of Z_{p}. Let us describe the x_{n}’s more explicitly. Given x_{1} such that 0 ≤ x_{1} ≤ p − 1, let c_{0} = x_{1}. Note that x_{2} then needs to satisfy the conditions 0 ≤ x_{2} ≤ p^{2} − 1 and x_{2} ≡ x_{1} mod p, so we can take x_{2} = x_{1} + c_{1}p for some 0 ≤ c_{1} ≤ p − 1. Combining, x_{2} = c_{0} + c_{1}p. Continuing this, we get x_{n} = c_{0} + c_{1}p + c_{2}p^{2} + ⋯ + c_{n−1}p^{n−1} with 0 ≤ c_{i} ≤ p − 1. As n → ∞, we get x = c_{0} + c_{1}p + c_{2}p^{2} + · · ·. Since Q_{p} = Z[1/p], we get an “p-expansion” of x ∈ Q_{p} as x = c_{−N}p^{−N} + c_{−N+1}p^{−N+1} + ⋯ + c_{0}p^{0} + c_{1}p + c_{2}p^{2} + · · ·.

REMARK. Using the preceding proposition, it is easy to prove that Z_{p} is compact. Indeed, Z/p^{Z} is finite for every n and therefore compact. Since Z_{p} is a closed subset of the ∏_{n} Z/p^{n}Z which is compact, Z_{p} is compact.

1.2 Extensions of Q_{p}

Recall that, given a number field F, an non-archimedean absolute value || || corresponds uniquely to a prime ideal 𝔭. The completion of F with respect to || || is denoted by F_{𝔭} and defines a finite extension of Q_{p}.

Let K be a finite extension of Q_{p} which comes equipped with an absolute value || ||_{K} which extends || ||_{Q_{p}} on Q_{p}. If such || ||_{K} exists, then K is complete with respect to || ||_{K} and is unique (up to equivalence). But does it exists? Yes! It is

||x||_{K} = (||N_{K/Q_{p}}(x)||_{Q_{p}})^{1/[K:Q_{p}]}.

It extends || ||_{Q_{p}} since, for x ∈ Q_{p}, (||N_{K/Q_{p}}(x)||_{Q_{p}})^{1/[K:Q_{p}]} = (||x^{[K:Q_{p}]}||_{Q_{p}})^{1/[K:Q_{p}]} = ||x||_{Q_{p}}, and it is non-archimedean.

Define (normalise) the valuation v_{K} : K → R_{≥0} by ||x||_{K} = (1/#k(v_{K}))^{v_{K}(x)} = (1/p^{f_{K}})^{v_{K}(x)}. It is normalised so that it sends a uniformiser ϖ_{K} to 1. Note that Gouvea normalise the valuation by defining v′_{K} by ||x||_{K} = (1/p)^{v′_{K}(x)} which sends ϖ_{K} to 1/e_{K} where e_{K} is the ramification index (consequently v_{K} = e_{K}v′_{K}).

1.3 Archimedean absolute values on F

An absolute value is archimedean if it is not non-archimedean. The only archimedean absolute values are essentially the ordinary abosulte value on C and the valuations which induces on its subfields, e.g., R.

THEOREM (Ostrowski) A field complete with respect to an archimedean absolute value is isomorphic to either R or C. Proof. See Theorem 1.1, Chapter 3, Cassels.

1.4 Nonarchimedean absolute values on F

Let F be a number field, O_{F} be the ring of integers, and 𝔭 be a prime ideal of F. Given these, we have

(1-1) the completion F^{∧} of F with respect to the (equivalent class of) the (non-archimedean) absolute value || || “arising from” the discrete valuation v associated to 𝔭 (it is equipped with the metric topology);

(1-2) the closure O_{F}^{∧} of O_{F} with respect to the restriction of the absolute value to O_{F}; (1-3) the closure (𝔭^{r})^{∧} in F^{∧} of 𝔭^{r} with respect to the topology defined by the absolute value on F^{∧}

for every r ∈ Z_{≥0};

(2-2) the discrete valuation ring O_{F^{∧}} = {x ∈ F^{∧}| v^{∧}(x) ≥ 0} = {x ∈ F^{∧} | ||x||^{∧} ≤ 1} where v^{∧} (resp. || ||^{∧}) is a unique extension of v (resp. || ||) on F to F^{∧} (See Theorem 11, p.77 of Fröhlich-Taylor);

