Henselian Rings and Equicharacteristic Cohen Structure — Examples

Start with $g_1=T$ and $h_1=T-1$. Then $f-g_1{h}_1=-t$. A residue Bezout identity is $1=1\cdot T+(-1)\cdot (T-1)(\mathrm{m}\mathrm{o}\mathrm{d}t),$ so [L1] applies.

Choose constant corrections $u_1=t$ and $v_1=-t$. Then $u_1{h}_1+v_1{g}_1=t(T-1)-tT=-t=f-g_1{h}_1,$ so $g_2=T+t,h_2=T-1-t$ satisfies $g_2{h}_2=T^2-T-t-t^2\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2)\mathrm{.}$ This is the first explicit correction step from [L2].

Now $f-g_2{h}_2=t^2$. Choose $u_2=-t^2$ and $v_2=t^2$. Then $u_2{h}_2+v_2{g}_2\equiv -t^2(T-1)+t^{2}T=t^2(\mathrm{m}\mathrm{o}\mathrm{d}{t}^3),$ so $g_3=T+t-t^2,h_3=T-1-t+t^2$ satisfies $g_3{h}_3\equiv f(\mathrm{m}\mathrm{o}\mathrm{d}{t}^3)\mathrm{.}$

Thus the factorization is lifted explicitly modulo $t$, modulo $t^2$, and modulo $t^3$. The computation makes the abstract correction lemma concrete.

By [L1], the complete local ring $k\mathrm{⟦}t\mathrm{⟧}$ is Henselian. Modulo $t$, one has $\bar{f}=T(T-1)$, so $\bar{a}=1$ is a simple root. By [L2], there is a unique root $a\in k\mathrm{⟦}t\mathrm{⟧}$ with $a\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}t)$ and $f(a)=0$. The first two correction stages give $a\equiv 1+t(\mathrm{m}\mathrm{o}\mathrm{d}{t}^2),a\equiv 1+t-t^2(\mathrm{m}\mathrm{o}\mathrm{d}{t}^3)\mathrm{.}$

The residue factorization $\bar{f}=T(T-1)$ is monic and coprime, so [L3] gives a factorization $f=(T-b)h(T)$ with $b\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}t)$. Evaluating at $T=b$ gives $f(b)=0$, so $b$ is another lift of the same simple residue root.

By uniqueness in [L2], one has $a=b$. Equivalently, the root obtained from the lifted linear factor is exactly the same root obtained from the simple-root lifting procedure.

Modulo $2$, the polynomial becomes $\bar{f}(T)=T^2-1=(T-1{)}^2,$ so the residue root $\bar{1}$ has multiplicity $2$. Equivalently, ${\bar{f}}^{\mathrm{\prime }}(\bar{1})=2\bar{1}=0$.

In ${\mathbf{Z}}_2$, both $1$ and $-1$ satisfy $f(T)=0$, and both reduce to $\bar{1}$ modulo $2$. Hence the residue root $\bar{1}$ has at least two lifts.

Therefore the derivative-unit hypothesis in [L1] cannot be dropped: a multiple residue root need not lift uniquely even in a complete local ring.

For every $n\ge 1$ one has $I^n=(t^n)k\mathrm{⟦}t\mathrm{⟧}\times (t^n)k\mathrm{⟦}t\mathrm{⟧},$ so $A\mathrm{/}{I}^n\cong k\mathrm{⟦}t\mathrm{⟧}\mathrm{/}(t^n)\times k\mathrm{⟦}t\mathrm{⟧}\mathrm{/}(t^n)\mathrm{.}$ Thus $A$ is the inverse limit of the quotients $A\mathrm{/}{I}^n$ coordinatewise, and ${\bigcap }_{n\ge 1}{I}^n=0\times 0,$ again coordinatewise. Hence $A$ is $I$-adically complete and separated, so [L1] shows that $(A,I)$ is Henselian.

The residue ring is $A\mathrm{/}I\cong k\times k$, and $(1,0)$ is an idempotent there. The obvious lift in $A$ is $e=(1,0)$, which indeed satisfies $e^2=e$. By [L2], that lift is the unique idempotent of $A$ reducing to $(1,0)$. Its complement is $1-e=(0,1)$.

