Exercise 6, § 8

Let K be a field, v a valuation on K and A the ring of v.

(a) Suppose that A is Henselian (Chapter III, § 4, Exercise 3) in which case we also say, by an abuse of language, that K is Henselian for v. Then, for every algebraic extension L of K and every valuation v' on L extending v, the ring A' of v' is Henselian (Chapter III, § 4, Exercise 4).

(b) If K is complete with v and v is of height 1, A is Henselian. Give an example where K is complete with v and v is of height 2, but A is not Henselian (cf. Exercise 1).

(c) If K is linearly compact with v, show that K is Henselian. (In the notation condition (H) of Chapter III, § 4, Exercise 3, let $\mathscr{L}$ denote the set of prime ideals of A (totally) ordered by inclusion and $\mathfrak{L}$ the set of ordered pairs ($\mathfrak{B}_+$), where $\mathfrak{B}$ is a well ordered subset (with respect to the relation $\supset$) of $\mathscr{L}$ and $\phi : p \mapsto (Q_p, Q'_p)$ a mapping from $\mathfrak{B}$ to the set $A[X] \times A[X]$, with the following properties: (1) Q, and $Q'_p$ are monic of respective degrees $\deg(\overline{Q})$ and $\deg(\overline{Q'})$; (2) if p $\supset q$ are two elements of $\mathfrak{B}$, the coefficients of the polynomials $Q - Q_p$ and $Q'_p - Q'_a$ belong to p; (3) the coefficientsof $P - Q_p Q'_p$ belong to p; (4) $\tilde{f}(Q_p) = \overline{Q}$, $\tilde{f}(Q'_p) = \overline{Q}'$. Define on $\mathscr{L}$ an ordering by setting $(\mathfrak{B}, \phi) \leq (\mathfrak{B}', +')$ when $\mathfrak{B} \subset \mathfrak{B}'$ and $\phi'$ is an extension of $\phi$ to $\mathfrak{B}'$.Prove that $\mathscr{L}$ admits a maximal element $(\mathfrak{B}_0, \phi_0)$ and that $\mathfrak{B}_0$ has a last element equal to (0): obtain a contradiction from the opposite by considering two cases, according to whether or not $\mathfrak{B}_0$ has a last element; in the former case, if p is this last element, consider an element c $\in A$ such that $v(c)$ is the least value taken by v on the set of coefficientsof $P - Q_p Q'_p$ and the least prime ideal p' of A containing c; if $p' = p$, use Exercise 2 (a) of § 4 and the fact that a quotient of a linearly compact module by a closed submodule is linearly compact and hence complete. If on the contrary $\mathfrak{B}_0$ has no last element, use directly the hypothesis that A is linearly compact.)

(d) In the notation of Exercise 5, show that for $A$ to be Henselian, it is necessary and sufficient that $A'$ and $A$ be so. (Note that, if $P \in A'[X]$, there exists $a \in A$ such that $a^n P(X/a) \in A[X]$ (where $n$ is the degree of $P$); to verify axiom (H) of Chapter III, § 4, Exercise 3 for $A'$, attention may be confined to the polynomials of $A[X]$; use the fact that $A' = A_p$, where $p$ is a prime ideal of $A$ contained in the maximal ideal $m$, and note that, if two polynomials in $(A/p)[X]$ are strongly relatively prime, so are their images in $(A/m)[X]$.)

(e) Suppose that $A$ is Henselian; then, for every algebraic extension $L$ of $K$, there exists (up to equivalence) only one valuation on $L$ extending $v$. (If $P \in A[X]$ is a monic irreducible polynomial, $x_i$ ($1 \leq i \leq m$) the distinct roots of $P$ in its splitting field $N$, show that, for every valuation $v'$ on $N$ extending $v$, the $v'(x_i)$ are all equal for $1 \leq i \leq m$.) Conversely, if the valuation ring $A$ has this property, it is Henselian (observe that, in the integral closure $A'$ of $A$ in $L$, there can only exist one maximal ideal lying above $m$ (Chapter V, § 1, Exercise 13) and conclude that, if $P \in A[X]$ is a monic irreducible polynomial its image in $(A/m)[X]$ cannot be a product of two relatively prime polynomials).