Henselian Rings and Equicharacteristic Cohen Structure

By [L1], one clause in the definition of a Henselian pair is exactly the containment $I\subseteq J(A)$.

Therefore the defining ideal of a Henselian pair lies in the Jacobson radical.

Because $(g_0,h_0)=(1)$ in $(A\mathrm{/}I)[T]$, there exist $a_0,b_0\in (A\mathrm{/}I)[T]$ such that $a_0{g}_0+b_0{h}_0=1.$

Lift the coefficients of $a_0$ and $b_0$ to polynomials $a,b\in A[T]$. Reducing coefficientwise modulo $I$ gives $\overline{ag+bh}=\bar{a}\bar{g}+\bar{b}\bar{h}=a_0{g}_0+b_0{h}_0=1.$ Hence $ag+bh\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$.

Thus a coprime residue factorization always admits a lifted Bezout relation modulo $I$.

Since $\mathrm{deg}u<m$, [L1] shows that the coefficient of $T^m$ in $u$ is $0$. Hence the coefficient of $T^m$ in $g+u$ is the coefficient of $T^m$ in $g$, namely $1$. Therefore $g+u$ is monic of degree $m$.

Since $\mathrm{deg}v<n$, [L1] shows that the coefficient of $T^n$ in $v$ is $0$. Hence the coefficient of $T^n$ in $h+v$ is the coefficient of $T^n$ in $h$, namely $1$. Therefore $h+v$ is monic of degree $n$.

Hence degree-bounded corrections preserve the prescribed monicity and degrees throughout Hensel lifting.

Put $M=I^r\mathrm{/}{I}^{r+1}$, viewed as an $A\mathrm{/}I$-module, and write $\epsilon \in M[T]$ for the class of $e=f-gh$. Since $f$ and $gh$ are monic of the same degree $m+n$, one has $\mathrm{deg}\epsilon <m+n$. Multiplying $ag+bh\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$ by $\epsilon$ gives $\epsilon =(a\epsilon )g+(b\epsilon )h\text{in }M[T]\mathrm{.}$

Divide $b\epsilon \in M[T]$ by the monic polynomial $g$ to write $b\epsilon =qg+\bar{u},\mathrm{deg}\bar{u}<m\mathrm{.}$ Substitution in step 1.1 gives $\epsilon =\bar{u}h+\bar{v}g,\bar{v}:=a\epsilon +qh\mathrm{.}$ The polynomial $\epsilon -\bar{u}h=\bar{v}g$ has degree less than $m+n$; because $g$ is monic of degree $m$, this forces $\mathrm{deg}\bar{v}<n$. Lift $\bar{u},\bar{v}$ coefficientwise to polynomials $u,v\in I^r[T]$ with the same degree bounds.

Set $g^{\mathrm{\prime }}=g+u$ and $h^{\mathrm{\prime }}=h+v$. Then $g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}=gh+uh+vg+uv\mathrm{.}$ By step 2.1, the first three terms agree with $f$ modulo $I^{r+1}$, while $uv\in I^{2r}[T]\subseteq I^{r+1}[T]$ because $r\ge 1$. Hence $f-g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}\in I^{r+1}[T]$. By [L2], $g^{\mathrm{\prime }}$ and $h^{\mathrm{\prime }}$ remain monic of degrees $m,n$.

Therefore one Hensel correction step improves a lift modulo $I^r$ to a lift modulo $I^{r+1}$ without changing the prescribed degrees.

Fix a coefficient index $j$. If $s>r$, then the coefficient of $T^j$ in $g_s-g_r$ is a sum of coefficients from the increments $g_{k+1}-g_k$ for $k=r,\dots ,s-1$. By [L1], each summand lies in $I^k\subseteq I^r$, so the whole difference lies in $I^r$. Thus the $j$th coefficients of the $g_r$ form an $I$-adic Cauchy sequence.

The same argument applied to the increments $h_{k+1}-h_k\in I^k[T]$ shows that each coefficient sequence of the $h_r$ is also $I$-adically Cauchy.

Hence the iterative Hensel corrections are coefficientwise Cauchy.

By completeness and [L2], each coefficient sequence of $g_r$ and of $h_r$ has a limit in $A$. Since the degrees are fixed and the leading coefficients are always $1$, these limits assemble into monic polynomials $g,h\in A[T]$ of the same degrees.

Fix a coefficient index $j$ of the product. Only finitely many coefficient pairs contribute to the $T^j$-coefficient of $g_r{h}_r$, so ordinary continuity of finite sums and products shows that the $T^j$-coefficient of $g_r{h}_r$ converges to the $T^j$-coefficient of $gh$.

