Kahler Differentials Conormal Sequences and Infinitesimal Lifting — Examples

By [F1] the module ${\mathrm{\Omega }}_{P\mathrm{/}k}$ is free with basis $\mathrm{d}x,\mathrm{d}y$ over $P$, and $\mathrm{d}f=(\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x)\mathrm{d}x+(\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y)\mathrm{d}y$ for every $f\in P$.

We compute the partial derivatives of the powers of the variables: $\mathrm{\partial }(x^n)\mathrm{/}\mathrm{\partial }x=n{x}^{n-1}$ for every $n\ge 0$, where the case $n=0$ reads $\mathrm{\partial }(1)\mathrm{/}\mathrm{\partial }x=0$, and $\mathrm{\partial }(y^m)\mathrm{/}\mathrm{\partial }x=0$ for every $m\ge 0$. Indeed, $\mathrm{\partial }(1)\mathrm{/}\mathrm{\partial }x=0$ because $1\in k$ is annihilated by a $k$-derivation [F2], and if $\mathrm{\partial }(x^{n-1})\mathrm{/}\mathrm{\partial }x=(n-1)x^{n-2}$ then the Leibniz rule [F2] and $\mathrm{\partial }x\mathrm{/}\mathrm{\partial }x=1$ [F1] give $\mathrm{\partial }(x^n)\mathrm{/}\mathrm{\partial }x=\mathrm{\partial }(x\cdot x^{n-1})\mathrm{/}\mathrm{\partial }x=x^{n-1}+x(n-1)x^{n-2}=n{x}^{n-1}$; the same induction with $\mathrm{\partial }y\mathrm{/}\mathrm{\partial }x=0$ gives $\mathrm{\partial }(y^m)\mathrm{/}\mathrm{\partial }x=0$. Interchanging the roles of $x$ and $y$ gives $\mathrm{\partial }(x^n)\mathrm{/}\mathrm{\partial }y=0$ and $\mathrm{\partial }(y^m)\mathrm{/}\mathrm{\partial }y=m{y}^{m-1}$.

Multiplying out with the Leibniz rule: $\mathrm{\partial }(x^a{y}^b)\mathrm{/}\mathrm{\partial }x=y^b\mathrm{\partial }(x^a)\mathrm{/}\mathrm{\partial }x+x^a\mathrm{\partial }(y^b)\mathrm{/}\mathrm{\partial }x=a{x}^{a-1}{y}^b$, the term being $0$ when $a=0$, and likewise $\mathrm{\partial }(x^a{y}^b)\mathrm{/}\mathrm{\partial }y=x^a\mathrm{\partial }(y^b)\mathrm{/}\mathrm{\partial }y=b{x}^a{y}^{b-1}$, the term being $0$ when $b=0$.

Substituting step 2.1 into the formula of step 1.1 gives $\mathrm{d}(x^a{y}^b)=a{x}^{a-1}{y}^b\mathrm{d}x+b{x}^a{y}^{b-1}\mathrm{d}y$ for all $a,b\ge 0$, with the integer coefficients evaluated in $k$: if $k$ has characteristic $p$ and $p\mid a$, then $a=0$ in $k$ and the $\mathrm{d}x$-term vanishes, while the class $x^{a-1}{y}^b$ remains meaningful for $a\ge 1$ and the case $a=0$ is handled as $0$. Since $\mathrm{d}x,\mathrm{d}y$ form a basis of the free module ${\mathrm{\Omega }}_{P\mathrm{/}k}$, the formula determines $\mathrm{d}$ on every monomial and, by additivity and $k$-linearity of the universal derivation, on all of $P$; in particular $\mathrm{d}x$ and $\mathrm{d}y$ are $k$-linearly independent elements of ${\mathrm{\Omega }}_{P\mathrm{/}k}$.

Apply [F1] with $A=k$, $n=2$, $r=1$ and $f_1=f$: the ideal $I=(f)$ is generated by the single element $f$, and ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is the cokernel of the $B$-linear map $B^1\to B^2$ with the single column $(f_x,f_y)$. Identifying $B^2=B\mathrm{d}x\oplus B\mathrm{d}y$ by the standard basis, the image is the cyclic submodule generated by $f_x\mathrm{d}x+f_y\mathrm{d}y$, so ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong (B\mathrm{d}x\oplus B\mathrm{d}y)\mathrm{/}B(f_x\mathrm{d}x+f_y\mathrm{d}y)$.