(2-1) the field of fractions of O_{F^{∧}};

(2-3) the r-th power 𝔪^{r} of the maximal ideal 𝔪 = {x ∈ F^{∧} | v^{∧}(x) > 0} = {x ∈ F^{∧} | ||x||^{∧} < 1} of O_{F}^{∧} for every r ∈ Z_{≥0};

(3-2) the inverse limit ^{∧}O_{F} of the inverse system {O_{F} /𝔭^{n}}_{n∈{1,2,...}} together with O_{F} -module homomorphisms θ_{n_{1},n_{2}} : O_{F} /𝔭^{n_{2}} → O_{F} /𝔭^{n_{1}} for n_{1} ≤ n_{2} canonically defined by 𝔭^{n_{1}} ⊇ 𝔭^{n_{2}};

(3-1) the field of fractions ^{∧}F of ^{∧}O_{F} (it is equipped with the inverse limit topology); (3-3) ker(^{∧}O_{F} → O_{F} /𝔭^{r}) for every r ∈ Z_{≥0}. Then (1-∗)≃ (2-∗)≃ (3-∗) for ∗ = 1, 2, 3. Note that F^{∧} ≃ ^{∧}F (and hence for ∗ = 2, 3) is an isomorph-

ism of topological spaces. The metric topology on F^{∧} defined by || ||^{∧} is Hausdorff; the inverse limit topology on ^{∧}F is defined by assuming a fundamental system of neighbourhoods of 0 to be {𝔭^{n}}_{n∈{1,2,...}} in the sense that G ⊆ F is open if and only if G contains 𝔭^{n} for some n, which makes it a totally disconnected topological space.

1.5 C_{p}

Since the algebraic closure \overline{Q}_{p} of Q_{p} is the union of all finite algebraic extensions of Q_{p}, there is a unique extension of the p-adic valuation on Q_{p} to \overline{Q}_{p}. LEMMA The algebraic closure \overline{Q}_{p} of Q_{p} is NOT complete. The completion C_{p} of \overline{Q}_{p} is algebraically closed. Cassels p.149

2 Finding rational points

HENSEL’S LEMMA. Let F be a field complete with respect to a non-archimedean absolute value || ||. Let O denote {x ∈ F |||x|| ≤ 1}. Let f(X) = c_{0} + c_{1}X + ⋯ + c_{n}X^{n} be a polynomial O[X]. Suppose that there exists x_{0} ∈ O which satisfies ||f(x_{0})|| < ||f′(x_{0})||^{2} where by f′ we mean the formal derivative of f. Then there exists x ∈ O such that f(x) = 0.

Proof. Define f_{i}(X) by f(X + Y) = f(X) + f_{1}(Y)Y + f_{2}(X)Y^{2} + ⋯ + f_{n}(X)Y^{n}. In particular, f_{1}(X) = f′(X). Think of f(x_{0}) and f_{1}(x_{0}) as elements of F, let y_{0} be an element of F defined by f(x_{0}) + y_{0}f_{1}(x_{0}) = 0. Firstly observe that ||y_{0}|| = ||f(x_{0})||/||f_{1}(x_{0})||. Secondly ||y_{0}f_{1}(x_{0})|| = ||f(x_{0})|| < ||f_{1}(x_{0})||^{2} implies that ||y_{0}|| < ||f_{1}(x_{0})||. In particular, y_{0} ∈ O.

\underline{||f(x_{0} + y_{0})|| < ||f(x_{0})||}: by the definition of the f_{i}(X), ||f(x_{0} + y_{0})|| = ||f_{2}(x_{0})y_{0}^{2} + f_{3}(x_{0})y_{0}^{3} ⋯ || ≤ max_{i≥2}||f_{i}(x_{0})y_{0}^{i}|| ≤ max_{i≥2}||y_{0}^{i}|| ≤ ||y_{0}||^{2} = ||f(x_{0})||^{2}/||f_{1}(x_{0})||^{2} = ||f(x_{0})||(||f(x_{0})||/||f_{1}(x_{0})||^{2}) < ||f(x_{0})||.

\underline{||f_{1}(x_{0} + y_{0})|| = ||f_{1}(x_{0})||}: ||f_{1}(x_{0} + y_{0}) − f_{1}(x_{0})|| ≤ ||(x_{0} + y_{0}) − x_{0}|| = ||y_{0}|| < ||f_{1}(x_{0})||. Therefore ||f_{1}(x_{0} + y_{0})|| = ||f_{1}(x_{0} + y_{0}) − f_{1}(x_{0}) + f_{1}(x_{0})|| = ||f_{1}(x_{0})||. Let x_{1} = x_{0} + y_{0}.

