Exercise 14, § 2

(a) Let $A$ be a subring of the field $K$ and $x$ an element $\neq 0$ of $K$. Show that, if $x$ is not integral over $A$, there exists a maximal ideal of $A[x^{-1}]$ which contains $x^{-1}$ and that, for every maximal ideal $m$ of $A[x^{-1}]$ containing $x$, the ideal $A \cap m$ is maximal in $A$ (observe that $x$ does not belong to the ring $A[x^{-1}]$).

(b) Let $A$ be a local integral domain, $m$ its maximal ideal, $K$ a field containing $A$ and $k$ the residue field of $A$. Let $x \in K$ be such that $x \notin A$ and suppose that $A$ is integrally closed in the ring $A[x]$. Show that there exists an $A$-algebra homomorphism $A[x] \to k[X]$ which extends the canonical homomorphism $A \to k$ and which maps $x$ to $X$. (It is necessary to prove that, if $\sum_{i=0}^n a_i x^i = 0$, where $a_i \in A$ for all $i$, then all the $a_i$ belong to $m$. Otherwise there would be a relation $a_n x^{n-p} + \ldots + a_p = -(a_{p+1} x^{-1} + \ldots + a_0 x^{-p})$, where $a_p$ is invertible in $A$; using Exercise 5 of § 1, show that the common value $b$ of the two sides belongs to $A$ and, using (a), show that necessarily $b \in m$; conclude that $x^{-1}$ would be integral over $A$, which is absurd.)

(c) Under the hypotheses of (b), show that the ideal $mA[x]$ of $A[x]$ is prime and not maximal.