Multiplication by $e$ and $1-e$ gives $A=eA\times (1-e)A=(k\mathrm{⟦}t\mathrm{⟧}\times 0)\times (0\times k\mathrm{⟦}t\mathrm{⟧})\mathrm{.}$ Thus the lifted idempotent exactly recovers the original product decomposition.

By [L1], $k\mathrm{⟦}t\mathrm{⟧}$ is local with maximal ideal $(t)$. Its $t$-adic topology is complete by construction of the formal power-series ring.

By [L2], this local domain is a discrete valuation ring. Applying [L3] to the complete local ring $k\mathrm{⟦}t\mathrm{⟧}$ shows that it is Henselian.

Therefore every complete discrete valuation ring modeled by $k\mathrm{⟦}t\mathrm{⟧}$ is Henselian.

In $A$, every class with nonzero constant term is a unit, so the nonunits are exactly the classes divisible by $x$. Thus $A$ is local with maximal ideal $(x)$, and $(x{)}^n=0$.

The descending chain of ideals in $A$ is finite because every ideal is one of $(1),(x),\dots ,(x^{n-1}),(0)$, so $A$ is Artinian. Therefore [L2] applies and shows that $A$ is Henselian.

Hence the truncated polynomial local ring $k[x]\mathrm{/}(x^n)$ is a concrete Artinian Henselian ring.

By [L1] and [L2], the ring $A={\mathbf{Z}}_{(7)}$ is local with residue field ${\mathbf{F}}_7$. In that field, $\bar{f}(3)=3^2-2=7\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}7),{\bar{f}}^{\mathrm{\prime }}(3)=2\cdot 3=6\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}7),$ so $3$ is a simple residue root.

Suppose $a\mathrm{/}b\in {\mathbf{Z}}_{(7)}$ with $7\nmid b$ satisfies $(a\mathrm{/}b{)}^2=2$. Then $a^2=2b^2$ in $\mathbf{Z}$. The $2$-adic valuation of the left side is even, while the valuation of the right side is odd, impossible. Hence $2$ has no square root in $\mathbf{Q}$, and therefore no root in ${\mathbf{Z}}_{(7)}$.

The simple residue root from step 1.1 does not lift, so [L3] shows that $A$ cannot be Henselian.

The canonical quotient map $k\mathrm{⟦}X,Y\mathrm{⟧}\twoheadrightarrow A,X\mapsto x,Y\mapsto y,$ has kernel containing the principal ideal $(Y^2-X^3)$ by construction.

Conversely, by definition $A$ is exactly the quotient by that relation, so the kernel is $(Y^2-X^3)$. Therefore $A\cong k\mathrm{⟦}X,Y\mathrm{⟧}\mathrm{/}(Y^2-X^3)\mathrm{.}$ The maximal ideal is generated by the classes of $x$ and $y$, and the relation has no linear term, so the ring is singular at that point.

This is an explicit instance of [L1]: the cusp local ring is a concrete complete local power-series quotient.

By [L1], $A$ is local with maximal ideal $(t)$ and residue field $A\mathrm{/}(t)\cong k(u)$. The standard inclusion of $k(u)$ into $A$ is therefore a coefficient field, in line with [L2].

Consider the subfield $K^{\mathrm{\prime }}=k(u+t)\subseteq A$. For every nonzero polynomial $q(Z)\in k[Z]$, the residue of $q(u+t)$ modulo $t$ is $q(u)$, which is nonzero in $k(u)$. Hence $q(u+t)$ is a unit of $A$, so every rational function in $u+t$ lies in $A$ and $K^{\mathrm{\prime }}$ is indeed a subfield. Its residue image is again $k(u)$ because $u+t\mapsto u$.

The two coefficient fields are distinct: if $u+t$ lay in the constant field $k(u)$, then subtracting $u$ would place $t$ in $k(u)$, but every nonzero element of $k(u)$ is a unit in $A$ whereas $t$ lies in the maximal ideal. Thus $k(u+t)\mathrm{\ne }k(u)$.

Therefore coefficient fields in a complete equicharacteristic local ring need not be canonical.