Summary so far: given (1) x_{0} ∈ O such that ||f(x_{0})|| < ||f_{1}(x_{0})||^{2}, define (2) y_{0} ∈ O by f(x_{0}) + y_{0}f_{1}(x_{0}) = 0 and let (3) x_{1} = x_{0} + y_{0}. It satisfies ||f(x_{1})|| < ||f(x_{0})|| and ||f_{1}(x_{1})|| = ||f_{1}(x_{0})||. Combining, we get ||f(x_{1})|| < ||f(x_{0})|| < ||f_{1}(x_{0})||^{2} = ||f_{1}(x_{1})||^{2}.

Inductively define x_{n} by: given (1) x_{n−1} ∈ O such that ||f(x_{n−1})|| < ||f_{1}(x_{n−1})||^{2}, define (2) y_{n−1} by f(x_{n−1})+y_{n−1}f_{1}(x_{n−1}) = 0 and deduce y_{n−1} ∈ O from the inequality which x_{n−1} satisfies in (1), and let

(3) x_{n} = x_{n−1} + y_{n−1}. Then ||f(x_{n})|| < ||f(x_{n−1})||, ||f_{1}(x_{n})|| = ||f_{1}(x_{n−1})|| and ||f(x_{n})|| < ||f_{1}(x_{n})||^{2}. \underline{||f(x_{n})||/||f_{1}(x_{n})||^{2} ≤ (||f(x_{0})||/||f_{1}(x_{0})||^{2})^{2^{n}}}: by definition, f(x_{n}) = f(x_{n−1}+y_{n−1}) = f_{2}(x_{n−1})y_{n−1}^{2}+

f_{3}(x_{n−1})y_{n−1}^{3} ⋯ and hence ||f(x_{n})|| ≤ ||y_{n−1}||^{2}. It then follows that ||f(x_{n})||/||f_{1}(x_{n})||^{2} ≤ ||y_{n−1}||^{2}/||f_{1}(x_{n})||^{2} = ||y_{n−1}||^{2}/||f_{1}(x_{n−1})||^{2} = (||y_{n−1}||/||f_{1}(x_{n−1})||)^{2} = (||f(x_{n−1})||/||f_{1}(x_{n−1})||^{2})^{2}. By induction, we then deduce ||f(x_{n})||/||f_{1}(x_{n})||^{2} ≤ (||f(x_{0})||/||f_{1}(x_{0})||^{2})^{2^{n}}.

\underline{||f(x_{n})|| → 0} as n → ∞: ||f(x_{n})|| ≤ (||f(x_{0})||/||f_{1}(x_{n})||^{2})^{2^{n}}||f_{1}(x_{n})||^{2} = (||f(x_{0})||/||f_{1}(x_{n})||^{2})^{2^{n}}||f_{1}(x_{0})||^{2} → 0 as ||f(x_{0})||/||f_{1}(x_{0})||^{2} < 1.

The x_{n} then form a Cauchy sequence since ||x_{n+1}−x_{n}|| = ||y_{n}|| = ||f(x_{n})||/||f_{1}(x_{n})|| = ||f(x_{n})||/||f_{1}(x_{0})|| → 0 as n → ∞. Since F is complete with respect to || ||, it converges to x ∈ F such that f(x) = 0.

APPLICATIONS OF HENSEL’S LEMMA.

LEMMA. Suppose p ≠ 2. Let y ∈ Z_{p} with ||y|| = 1. Suppose that there exists x_{0} ∈ Z_{p} such that ||x_{0}^{2} − y|| < 1. Then y = x^{2} for some x ∈ Z_{p}. Proof. Let f(X) = X^{2} − y ∈ Z_{p}[X]. Its formal derivative f′(X) is 2X. Then ||f′(x_{0})|| = ||2||||x_{0}|| = 1||x_{0}|| = 1 since ||x_{0}||^{2} = ||x_{0}^{2}|| = ||(x_{0}^{2} − y) + y|| = ||y|| = 1 (note that ||x_{0}^{2} − y|| < 1) and hence ||x_{0}|| = 1. Therefore ||f(x_{0})|| < 1 = ||f′(x_{0})||^{2}. By Hensel’s lemma, there exists x ∈ Z_{p} such that f(x) = 0, in other words, x^{2} = y. □

COROLLARY. Q_{p}^{×}/Q_{p}^{×2} has order 4 and exponent 2. More precisely Q_{p}^{×}/Q_{p}^{×,2} = {1, p, u, up} where x is a (any) quadratic non-residue.

