Let $p$ be a prime number, $x$ be an indeterminate over $\mathbb{Q}$, and set $A:= \mathbb{Z}_{(p)}$ (the localization of $\mathbb{Z}$ at $p$), $B:=\mathbb{Q}[[x]]$ (power series over $\mathbb{Q}$). How can we show that the ring $S:= A + xB$ is a 2-dimensional valuation domain whose maximal ideal $N= p\mathbb{Z}_{(p)}+xB$ is principal?


The ring $S$ is the subring of $\mathbb Q[[x]]$ consisting of

$S=\lbrace \sum_{i \ge 0} a_i x^i: a_0 \in \mathbb Z_{(p)} \rbrace.$

It's obviously an integral domain.

The ideal

$M:= p \mathbb Z_{(p)}+x\mathbb Q[[x]] = \lbrace \sum_{i \ge 0} a_i x^i: a_0 \in \mathbb Z_{(p)}, p\mid a_0 \rbrace \subset S$

is the principal ideal $pS$. Namely, $pS \subseteq M$ is clear, for the other inclusion note that for a general $y=\sum_{i \ge 0} a_i x^i \in \mathbb Q[[x]]$ we can write

$$xy= x\sum_{i \ge 0} a_i x^i = p \cdot (\sum_{i \ge 0}\frac1p a_i x^{i+1})= p \cdot (\sum_{i \ge 1}\frac1p a_{i-1} x^{i}) \in pS,$$ hence $pS$ contains $x\mathbb Q[[x]]$ and a fortiori $p \mathbb Z_{(p)}+x\mathbb Q[[x]]$.

I claim that it's easily checked that

$$v(\sum_{i \ge 0} a_i x^i) := (j :=\min\lbrace i: a_i \neq 0\rbrace, v_p(a_j))$$

(where the first component is often called the order of a power series, while $v_p$ is the $p$-adic valuation) defines a two-dimensional valuation $v: S\setminus \lbrace 0 \rbrace \rightarrow \mathbb Z_{\ge 0} \oplus \mathbb Z$, where the latter is given the lexicographic order.

Note in particular that $M$ consists of $0$ and those elements $s \in S$ with $v(s) > (0,0)$ or equivalently $v(s) \ge (0,1)$, and once we checked $v$ is a valuation, this must be the unique maximal ideal of $S$. Its quotient is $S/pS \simeq \mathbb F_p$.

Note that we have $$M^n = \lbrace 0 \rbrace \cup \lbrace s: v(s) \ge (0, n)\rbrace,$$ with quotients $S/M^n \simeq \mathbb Z_{(p)}/p^n \mathbb Z_{(p)} \simeq \mathbb Z/ p^n $, and underneath the infinite chain of $M^n$'s there lies the prime ideal $$I := x\mathbb Q[[x]]=\lbrace 0 \rbrace \cup \lbrace s: v(s) \in (\mathbb Z_{\ge 1}, \mathbb Z)\rbrace$$ with quotient $S/I \simeq \mathbb Z_{(p)}$. The further interesting ideals $$I_r := \lbrace 0 \rbrace \cup \lbrace s: v(s) \in (\mathbb Z_{\ge r}, \mathbb Z)\rbrace$$

are not prime for $r \ge 2$, as the image of $x$ is a zero-divisor in $S/I_r$.

  • $\begingroup$ A remark: the prime ideal you mentioned is xQ[[x]]. $\endgroup$ – user26857 Jul 12 '19 at 6:43

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