The derivatives of the powers of $x$: by the Leibniz rule [F3] and $\mathrm{\partial }x\mathrm{/}\mathrm{\partial }x=1$, $\mathrm{\partial }y\mathrm{/}\mathrm{\partial }x=0$ [F2], induction on $n\ge 0$ gives $\mathrm{\partial }(x^n)\mathrm{/}\mathrm{\partial }x=n{x}^{n-1}$ and $\mathrm{\partial }(y^m)\mathrm{/}\mathrm{\partial }x=0$ for all $m\ge 0$, with the case $n=0$ read as $\mathrm{\partial }(1)\mathrm{/}\mathrm{\partial }x=0$.

The case $f=0$: then $B=P$ and $f_x=f_y=0$, so the relation submodule in step 1.1 is $B\cdot 0=0$ and ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong B^2$, which is the free module on $\mathrm{d}x,\mathrm{d}y$ already recorded in [F2].

Suppose $k$ has characteristic $p>0$ and $f=x^p$. Then step 1.2 gives $f_x=\mathrm{\partial }(x^p)\mathrm{/}\mathrm{\partial }x=p{x}^{p-1}=0$ and $f_y=\mathrm{\partial }(x^p)\mathrm{/}\mathrm{\partial }y=0$ because $p=0$ in $k$, so again the relation submodule of step 1.1 vanishes and ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong B^2$ even though $B=k[x,y]\mathrm{/}(x^p)$ is non-reduced: the presentation records no relation at all. If instead $2\mathrm{\ne }0$ in $k$ and $f=x^2$, the relation is $B\cdot (2x\mathrm{d}x)$ and $2x\mathrm{\ne }0$ in $B$: the ideal $(x^2)$ contains no nonzero polynomial of degree one, so it cannot contain $2x$. Thus the relation is a nonzero cyclic submodule even when $2$ is a zero divisor in $k$; in either case the displayed presentation of step 1.1 is the one computed, and no hypothesis on $f$ beyond $f\in k[x,y]$ was used.

Combining steps 1.1 with the cases 2.1 and 2.2, for every $f\in k[x,y]$ the module ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is the quotient displayed in the Example, the relation $f_x\mathrm{d}x+f_y\mathrm{d}y$ being determined by the partial derivatives of $f$ and possibly vanishing; since [F1] requires no flatness, smoothness or non-vanishing hypothesis, the presentation holds without any regularity assumption on $f$, and the map $P\to B$ is the surjection onto the quotient by $(f)$ in all cases.

Apply [F1] with $A=k$, $n=1$, $r=1$, $P=k[t]$, $f_1=t^2$ and $B=k[t]\mathrm{/}(t^2)$: the derivative is $f^{\mathrm{\prime }}=2t$, whose class in $B$ is $2ϵ$, so the cokernel of multiplication by $2ϵ$ on $B$ is ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong B\mathrm{/}(2ϵ)$. Writing the image of the free generator for $B^1$ as $\mathrm{d}ϵ$, this reads ${\mathrm{\Omega }}_{B\mathrm{/}k}=B\mathrm{d}ϵ\mathrm{/}(2ϵ\mathrm{d}ϵ)$, the submodule $B(2ϵ\mathrm{d}ϵ)$ corresponding to the ideal $(2ϵ)\subseteq B$ under the identification $B\mathrm{d}ϵ\cong B$.

Suppose $2=0$ in $k$. Then $2ϵ=0\in B$, so the ideal $(2ϵ)$ is the zero ideal [step 1.1], and ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong B\mathrm{/}(0)=B$: the assignment $b\mapsto b\mathrm{d}ϵ$ is a $B$-module isomorphism $B\to {\mathrm{\Omega }}_{B\mathrm{/}k}$ with inverse induced by $\mathrm{d}ϵ\mapsto 1$, so $\mathrm{d}ϵ$ is a basis of ${\mathrm{\Omega }}_{B\mathrm{/}k}$ over $B$ and ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is free of rank one.

Suppose $k$ is a field of characteristic $\mathrm{\ne }2$. Then $2\mathrm{\ne }0$ in $k$, so $2$ is a unit of $k$ and hence of $B$, and $(2ϵ)=(ϵ)$: the two generators differ by the unit $2$ [step 1.1]. Hence ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong B\mathrm{/}(ϵ)$. By [F3], applied to the surjection $B\to k$ with $ϵ\mapsto 0$ whose kernel is $(ϵ)$, one has $B\mathrm{/}(ϵ)\cong k$, so ${\mathrm{\Omega }}_{B\mathrm{/}k}\cong k$ is one-dimensional over $k$ with basis image of $\mathrm{d}ϵ$, and $ϵ\mathrm{d}ϵ=0$ holds in ${\mathrm{\Omega }}_{B\mathrm{/}k}$ while $\mathrm{d}ϵ\mathrm{\ne }0$.