LEMMA. Suppose p = 2. Let x ∈ Z_{2} such that x ≡ 1 mod 8. Then x = y^{2} for some y ∈ Z_{2}. Proof. Let f(X) = x^{2} − y. Then y ≡ 1 mod 8 implies that ||f(1)|| = ||1 − y|| ≤ (1/2)^{3}. OTOH, ||f′(1)|| = ||2|| = 1/2. Therefore ||f(1)|| ≤ 1/8 < 1/4 = ||f′(1)||. Apply Hensel’s lemma. □

COROLLARY. Q_{2}^{×}/Q_{2}^{×2} has order 8 and exponent 2.

LEMMA. Suppose p ≠ 3. Let y ∈ Z_{p} with ||y|| = 1. Suppose that y ≡ x_{0} mod p for some x_{0}ınZ_{p}. Then there exists x ∈ Z_{p} such that y = x^{3}. Proof. Let f(X) = X^{3} − y. Then ||f(x_{0})|| = ||x_{0}^{3} − y|| < 1 while ||f′(x_{0})|| = ||3x_{0}^{2}|| = ||3||||x_{0}^{2}|| = 1 since ||x_{0}||^{3} = ||x_{0}^{3}|| = ||(x_{0}^{3} − y) + y|| = ||y|| = 1 and hence ||x_{0}|| = 1. □

LEMMA. Let p = 3. Then y ∈ Z_{3}^{×} is a cube, i.e., there exists x ∈ Z_{3} such that y = x^{3} if and only if y ≡ ±1 mod 9. Proof. Exercise. □

LEMMA. (X^{2} − 2)(X^{2} − 17)(X^{2} − 34) = 0 has a solution in Q_{p} for every prime p and Q_{∞} = R but does not have a solution in Q. Proof. Clearly there is no solution in Q. OTOH, clearly there is a solution x = \sqrt{2} or \sqrt{17}, for example, in R. Note that 2 ∈ Q_{17}^{×2}. This follows since 2 ∈ Z ↪ Z_{17}, ||2||_{Q_{17}} = 1 and 17|(6^{2} − 2), i.e., “x_{0}”= 6. In which case, there is x ∈ Q_{17}^{×} such that x^{2} = 2, i.e., x^{2} − 2 = 0 and x is a solution in Q_{17}. On the other hand 17 ∈ Q_{2}^{×2}. This follows since 17 ≡ 1 mod 8. Therefore there is a solution in Q_{2}. If p ≠ 2, 17, then at least one, y, of {2, 17, 34} is quadratic residue. This is because the quadratic reciprocity is ‘multiplicative”. In which case, there is x_{0} ∈ Z ↪ Z_{p} such that p|x^{2} − y, i.e., ||x_{0}^{2} − y|| < 1. Since ||y||_{Q_{p}} = 1, it follows from lemma above that there exists x ∈ Z_{p} such that y = x^{2}. x is a solution in Q_{p}. □

2.1 Newton polygons

Let F be a field complete with respect to a non-archimedean absolute value || ||. Let f(X) = f_{0} + f_{1}X + ⋯ + f_{n}X^{n} ∈ F [X]. By computing Newton polygons, one can understand something about the roots of f(X). Assume f_{0} ≠ 0 and f_{n} ≠ 0. Plot in R^{2} the pairs (i, log||f_{i}||) for f_{i} ≠ 0 with “log base” > 1 (so that x ≤ y if and only if log x ≤ log y). Then the Newton polygon of f is most simply described as the upper boundary of the convex cover of the pairs.