For every $r$, the coefficient of $T^j$ in $f-g_r{h}_r$ lies in $I^r$ by hypothesis. Passing to the limit in step 2.1 shows that the coefficient of $T^j$ in $f-gh$ lies in every $I^r$. By separatedness and [L2], that coefficient is $0$. Since this holds for every $j$, one has $f=gh$.

Therefore the coefficientwise limits of the iterative factors multiply back to the original polynomial.

The congruences modulo $I$ hold by hypothesis, so the claim is true for $r=1$.

Assume $g\equiv g^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^r[T])$ and $h\equiv h^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^r[T])$ for some $r\ge 1$. Write $g^{\mathrm{\prime }}=g+u$ and $h^{\mathrm{\prime }}=h+v$ with $u,v\in I^r[T]$. Since each pair consists of monic polynomials of the same degree, $\mathrm{deg}u<m$ and $\mathrm{deg}v<n$. From $gh=(g+u)(h+v)$ we get $uh+vg+uv=0.$ Modulo $I^{r+1}$ the term $uv$ vanishes, so $uh+vg\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{I}^{r+1})\mathrm{.}$

Put $M=I^r\mathrm{/}{I}^{r+1}$, and let $\bar{u},\bar{v}\in M[T]$ be the classes of $u,v$. Step 2.1 gives $\bar{u}h+\bar{v}g=0$. Choose $a,b$ with $ag+bh\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$ by [L1]. Modulo the monic polynomial $g$, multiplication by $h$ is invertible with inverse $b$, so $\bar{u}h=0$ implies $\bar{u}=0$ modulo $g$. Since $\mathrm{deg}\bar{u}<m=\mathrm{deg}g$, this gives $\bar{u}=0$. The equation then becomes $\bar{v}g=0$, and multiplication by the monic polynomial $g$ is injective on $M[T]$, so $\bar{v}=0$. Hence $u,v\in I^{r+1}[T]$, proving the two congruences modulo $I^{r+1}$.

By induction, the two lifted factorisations agree modulo every power $I^r$.

Suppose $f=gh=g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ are two monic lifts of the given residue factorization. By [L2], they agree modulo $I^r[T]$ for every $r\ge 1$.

In the present page's convention, [L1] already includes uniqueness of the lifted factorization. Therefore the two lifts must coincide. Step 1.1 records the explicit congruence mechanism that later examples use.

Hence a coprime monic residue factorization has at most one Hensel lift.

Since $\bar{f}(\bar{a})=0$, polynomial division by $T-\bar{a}$ gives a factorization $\bar{f}=(T-\bar{a})\bar{h}$ for some $\bar{h}\in k[T]$.

Differentiating the identity of step 1.1 and evaluating at $\bar{a}$ yields ${\bar{f}}^{\mathrm{\prime }}(\bar{a})=\bar{h}(\bar{a})\mathrm{.}$ The left side is nonzero by hypothesis, so $\bar{h}(\bar{a})\mathrm{\ne }0$. Therefore $T-\bar{a}$ does not divide $\bar{h}$, which is equivalent to $(T-\bar{a},\bar{h})=(1)$ in $k[T]$.

Thus a simple residue root determines a coprime residue factorization.

By [L1], write $\bar{f}=(T-\bar{a})\bar{h}$ with $T-\bar{a}$ and $\bar{h}$ coprime. Since $(A,\mathfrak{m})$ is Henselian, [L2] gives a lifted factorization $f=(T-a)h$ with $a\in A$ reducing to $\bar{a}$. Evaluating at $T=a$ gives $f(a)=0$.

If $a^{\mathrm{\prime }}$ is another lift of $\bar{a}$ with $f(a^{\mathrm{\prime }})=0$, then $f=(T-a^{\mathrm{\prime }})h^{\mathrm{\prime }}$ for some monic $h^{\mathrm{\prime }}$, and this is another lift of the same residue factorization. By [L2], the lifted factorization is unique, so $T-a=T-a^{\mathrm{\prime }}$ and hence $a=a^{\mathrm{\prime }}$.

Therefore factor lifting implies unique lifting of every simple residue root.

Because ${\bar{e}}^2=\bar{e}$, one has $T^2-T=(T-\bar{e})(T-(1-\bar{e}))$ in $(A\mathrm{/}I)[T]$. The two factors are monic, and their difference is $1-2\bar{e}$, which is a unit because $\bar{e}(1-\bar{e})=0$ forces every prime quotient to send $\bar{e}$ to $0$ or $1$. Hence the factors are coprime.