In the case of step 2.2 the module ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is not free over $B$: it has $k$-dimension $1$, whereas a free $B$-module of rank one has $k$-dimension equal to ${\dim }_{k}B=2$, since $\{1,ϵ\}$ is a $k$-basis of $B$ (every class in $k[t]\mathrm{/}(t^2)$ is uniquely $c+dϵ$ with $c,d\in k$). In the case of step 2.1 the dimension count is reversed and ${\mathrm{\Omega }}_{B\mathrm{/}k}$ is free of rank one, so the two characteristics genuinely give different answers, and the presentation of step 1.1 is the common source of both. By [F2] these computations are those of the relative differentials of the dual-numbers scheme over $k$ when $k$ is a field.

By [F2] there is $\alpha \in L$ with $L=k(\alpha )$; since $L\mathrm{/}k$ is separable, [F5] makes $\alpha$ separable over $k$, so its minimal polynomial $m\in k[x]$ is monic, irreducible and separable.

By [F4] the separability of the irreducible polynomial $m$ says $m^{\mathrm{\prime }}\mathrm{\ne }0$, where $m^{\mathrm{\prime }}$ is the formal derivative; since $\mathrm{deg}{m}^{\mathrm{\prime }}<\mathrm{deg}m=n$, the minimality statement of [F3] shows $m^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$: a vanishing of $m^{\mathrm{\prime }}$ at $\alpha$ would force $m\mid m^{\mathrm{\prime }}$ and hence $m^{\mathrm{\prime }}=0$.

Because $\mathrm{d}$ is a $k$-derivation [F1] and $m(\alpha )=0$, applying $\mathrm{d}$ to the relation $m(\alpha )=c_0+c_1\alpha +\cdots +c_n{\alpha }^n$ with $c_n=1$ and using additivity, $k$-linearity and the Leibniz rule gives $0=\mathrm{d}(m(\alpha ))={\sum }_i{c}_i\mathrm{d}({\alpha }^i)=m^{\mathrm{\prime }}(\alpha )\mathrm{d}\alpha$, with $m^{\mathrm{\prime }}(\alpha )\in L$ the value at $\alpha$ of the formal derivative. Since $m^{\mathrm{\prime }}(\alpha )\mathrm{\ne }0$ by step 2.1 and $L$ is a field, $m^{\mathrm{\prime }}(\alpha )$ is invertible in $L$, so $\mathrm{d}\alpha =0$.

It follows that $\mathrm{d}$ vanishes on all of $L$: by [F6] every $f\in L$ equals $p(\alpha )$ for a polynomial $p\in k[x]$, and the Leibniz rule over the expansion of $p(\alpha )$ in powers of $\alpha$ gives $\mathrm{d}p(\alpha )=p^{\mathrm{\prime }}(\alpha )\mathrm{d}\alpha =0$, since $\mathrm{d}\alpha =0$ [F1, step 3.1].

By [F1] the module ${\mathrm{\Omega }}_{L\mathrm{/}k}$ is generated over $L$ by the elements $\mathrm{d}f$ with $f\in L$, and step 4.1 shows that each such generator is zero; hence ${\mathrm{\Omega }}_{L\mathrm{/}k}=0$.

The class $\alpha$ satisfies ${\alpha }^p=a$ by construction, and $L=k(\alpha )$, since $L$ is generated as a $k$-algebra by $\alpha$. If $f=X^p-a$ is irreducible, then $L$ is a field of degree $p$ over $k$ and $\alpha$ is a primitive element; the next steps establish the irreducibility.

Assume for contradiction that $f$ is reducible. Since $k[X]$ is a principal ideal domain [F1], a reducible nonzero non-unit factors into irreducibles, so $f$ has a monic irreducible factor $m$ of degree $d$ with $1\le d\le p-1$. Let $A=k[X]\mathrm{/}(m)$, a field by [F2], and let $x$ denote the class of $X$ in $A$; then $m(x)=0$, and $x^p=a$ because $m$ divides $f=X^p-a$ in $k[X]\subseteq A[X]$.

In $A[X]$ the binomial theorem [F3] together with $p\mid \left(\genfrac{}{}{0px}{}{p}{i}\right)$ for $0<i<p$ [F4] gives the Frobenius identity $(X-x{)}^p=X^p-x^p=X^p-a$, the intermediate coefficients vanishing in $A$ of characteristic $p$; hence $m$, considered in $A[X]$, divides $(X-x{)}^p$.

