Bourbaki

Exercise 1, § 6

Commutative Algebra, VI, 459

Bourbaki marks this one as harder than the rest.

Every element z of the p-adic field Qp\mathbf{Q}_p (pp a prime number) may be written uniquely as ph∑k=0∞ckpkp^h \sum_{k=0}^{\infty} c_k p^k, where h∈Zh \in \mathbf{Z}, c0≠0c_0 \neq 0, 0≤ck<p0 \leq c_k < p for all k≥0k \geq 0 ("p-adic expansion" of z).

(a) Show that, in order that z∈Qz \in \mathbf{Q}, it is necessary and sufficient that there exist two integers m≥0, n≥1m \geq 0,\ n \geq 1 such that ck+n=ckc_{k+n} = c_k for all k≥mk \geq m. (Observe that, if z=a/b∈Qz = a/b \in \mathbf{Q} where bb is not divisible by pp, then

a−b∑k=0nckpk=an+1pn+1 a - b \sum_{k=0}^n c_k p^k = a_{n+1} p^{n+1}

where an+1a_{n+1} is an integer and the sequence ∣ak∣|a_k| (ordinary absolute value) is bounded.)

(b) Suppose that the increasing sequence (kn)(k_n) of integers kk such that ck≠0c_k \neq 0 is such that lim sup⁡n→m(kn+1/kn)=+∞\limsup_{n \to m} (k_{n+1}/k_n) = +\infty. Show that zz is transcendental over Q\mathbf{Q}. (For x∈Qx \in \mathbf{Q}, let ∣x∣|x| and ∣x∣p|x|_p respectively denote the usual absolute value and a pp-adic absolute value. Suppose that P∈Z[X]P \in \mathbf{Z}[X] is an irreducible polynomial such that P(z)=0P(z) = 0; if zn=pn∑k=0∞ckpkz_n = p^n \sum_{k=0}^\infty c_k p^k, show that ∣P(zn)/(z−zn)∣p|P(z_n)/(z - z_n)|_p tends to a limit ≠0\neq 0 as nn tends to +∞+\infty, using Taylor's formula; then obtain a contradiction from the hypothesis considering the usual absolute values ∣P(zn)∣|P(z_n)|.)

Solution

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Transcribed from the printed volume. Not translated.