# A 2-dimensional valuation domain

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 $$S/I_r$$.

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