By [L1], this residue factorization lifts uniquely to $T^2-T=(T-e)(T-(1-e))$ for some $e\in A$ lifting $\bar{e}$. Evaluating at $T=e$ yields $e^2-e=0$, so $e$ is idempotent.

If $e^{\mathrm{\prime }}$ is another lifted idempotent, then $(T-e^{\mathrm{\prime }})(T-(1-e^{\mathrm{\prime }}))$ is a second lift of the same residue factorization. By [L1], the factorization is unique, so $e^{\mathrm{\prime }}=e$.

Therefore idempotents lift uniquely in a Henselian pair.

The residue ring $A\mathrm{/}\mathfrak{m}$ is a field, so its only idempotents are $0$ and $1$. For $\bar{e}\in \{0,1\}$, the polynomial $p(T)=T^2-T$ satisfies $p(\bar{e})=0$ and $p^{\mathrm{\prime }}(\bar{e})=2\bar{e}-1\in \{-1,1\},$ hence $\bar{e}$ is a simple residue root.

By the assumed simple-root lifting property, there is a unique lift $e\in A$ of $\bar{e}$ with $p(e)=0$. The equation $p(e)=0$ is exactly $e^2=e$, so $e$ is idempotent.

Therefore the simple-root form lifts residue idempotents uniquely. The role of [L1] is only to identify this as the same mechanism already proved for Henselian local rings.

The hypothesis of this item is exactly the simple-root lifting condition in [L1], and conclusion (3) of [L1] is exactly the coprime monic factor-lifting property. Therefore the cited source yields the required factorization statement for $(A,\mathfrak{m})$.

Therefore unique lifting of simple residue roots recovers unique lifting of coprime monic factorizations.

If $A$ is Henselian, then [L3] identifies $(A,\mathfrak{m})$ as a Henselian pair, and [L1] gives unique lifting of every simple residue root.

Conversely, assume every simple residue root lifts uniquely. Then [L2] gives the coprime monic factor-lifting property for $(A,\mathfrak{m})$. Since $\mathfrak{m}$ is the unique maximal ideal of the local ring, one has $\mathfrak{m}\subseteq J(A)$, so [L3] shows that $A$ is Henselian.

Therefore a local ring is Henselian exactly when every simple residue root lifts uniquely.

By [L1], conditions (1) and (2) are equivalent.

By [L3], condition (1) says exactly that the pair $(A,\mathfrak{m})$ is Henselian. Then [L2] identifies this with the finite-algebra idempotent bijection in condition (3). Hence conditions (1) and (3) are equivalent.

Since (1) is equivalent to both (2) and (3), all three conditions are equivalent.

Let $x\in I$ and $r\in A$. Then $rx\in I$, so $1-rx\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$. By [L1], the element $1-rx$ is a unit. This is the Jacobson-radical criterion for $x$, so $x\in J(A)$. Hence $I\subseteq J(A)$.

Let $f\in A[T]$ be monic and let $\bar{f}=g_0{h}_0$ be a coprime monic factorization in $(A\mathrm{/}I)[T]$. Choose monic lifts $g_1,h_1$ of $g_0,h_0$. By [L2], choose $a,b\in A[T]$ with $a{g}_1+b{h}_1\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$. Repeatedly applying [L3] produces monic pairs $(g_r,h_r)$ with $f-g_r{h}_r\in I^r[T]$ for every $r\ge 1$.

By [L4], the coefficient sequences of $(g_r)$ and $(h_r)$ are Cauchy and converge to monic polynomials $g,h\in A[T]$ with $f=gh$. Their reductions are still $g_0,h_0$. Thus the required lifted factorization exists.

If $f=g^{\mathrm{\prime }}{h}^{\mathrm{\prime }}$ is another monic lift of the same residue factorization, then [L5] gives $g\equiv g^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^r[T])$ and $h\equiv h^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}{I}^r[T])$ for every $r\ge 1$. Separatedness forces $g=g^{\mathrm{\prime }}$ and $h=h^{\mathrm{\prime }}$. Hence the lift is unique.

Steps 1.1-3.1 verify both clauses of the definition, so $(A,I)$ is a Henselian pair.

By [L1], the pair $(A,\mathfrak{m})$ is Henselian because $A$ is complete and separated for the $\mathfrak{m}$-adic topology.

By definition, a local ring is Henselian exactly when its maximal-ideal pair is Henselian. Therefore $A$ is Henselian.