On the other hand $m(x)=0$, so the division algorithm in the field $A$ [F5] gives $m=q(X-x)+m(x)=q(X-x)$, that is, $X-x$ divides $m$. In the principal ideal domain $A[X]$ [F1], the degree-one polynomial $X-x$ is irreducible (a factorization would have to split the degree $1$ into two nonnegative degrees, forcing a degree-$0$ factor, which is a unit of $A[X]$), so the only monic factor of $(X-x{)}^p$ of degree $d$ is $(X-x{)}^d$; since $m$ is monic of degree $d$, we get $m=(X-x{)}^d$.

Comparing the coefficient of $X^{d-1}$ in the identity $m=(X-x{)}^d$ of step 4.1 gives: the coefficient of $X^{d-1}$ in $(X-x{)}^d$ is $-dx$, and the coefficient of $X^{d-1}$ in $m\in k[X]$ lies in $k$, so $dx\in k$. Since $1\le d\le p-1$ and $k$ has characteristic $p$, the class of $d$ in $k$ is nonzero and invertible, so $x=d^{-1}(dx)\in k$; then $a=x^p\in k^p$, contradicting the hypothesis $a\mathrm{\notin }{k}^p$. Hence $f=X^p-a$ is irreducible, $L$ is a field with $[L:k]=p$, and $L=k(\alpha )$.

Now compute the differentials. Apply [F6] with $A=k$, $n=r=1$, $P=k[X]$, $q=f=X^p-a$ and $B=L$: the derivative is $f^{\mathrm{\prime }}=p{X}^{p-1}=0$, because $p=0$ in $k$, so the relation submodule $L\cdot f^{\mathrm{\prime }}(\alpha )$ is zero and ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\mathrm{/}(0)$ with the image of the basis vector written $\mathrm{d}\alpha$. Thus ${\mathrm{\Omega }}_{L\mathrm{/}k}\cong L\mathrm{d}\alpha \cong L$ as $L$-modules, in particular ${\mathrm{\Omega }}_{L\mathrm{/}k}\mathrm{\ne }0$ because the field $L$ is nonzero.

The extension is finite, algebraic and purely inseparable: every element of $L=k(\alpha )$ is a polynomial ${\sum }_i{c}_i{\alpha }^i$ with $c_i\in k$ by step 5.1, and its $p$-th power is ${\sum }_i{c}_i^p{a}^i\in k$ because ${\alpha }^p=a$ and the Frobenius map is additive in characteristic $p$ [F3]. So every element of $L$ has its $p$-th power in $k$, and the elementwise criterion of [F7] makes $L\mathrm{/}k$ purely inseparable; by step 5.1 it is finite of degree $p$.

Combining steps 6.1 and 6.2: ${\mathrm{\Omega }}_{L\mathrm{/}k}$ is a free $L$-module of rank one and hence nonzero, while $L\mathrm{/}k$ is a finite algebraic purely inseparable extension. This refutes the displayed statement and shows that the vanishing of $\mathrm{\Omega }$ for finite separable extensions cannot be extended to all finite algebraic extensions; the obstruction is precisely the vanishing derivative $f^{\mathrm{\prime }}=0$ of the inseparable polynomial $X^p-a$.

Apply [F1] with $A=k$: the sequence $I\mathrm{/}{I}^2\stackrel{\alpha }{\to }B{\otimes }_P{\mathrm{\Omega }}_{P\mathrm{/}k}\to {\mathrm{\Omega }}_{B\mathrm{/}k}\to 0$ is exact, $\alpha$ sends the class of $i\in I$ to $1\otimes \mathrm{d}i$, and the statement of [F1] explicitly leaves injectivity of $\alpha$ open.

The powers of the ideal: $I=x^{2}P$ and $I^2=x^{4}P$, since $I^2$ is generated by the products of two elements of $I$, and $x^{2}f\cdot x^{2}g=x^{4}fg$. Hence $I\mathrm{/}{I}^2=(x^2)\mathrm{/}(x^4)$. The class $[x^3]$ is nonzero: if $x^3\in (x^4)$, there would be $g\in P$ with $x^3=x^{4}g$, so $\mathrm{deg}(x^3)=3$ equals $\mathrm{deg}(x^{4}g)=4+\mathrm{deg}g$ for $g\mathrm{\ne }0$ and is impossible for $g=0$, by the degree rule of [F4] applied in the integral domain $P=k[x]$.