Suppose that the (s + 1) pairs (i_{0}, log||f_{0}||), …, (i_{r}, log||f_{i_{r}}||), …, (i_{s}, log||f_{i_{s}}||), with i_{0} = 0, i_{s} = n, and 0 = i_{0} < ⋯ < i_{r} < ⋯ < i_{s} = n form the Newton polygon of f, we let

slope_{r} =^{def} (log||f_{i_{r}}|| − log||f_{i_{r−1}}||)/i_{r} − i_{r−1}

for 1 ≤ r ≤ s. Then slope_{1} > slope_{2} > ⋯ > slope_{s}. In which case, we say *f* is of type (l_{1}, slope_{1}; l_{2}, slope_{2}; , …,; l_{s}, slope_{s}) where l_{1} = i_{1} − i_{0} = i_{1}, l_{2} = i_{2} − i_{1}, …, l_{s} = i_{s} − i_{s−1} = n − l_{s−1}. If s = 1, in which case the Newton polygon is (0, log||f_{0}||)−(n, log||f_{n}||) with a single slope, say *f* is basic.

USEFUL ANALYSIS (VARIATIONS OF ABSOLUTE VALUE): Let F be a field with a nonarchimedean absolute value || ||. Let c > 0 and define, for f = c_{0} + c_{1}X + c_{2}X^{2} + ⋯ + c_{n}X^{n} ∈ F [X], ||f||_{c} = max_{0≤i≤n}||c_{i}||c^{i}. For h = f/g ∈ k(X), define ||h||_{c} = ||f||_{c}/||g||_{c}. Then || ||_{c} define a nonarchimedean absolute value. See Lemma 1.1, Chapter 6, Cassels.

More generally, for f(X_{1}, …, X_{n}) = ∑_{i_{1},...,i_{n}} f_{i_{1},...,i_{n}}X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}} ∈ F [X_{1}, …, X_{n}], define ||f||_{c_{1},...,c_{n}} = max_{i_{1},...,i_{n}}c_{1}^{i_{1}} ⋯ c_{n}^{i_{n}}||f_{i_{1},...,i_{n}}||. Extend it similarly to F (X_{1}, …, X_{n}). It defines a non-archimedean absolute value. Since F (X_{1}, …, X_{i}) = F (X_{1}, …, X_{i−1})(X_{i}), this follows by induction on 1 ≤ i ≤ n.

REMARK. Suppose that the Newton polygon of f ∈ F [X] is the set of points in R^{2}

x_{0} = (i_{0}, log||f_{0}||), …, x_{r} = (i_{r}, log||f_{i_{r}}||), …, x_{s} = (i_{s}, log||f_{i_{s}}||).

Suppose s > slope_{1}. Every point, except x_{0} is below the line of slope s passing through x_{0}:

log||f_{i}|| < s(i − i_{0}) + log||f_{i_{0}}||.

If s = −logc, then ||f_{i}|| < c^{−(i−i_{0})}||f_{i_{0}}||, i.e., ||f_{i}||c^{i} < ||f_{i_{0}}||c^{i_{0}}. Therefore “||f||_{c} is attained precisely at i_{0}, i.e., ||f||_{c} = ||f_{i_{0}}||c^{i_{0}} = ||f_{0}||c^{0}(which is also ||f_{0}||_{c}). Note that ||f(X) − f_{0}||_{c} < ||f_{0}||. So if ||x|| ≤ c, then ||f(x) − f_{0}||_{c} ≤ ||f(c) − f_{0}||_{c} ≤ max_{i≥2}||f_{i}||c^{i} = ||f(X) − f_{0}||_{c} < ||f_{0}||. This implies that f(X) does not have roots of absolute value || || ≤ c. Indeed, if ||x|| ≤ c and f(x) = 0 then ||0 − f_{0}||_{c} = ||f_{0}|| < ||f_{0}|| and contradiction!

Suppose s = slope_{1} and it is of the form −logc. Then every point is either on or below the line of slope s passing through x_{0}:

log||f_{i}|| ≤ s(i − i_{0}) + log||f_{i_{0}}||

where the equality holds for i = i_{0} and i = i_{1} (in which case ||f_{i_{1}}||c^{i_{1}} = ||f_{i_{0}}||c^{i_{0}}), possibly more at i ∈ [i_{0}, i_{1}] with x_{i} on the line while the (strict) inequality holds for every i > i_{1} in which case ||f_{i}||c^{i} < ||f_{i_{0}}||c^{i_{0}} = ||f_{0}|| as in the case above. Also, one can deduce that ||f(X) − ∑_{i∈[i_{0},i_{1}]} f_{i}X^{i}||_{c} < ||f||_{c} (subtracting from f(X) the terms which do and “may” attain the maximum, i.e, the RHS, the || || of the remaining terms, i.e., the LHS is certainly strictly less that the RHS).