Choose $N\ge 1$ with $I^N=0$. Then for every $r\ge N$, the quotients $A\mathrm{/}{I}^r$ stabilize at $A$, so $A$ is automatically complete and separated for the $I$-adic topology.

Applying [L1] to the ideal $I$ shows that $(A,I)$ is Henselian.

By [L1], there exists $N\ge 1$ with ${\mathfrak{m}}^N=0$. Since $\mathfrak{m}$ is the maximal ideal of a local ring, it lies in the Jacobson radical.

Therefore [L2] applies to the pair $(A,\mathfrak{m})$, so $(A,\mathfrak{m})$ is Henselian. Equivalently, the Artinian local ring $A$ is Henselian.

The quotient map $A\to A\mathrm{/}J$ is integral because every element of $A\mathrm{/}J$ satisfies a monic linear equation over the image of $A$. The integral-base-change lemma for Henselian pairs in Stacks tag 09XD therefore applies to the Henselian pair $(A,I)$ and shows that the quotient pair $(A\mathrm{/}J,(I+J)\mathrm{/}J)$ is again Henselian.

By [L1], every Henselian pair has the coprime monic factor-lifting property. Applying that definition to the pair from step 1.1 gives the claimed lifting property for $(A\mathrm{/}J,(I+J)\mathrm{/}J)$.

Hence Henselian factor lifting descends to quotients.

The quotient $A\mathrm{/}J$ is local with maximal ideal $(\mathfrak{m}+J)\mathrm{/}J$.

Since $(A,\mathfrak{m})$ is Henselian, the pair $(A,\mathfrak{m})$ is Henselian. Applying [L1] with $I=\mathfrak{m}$ shows that the quotient pair $(A\mathrm{/}J,(\mathfrak{m}+J)\mathrm{/}J)$ has the Henselian factor-lifting property.

Together with step 1.1, this is exactly the definition of a Henselian local ring for $A\mathrm{/}J$.

By [L1], the defining property of a coefficient field is precisely that the composite $K\hookrightarrow A\twoheadrightarrow A\mathrm{/}\mathfrak{m}$ is an isomorphism.

Therefore a coefficient field maps isomorphically to the residue field.

By [L2], the ring $A$ and its residue field $k$ have the same characteristic. If that characteristic is $0$, then the unique map $\mathbf{Z}\to A$ kills no nonzero integer, so it extends to an embedding $\mathbf{Q}\hookrightarrow A$. If the characteristic is $p>0$, then the image of $\mathbf{Z}\to A$ is a copy of ${\mathbf{F}}_p$ inside $A$.

Reducing these embedded prime fields modulo $\mathfrak{m}$ gives the prime subfield of $k$, because $A$ and $k$ have the same characteristic and prime subfields are unique by [L1].

Therefore the prime field of the residue field lifts canonically in the equicharacteristic case.

Let $\mathcal{S}$ be the set of subfields $K\subseteq A$ for which the residue map $K\to A\mathrm{/}\mathfrak{m}$ is injective. By [L1], $\mathcal{S}$ is nonempty. Order $\mathcal{S}$ by inclusion.

If $\mathcal{C}\subseteq \mathcal{S}$ is a chain, then ${\bigcup }_{K\in \mathcal{C}}K$ is again a subfield of $A$: closure under the field operations is inherited from some chain member containing the finitely many elements involved. Its residue map is still injective, because a nonzero element of the union already lies in one chain member where injectivity holds. Thus every chain in $\mathcal{S}$ has an upper bound in $\mathcal{S}$.

By [L2], the poset $\mathcal{S}$ has a maximal element. That is exactly a maximal residue-injective subfield of $A$.

Let $\bar{p}(T)\in \rho (K)[T]$ be the minimal polynomial of $\bar{u}$. Since $\bar{u}$ is separable over $\rho (K)$, one has ${\bar{p}}^{\mathrm{\prime }}(\bar{u})\mathrm{\ne }0$. Lift the coefficients of $\bar{p}$ through the residue isomorphism $\rho :K\to \rho (K)$ to a monic polynomial $p(T)\in K[T]\subseteq A[T]$.

By [L1], the simple residue root $\bar{u}$ of $\bar{p}$ lifts uniquely to some $u\in A$ with $p(u)=0$. Then $K[u]$ is an integral domain finite over $K$, and its fraction field $K^{\mathrm{\prime }}:=K(u)$ sits inside $A$ because every nonzero element of $K[u]$ has nonzero residue, hence is a unit in the local ring $A$. The residue image of $K^{\mathrm{\prime }}$ contains both $\rho (K)$ and $\bar{u}$.