Lemma 59.81.1. Let A be a ring. Let a, b \in A such that aA + bA = A and a \bmod bA is a root of unity. Then there exists a monogenic extension A \subset B and an element y \in B such that u = a - by is a unit.
Proof. Say a^ n \equiv 1 \bmod bA. In particular a^ i is a unit modulo b^ mA for all i, m \geq 1. We claim there exist a_1, \ldots , a_ n \in A such that
Namely, since 1 - a^ n \in bA we can find an element a_1 \in A such that 1 - a^ n - a_1 a^{n - 1} b \in b^2A using the unit property of a^{n - 1} modulo bA. Next, we can find an element a_2 \in A such that 1 - a^ n - a_1 a^{n - 1} b - a_2 a^{n - 2} b^2 \in b^3A. And so on. Eventually we find a_1, \ldots , a_{n - 1} \in A such that 1 - (a^ n + a_1 a^{n - 1}b + a_2 a^{n - 2}b^2 + \ldots + a_{n - 1} ab^{n - 1}) \in b^ nA. This allows us to find a_ n \in A such that the displayed equality holds.
With a_1, \ldots , a_ n as above we claim that setting
works. Namely, suppose that \mathfrak q \subset B is a prime ideal lying over \mathfrak p \subset A. To get a contradiction assume u = a - by is in \mathfrak q. If b \in \mathfrak p then a \not\in \mathfrak p as aA + bA = A and hence u is not in \mathfrak q. Thus we may assume b \not\in \mathfrak p, i.e., b \not\in \mathfrak q. This implies that y \bmod \mathfrak q is equal to a/b \bmod \mathfrak q. However, then we obtain
a contradiction. This finishes the proof. \square
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