Exercise 7, § 1

Let R be a ring. An ordered pair (A, p) consisting of a subring A of R and a prime ideal p of A is called maximal in R if there exists no ordered pair (A', p') consisting of a subring A' of R and a prime ideal p' of A' such that $A \subset A', A' \neq A$ and $p' \cap A = p$.

(a) For every subring B of R and every prime ideal q of B there exists an ordered pair (A, p), maximal in R, such that $A \supset B$ and $q = p \cap B$.

(b) For an ordered pair (A, p) to be maximal in R, it is necessary and sufficient that, for all $s \in R - A$, there exist a finite family of elements $c_i \in p$ ($1 \leq i \leq n$) such that the element $b = c_1 s + c_2 s^2 + \ldots + c_n s^n$ belongs to $A - p$ (use Chapter 11, § 2, no. 5, Corollary 3 to Proposition 11).

(c) Show that, if the ordered pair (A, p) is maximal in R, the product of two elements of $R - A$ belongs to $R - A$. (Ifs, s' are in $R - A$ and $ss' = a \in A$, consider the least integers $n, m$ such that there exists $c_i \in p$ ($1 \leq i \leq n$) for which $b = \sum_i c_i s^i \in A - p$ and $c'_j \in p$ ($1 \leq j \leq m$) for which

supposing for example $n \geq m$, consider the product $bb' \in A - p$ and obtain a contradiction from the definition of the integer $n$.)

(d) Under the hypotheses of (c), we write $S = R - A$; show that the ideal $p_1$ of A consisting of the $a \in A$ such that there exists $s \in S$ for which $sa \in A$ is contained in p (use (b)). A fortiori the ideals $p_0$ of A consisting of the $a \in A$ such that $sa \in A$ for all $s \in S$ is contained in p (Exercise 6 (b)). We write $A' = A/p_0, p' = p/p_0$ and $R' = R/p_0$; the ordered pair (A', p') is then maximal in the integral domain $R$ and, for all non-zero $a' \in A'$, there exists $s' \in S' = R' - A'$ such that $s'a' \in S'$.

(e) Let $(A, p)$ be a maximal ordered pair in the integral domain $R$ such that, for all non-zero $a \in A$, there exists $s \in S = R - A$ such that $sa \in S$. We write $S_0 = S \cup \{1\}$ and denote by $R$, the ring of fractions $S_0^{-1}R$; the canonical homomorphism $R \to R$, is injective and therefore identifies $R$ with a subring of $R_0$. Show that $R$, is a field identified with the field of fractions of $R$. Let $(B, q)$ be a maximal ordered pair in $R_0$ such that $B \supseteq A$ and $q \cap A = p$. Show that $B \cap R = A$; in other words, $A$ is the intersection of $R$ and a valuation ring of $R$, and $p$ is the intersection of $A$ and the maximal ideal of this valuation ring.

(f) Let $K_0$ be a field, $A$, the polynomial ring $K_0[X]$, $B$ the ring of formal power series $A_0[[Y]]$ and $R$ the ring of fractions $B_Y$ consisting of the formal power series $Y^{-h}f(Y)$, where $f \in B$ and $h \geq 0$. Let $P$ be the prime ideal $YB$ of formal power series without constant term, $A = K_0 + p$, which is a subring of $B$. Show that $A$ is the intersection of $R$ and a valuation ring $C$ of the field of fractions $K$ of $R$ and $p$ the intersection of $A$ and the maximal ideal of $C$; moreover, the field of fractions of $A$ is equal to that of $R$ and every $a \neq 0$ in $A$ is of the form $s^{-1}s'$, where $s, s'$ are in $S = R - A$. However, the ordered pair $(A, p)$ is not maximal in $R$.