4. Lifting in complete rings

1. Strongly relatively prime polynomials

Let $R$ be a commutative ring. Two elements $x, y$ of $R$ are called strongly relatively prime if the principal ideals $Rx$ and $Ry$ are relatively prime, in other words (Chapter II, § 1, no. 2) if $R x + R y = R$; it amounts to the same to say that there exist two elements $a, b$ of $R$ such that $ax + by = 1$.

Lemma 1 ("Euclid's Lemma"). Let $x, y$ be two strongly relatively prime elements of $R$; if $z \in R$ is such that $x$ divides $yz$, then $x$ divides $z$.

If $1 = ax + by$, then $z = x(az) + (yz)b$.

If $x$ and $y$ are strongly relatively prime in $R$, then

For polynomial rings there is the following result:

2. Restricted formal power series

Note that in such a ring, the ideals $3 \in \mathscr{B}$ are open and closed (General Topology, Chapter 111, § 2, no. 1, Corollary to Proposition 4 ). For all $3 \in \mathscr{B}$, the quotient topological ring $A/\mathfrak{J}$ is then discrete; for $8 \in \mathscr{B}$, $8' \in \mathscr{B}$, $\mathfrak{J}' \subset 8$, let

If $A$ is linearly topologized, the restricted formal power series in $A[[X_1, \ldots, X_p]]$ form a subring of $A[[X_1, \ldots, X_p]]$, denoted by $A\{X_1, \ldots, X_p\}$: for if

then $c''_{n_1 \ldots n_p} = \sum c_{r_1 \ldots r_p} c'_{s_1 \ldots s_p}$ for all systems $(r_k), (s_k)$ such that $r_k + s_k = n_k$ for $1 \leq k \leq p$; we conclude that if $n_k \geq 2m$, then $r_k \geq m$ or $s_k \geq m$ and hence, since $\mathfrak{z}$ is an ideal, $c''_{n_1 \ldots n_p} \in \mathfrak{g}$ so long as $n_k \geq 2m$ for at least one $k$, which establishes our assertion. Moreover, every derivative $\partial T / \partial X_i$ ($1 \leq i \leq p$) of a restricted formal power series is restricted, as follows immediately from the definition and the fact that the neighbourhoods $\mathfrak{z} \in \mathscr{B}$ are additive subgroups of $\mathbf{A}$.

If $\mathbf{A}$ is discrete, the ring of restricted formal power series is just the polynomial ring $\mathbf{A}[X_1, \ldots, X_n]$.

Let us always assume that $\mathbf{A}$ is linearly topologized and let $\mathscr{B}$ be a fundamental system of neighbourhoods of 0 in $\mathbf{A}$ consisting of ideals of $\mathbf{A}$; for all $\mathfrak{g} \in \mathscr{B}$, let $p_\mathfrak{g}: \mathbf{A} \to \mathbf{A}/\mathfrak{g}$ be the canonical homomorphism. By definition, for every restricted formal power series $T \in \mathbf{A}\{X_1, \ldots, X_n\}$,

Clearly

is an inverse system of rings (relative to the directed indexing set $\mathscr{B}$) and $(\bar{p}_\mathfrak{g})$ is an inverse system of homomorphisms $\mathbf{A}\{X_1, \ldots, X_n\} \to (\mathbf{A}/\mathfrak{g})[X_1, \ldots, X_p]$; as every polynomial is a restricted formal power series, $\bar{p}_\mathfrak{g}$ is surjective; its kernel $N_\mathfrak{g}$ is the ideal of $\mathbf{A}\{X_1, \ldots, X_n\}$ consisting of the restricted formal power series all of whose coefficients belong to $\mathfrak{g}$; we shall give $\mathbf{A}\{X_1, \ldots, X_n\}$ the (linear) topology for which the $N_\mathfrak{g}$ (for $\mathfrak{g} \in \mathscr{B}$) form a fundamental system of neighbourhoods of 0 (a topology which obviously depends only on that on $\mathbf{A}$). Then it follows from General Topology, Chapter III, § 7, no. 3, Proposition 2 that

is a strict morphism whose kernel is the closure of $\{0\}$ in $\mathbf{A}\{X_1, \ldots, X_n\}$ and whose image is dense in

3. Hensel's lemma

In a topological ring $A$, an element $x$ is called topologically nilpotent if $O$ is a limit of the sequence $(x^n)_{n \geq 0}$. If $A$ is a linearly topologized commutative ring, to say that $x$ is topologically nilpotent means that for every open ideal $\mathfrak{J}$ of $A$ the canonical image of $x$ in $A/\mathfrak{J}$ is a nilpotent element of that ring. If $\mathfrak{r}_g$ is the nil-radical of $A/\mathfrak{J}$, clearly $(\mathfrak{r}_g)$ is an inverse system of subsets and the set $t$ of topological nilpotent elements of $A$ is the inverse image of $\mathfrak{r} = \lim \mathfrak{r}_g$ under the canonical homomorphism $A \to \lim A/\mathfrak{J}$; it is therefore a closed ideal of $A$. If also $A$ is Hausdorff and complete, this ideal is contained in the Jacobson radical of $A$ and, for an element $x \in A$ to be invertible, it is necessary and sufficient that its class mod. $t$ be invertible in $A/t$ (§ 2, no. 13, Lemma 3).