Suppose slope_{1} > s > slope_{2} and s is of the form s = −logc. Then for every i > i_{1}, every point is below the line of slope s passing through x_{1}: for i > i_{1},

||f_{i}||c^{i} < ||f_{i_{1}}||c^{i_{1}}.

To sum up, define c_{i} by slope_{i} = −logc_{i} (so that c_{1} < c_{2} < ⋯ < c_{s}). Then there is no root of f with absolute value || || < c_{1}; there are roots with absolute value || || = c_{1}; there is no root with absolute value c_{1} < || || < c_{2} but there are some with || || = c_{2}, and so on. In some sense, the slopes of the Newton polygon of f “aligns” roots (if exist in F) precisely at “radii” corresponding to the slopes.

THEOREM (“Newton”!). Let F be a field complete with respect to a non-archimedean absolute value. Let f(X) ∈ F [X] of type (l_{1}, slope_{1}; l_{2}, slope_{2}; , …,; l_{s}, slope_{s}). Then f(X) = g_{1}(X) … g_{s}(X) where g_{r}(X) ∈ F [X] is basic of type (l_{r}, slope_{r}) for 1 ≤ r ≤ s.

Note that it does NOT assert that the g_{r} are irreducible.

REMARK. Suppose that the roots of f are in F. Suppose that f = f_{0} + f_{1}X + ⋯ + f_{n}X_{n} ∈ F [X], f_{0} ≠ 0, f_{n} ≠ 0 is basic of slope s. The every root of f have the absolute value || || = c where s = −logc. Indeed, as above, no root of f has absolute value < c, i.e., every root has || || ≥ c (∗). On the other hand, since f has slope s, log||f_{n}|| = sn + log||f_{0}||, i.e., ||f_{n}|| = c^{−n}||f_{0}||, i.e., ||f_{0}||/||f_{n}|| = c^{n} (∗∗). If we let x_{1}, …, x_{n} be the roots of f, then f(X) = C(X − x_{1}) ⋯ (X − x_{n}) for some C ∈ F, and f_{n} = C and f_{0} = (−1)^{n}(x_{1} … x_{n})C. Then ||f_{0}||/||f_{n}|| = ||x_{1} ⋯ x_{n}|| (∗ ∗ ∗). Combining (∗), (∗∗), (∗ ∗ ∗), ||x_{i}|| = c for 1 ≤ i ≤ n as required.

LEMMA. Suppose that f = f_{0} + f_{1}X + ⋯ + f_{n}X^{n}, g = g_{0} + g_{1}X + ⋯ + g_{m}X^{m} ∈ F [X] are basic of the same slope s. Then fg is basic of slope s.

Proof. Let c such that −logc = s. By the remark above the theorem, ||f||_{c} = ||f_{0}||_{c} = ||f_{n}X^{n}||_{c} and ||g||_{c} = ||g_{0}||_{c} = ||g_{m}X^{m}||_{c}. Therefore ||fg||_{c} = ||f_{0}g_{0}||_{c} = ||f_{n}g_{m}X^{n+m}||_{c}, i.e., ||fg||_{c} = ||(fg)_{0}||_{c} = ||(fg)_{n+m}X^{n+m}||_{c} which implies that fg is basic. □

LEMMA. Suppose that f is of type (l(f)_{1}, slope(f)_{1}; l(f)_{2}, slope(f)_{2}; , …,; l(f)_{s}, slope(f)_{s}) and g is pure of type (l(g)_{1}, slope(g)_{1}). Suppose moreover that slope(f)_{s} > slope(g)_{1}). Then fg is of type

(l(f)_{1}, slope(f)_{1}; l(f)_{2}, slope(f)_{2}; , …,; l(f)_{s}, slope(f)_{s}; l(g)_{1}, slope(g)_{1}).

Proof. Let c > 0 be such that −logc = slope(f)_{r} for some 1 ≤ r ≤ s. Then since slope(f)_{r} > slope(g)_{1}, ||g(X) − g_{0}||_{c} < ||g||_{c}. On the other hand, ||f(X) − ∑_{i∈[i_{r−1},i_{r}]} f_{i}X^{i}||_{c} < ||f||_{c} and therefore ||f(X)g(X) − g_{0} ∑_{i∈[i_{r−1},i_{r}]} f_{i}X^{i}||_{c} < ||fg||_{c}. Let d > 0 be such that − log d = slope(g)_{1}. Then similarly ||f(X)g(X) − f_{n}X^{n}g(X)||_{d} < ||fg||_{d}. These inequalities determine the type of fg. □

LEMMA. If f is irreducible, then f is basic.

