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
\[ 1 = a^ n + a_1 a^{n - 1}b + a_2 a^{n - 2}b^2 + \ldots + a_ n b^ n \]
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
\[ B = A[y]/(y^ n + a_1 y^{n - 1} + a_2 y^{n - 2} + \ldots + a_ n) \]
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
\[ 0 = y^ n + a_1 y^{n - 1} + a_2 y^{n - 2} + \ldots + a_ n = b^{-n}(a^ n + a_1 a^{n - 1}b + a_2 a^{n - 2}b^2 + \ldots + a_ nb^ n) = b^{-n} \]
a contradiction. This finishes the proof. $\square$
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