Note that if $A$ is a ring and $m$ a two-sided ideal of $A$, the elements of $m$ are topologically nilpotent with respect to the m-adic topology.

Theorem 1 (Hensel). Let $A$ be a complete Hausdorff linearly topologized commutative ring. Let $m$ be a closed ideal of $A$ whose elements are topologically nilpotent. Let $B = A/m$ be the quotient topological ring and $\varphi : A \to B$ the canonical mapping. Let $R$ be a restricted formal power series in $A\{X\}$, $\overline{P}$ a monic polynomial in $B[X]$ and $\overline{Q}$ a restricted formal power series in $B\{X\}$. Suppose that $\overline{\varphi}(R) = \overline{P} \cdot \overline{Q}$ and that $\overline{P}$ and $\overline{Q}$ are strongly relatively prime in $B\{X\}$. Then there exists a unique ordered pair $(P, Q)$ consisting of a monic polynomial $P \in A[X]$ and a restricted formal power series $Q \in A\{X\}$ such that (4)

The proof is divided into four steps. In the first three we assume that $A$ is discrete, in which case $R$ and $\overline{Q}$ are polynomials.

(I) $m^2 = 0$ Let $S, T$ be two polynomials of $A[X]$ such that $S$ is monic and $\bar{\varphi}(S) = \overline{P}$, $\bar{\varphi}(T) = \overline{Q}$; Proposition 2 of no. 1 shows that $S$ and $T$ are strongly relatively prime; hence (no. 1, Proposition 1) there exists a unique ordered pair of polynomials $(S', T')$ of $A[X]$ such that (5)

(2) $m$ is nilpotent Let $n$ be the smallest integer such that $m^n = 0$ and let us argue by induction on $n > 2$, the theorem having been shown for $n = 2$. Let $A = A/m^{n-1}$, $m' = m/m^{n-1}$; as ${m'}^{n-1} = 0$, there exists a unique ordered pair $(P', Q')$ of polynomials in $A'[X]$ such that $P'$ is monic, $R' = P'Q'$, $\psi(P') = \overline{P}$ and $\psi(Q') = \overline{Q}$, where $\psi$ denotes the canonical homomorphism $A' \to A'/m' = B$, $\theta$ the canonical homomorphism $A \to A'$ and $R' = \tilde{\theta}(R)$. On the other hand, as $(m^{n-1})^2 = 0$, there exists a unique ordered pair $(P, Q)$ of polynomials in $A[X]$ such that $P$ is monic and $R = PQ$, $\bar{\theta}(P) = \overline{P}$, $\bar{\theta}(Q) = \overline{Q}$; as $\phi = \psi \circ \theta$, this shows the existence and uniqueness of $P$ and $Q$ satisfying (4); moreover $P'$ and $Q'$ are strongly relatively prime by the induction hypothesis and hence so are $P$ and $Q$.

(3) A is discrete

Note that in this case $m$ is no longer necessarily nilpotent, but it is always a nilideal by hypothesis. Let $P_0$, $Q_0$ be two polynomials of $A[X]$ such that $\bar{\varphi}(P_0) = \overline{P}$, $\bar{\varphi}(Q_0) = \overline{Q}$ and $P_0$ is monic. Let us consider the ideal $n$ of $A$ generated by the coefficients of $R - P_0 Q_0$; it is finitely generated and contained in $m$, hence it is nilpotent (Chapter 11, § 2, no. 6, Proposition 15) and by definition, if $\psi : A \to A/n$ is the canonical mapping, then $\bar{\psi}(R) = \bar{\psi}(P_0) \bar{\psi}(Q_0)$. Moreover, $\bar{\psi}(P_0)$ and $\bar{\psi}(Q_0)$ are strongly relatively prime, as follows from the hypothesis on $\overline{P}$ and $\overline{Q}$ and Proposition 2 of no. 1 applied to the canonical homomorphism $A/n \to A/m$. By virtue of case (2), there therefore exists an ordered pair $(P, Q)$ of polynomials in $A[X]$ such that $P$ is monic and relations (4) hold. The fact that $\overline{P}$ and $\overline{Q}$ are strongly relatively prime implies also here that $P$ and $Q$ are strongly relatively prime in $A[X]$ by virtue of no. 1, Proposition 2, for $m$ is contained in the Jacobson radical of $A$. Suppose finally that $P_1$, $Q_1$ are two polynomials in $A[X]$ satisfying (4) and such that $P_1$ is monic and let $n_1$ be the finitely generated ideal of $A$ generated by the coefficients of $P - P_1$ and the coefficientsof $Q - Q_1$; as $n_1$ is contained in $m$, it is nilpotent and, if $\psi_1 : A \to A/n_1$ is the canonical mapping, $\bar{\psi}_1(P) = \bar{\psi}_1(P_1)$ and

(4) General case

Let $\mathscr{B}$ be a fundamental system of neighbourhoods of 0 in $A$ consisting of ideals of $A$. For all $\mathfrak{s} \in \mathscr{B}$, let $f_{\mathfrak{s}}$ be the canonical mapping $A \to A/\mathfrak{s}$, $\phi_{\mathfrak{s}}$ the canonical mapping