SUBLEMMA. Let || ||_{c} for some c. Let f(X) = f_{0} + f_{1}X + ⋯ + f_{n}X^{n} ∈ F [X] and suppose g(X) = g_{0} + g_{1}X + ⋯ + g_{m}X^{m} has ||g_{m}X^{m}||_{c} = ||g||_{c}. Let Q(X), R(X) ∈ F [X] be defined by f(X) = Q(X)g(X) + R(X) where degR(X) < m. Then ||Q||||g|| ≤ ||f|| and ||R|| ≤ ||f||.

Proof. The degree of Q is n − m. Let Q(X) = Q_{0} + Q_{1}X + ⋯ + Q_{n−m}X^{n−m}. The coefficients of Q(X) are determined by comparing LHS and RHS of f = Qg + R:

the coefficients of X^{n}: f_{n} = Q_{n−m}g_{m};

the coefficients of X^{n−1}: f_{n−1} = Q_{n−m−1}g_{m} + Q_{n−m}g_{m−1};

the coefficients of X^{n−2}: f_{n−2} = Q_{n−m−2}g_{m} + Q_{n−m−1}g_{m−1} + Q_{n−m}g_{m−2};

...

the coefficients of X^{n−j}: f_{n−j} = Q_{n−m−j}g_{m} + Q_{n−m−(j−1)}g_{m−1} + ⋯ + Q_{n−m}g_{m−j};

...

the coefficients of X^{n−(n−m)} = X^{m}: f_{n−(n−m)}(= f_{m}) = Q_{0}g_{m} + Q_{1}g_{m−1} + ⋯ + Q_{n−m}g_{0}.

By induction \underline{||Q_{n−m−j}||||g|| ≤ ||f||} for 0 ≤ j ≤ n − m: indeed,

j = 0: ||Q_{n−m}X^{n−m}||||g|| = ||Q_{n−m}X^{n−m}||||g_{m}X^{m}|| = ||Q_{n−m}g_{m}X^{n}|| = ||f_{n}X^{n}|| ≤ ||f||; j = 1: ||Q_{n−m−1}X^{n−m−1}||||g|| = ||Q_{n−m−1}X^{n−m−1}||||g_{m}X^{m}|| = ||Q_{n−m−1}g_{m}X^{n−1}|| = ||(f_{n−1} −

Q_{n−m}g_{m−1})X^{n−1}||. This is ≤ ||f|| because ||f_{n−1}X^{n−1}|| ≤ ||f||, and by the case j = 0 above, ||Q_{n−m}g_{m−1}X^{n−1}|| = ||Q_{n−m}X^{n−m}||||g_{m−1}X^{m−1}|| ≤ ||Q_{n−m}X^{n−m}||||g|| ≤ ||f||;

j = 2: ||Q_{n−m−2}X^{n−m−2}||||g|| = ||Q_{n−m−2}X^{n−m−2}||||g_{m}|| = ||Q_{n−m−2}g_{m}X^{n−2}|| = ||(f_{n−2} − Q_{n−m−1}g_{m−1}−Q_{n−m}g_{m−2})X^{n−2}||. Again this is ≤ ||f|| because ||f_{n−2}|| ≤ ||f||, ||Q_{n−m−1}g_{m−1}X^{n−2}|| = ||Q_{n−m−1}X^{n−m−1}||||g_{m−1}X^{m−1}|| ≤ ||Q_{n−m−1}X^{n−m−1}||||g|| ≤ ||f|| by j = 1, and ||Q_{n−m}g_{m−2}X^{n−2}|| = ||Q_{n−m}X^{n−m}||||g_{m−2}X^{m−2}|| ≤ ||Q_{n−m}X^{n−m}||||g|| ≤ ||f|| by j = 0.

Do this similarly, ||Q_{n−m−j}||||g|| ≤ ||f|| for 0 ≤ j ≤ n − m. Then ||Q||||g|| ≤ ||f|| and ||R|| = ||f − Qg|| ≤ max{||f||, ||Q||||g||} ≤ ||f|| since both ≤ ||f||. □

SUBLEMMA. Let || || = || ||_{c} for some c. Let f(X) = f_{0} + f_{1}X + ⋯ + f_{n}X^{n}. Suppose that there exists N with 0 < N < n such that ||f_{N}X^{N}||_{c} = ||f|| and ||f_{i}X^{i}|| < ||f|| for every i > N. Then f = gh for g, h ∈ F [X] and degg = N and degg = n − N.

Proof. ||f(X)−∑_{i=0}^{N} f_{i}X^{i}|| = C||f|| for some C < 1. We shall construct two sequences of polynomials G_{0}, G_{1}, … and H_{0}, H_{1}, …, in F [X] of degree N and of degree ≤ n − N respectively which eventually “converge” respectively to g and h we are looking for. Inductively define them as follows: suppose we have G_{j} ∈ F [X] of degree N and H_{j} ∈ F [X] of degree ≤ n − N satisfying ||f − G_{j}|| ≤ C||f|| and ||H_{j} − 1|| ≤ C. Define D_{j} by ||f − G_{j}H_{j}|| = D_{j}||f||.

G_{0} = ∑_{i=0}^{N} f_{i}X^{i}, H_{0} = 1 and D_{0} = C satisfy these.

\underline{D_{j} ≤ C}: D_{j}||f|| = ||f − G_{j}H_{j}|| = ||f − G_{j} + G_{j}(1 − H_{j})|| ≤ C||f||. This follows since ||f − G_{j}|| ≤ C||f|| and ||G_{j}(1 − H_{j})|| = ||G_{j}||||1 − H_{j}|| = ||f||||1 − H_{j}|| ≤ ||f||C where ||G_{j}|| = ||f|| follows since ||G_{j}|| = ||(G_{j} − f) + f|| = ||f|| as ||G_{j} − f|| ≤ C||f|| < ||f||.

Applying the preceding sublemma, with “f”=f − G_{j}H_{j} and “g”=G_{j}, we get Q_{j}(X), R_{j}(X) ∈ F [X] such that f − G_{j}H_{j} = Q_{j}G_{j} + R_{j}, degQ_{j} ≤ n − N, degR_{j} < N, ||Q_{j}|| ≤ D_{j} and ||R_{j}|| ≤ D_{j}||f||. Put G_{j+1} = G_{j} + R_{j} and H_{j+1} = H_{j} + Q_{j}.

\underline{D_{j+1} ≤ CD_{j} < D_{j}}: firstly, by definition, f −G_{j+1}H_{j+1} = R_{j}(1−H_{j})−R_{j}Q_{j}. Therefore D_{j+1}||f|| =^{def}

||f − G_{j+1}H_{j+1}|| = ||(H_{j} − 1)R_{j} + R_{j}Q_{j}|| ≤ max{||H_{j} − 1||||R_{j}||, ||R_{j}||||Q_{j}||} ≤ CD_{j}||f|| because ||H_{j} − 1||||R_{j}|| ≤ CD_{j}||f|| and ||R_{j}||||Q_{j}|| ≤ D_{j}D_{j}||f|| ≤ CD_{j}||f||.

As j → ∞, the G_{j} and H_{j} converge to give g, h ∈ F [X] such that f = gh. □

Proof of the theorem. Suppose that f is the product of irreducible polynomials f_{i}, i ∈ I. Then f_{i} is pure. It then follows that f is the product of g_{j}, j ∈ J with g_{j} pure of type (l_{j}, s_{j}) and s_{1} > s_{2} > …, > s_{n}. By induction, ∏_{j∈J} g_{j} is then of type (l_{1}, s_{1}; …,; l_{#J}, s_{#J}). Which is clearly the type of f.

THEOREM (“Weierstrass”) Weierstrass preparation theorem. Let f(x) = ∑_{n=0}^{∞} f_{n}x^{n} where f_{n} → ∞ as n → ∞. Define N by ||f_{N}|| = max_{n}||f_{n}||, i.e., ||f_{n}|| < ||f_{N}|| for all n > N. Then there is a polynomial g(x) = g_{0} + g_{1}x + ⋯ + g_{N}x^{N} and a power series h(x) = 1 + h_{1}x + h_{2}x^{2} ⋯ with h_{n} → ∞ such that f = gh. Furthermore ||g_{N}|| = max||g_{n}|| and ||h_{n}|| < 1 for all n > 0. Proof. See Theorem 5, Chapter 6 of Cassels. □