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Fix the Bernstein filtration on the Weyl algebra. Two good filtrations of a module are bounded shifts of one another: $F_{i}M\subseteq G_{i+c}M$ and conversely. If an operator of degree $d$ has symbol annihilating ${\mathrm{gr}}_{F}M$, its $r$th power lowers the expected filtration degree by $r$. For $r>2c$, comparison of filtrations forces the same power of its symbol to annihilate ${\mathrm{gr}}_{G}M$. Reversing the roles gives equality of the radical annihilator ideals and hence equality of support of a module. The algebra filtration is held fixed; changing between order and Bernstein filtrations need not preserve the same variety.

Choose a basis $x_1,x_2,x_3$ of the Lie algebra and filter its universal enveloping algebra by word length: $F_{0}R=\mathbb{C}$ and $F_{d}R$ is spanned by products of at most $d$ basis elements. The relations $x_j{x}_i-x_i{x}_j=[x_j,x_i]$ replace a degree-two commutator by degree one, so the degree-one symbols commute.

The Poincaré-Birkhoff-Witt theorem says the ordered monomials $x_1^{a_1}{x}_2^{a_2}{x}_3^{a_3}$ are a basis. Its relevant proof can be seen by reordering an inverted neighboring pair using the displayed relation. Reordering terminates because same-length terms have fewer inversions and bracket terms have smaller word length. Disjoint swaps commute; the only overlapping three-letter reorderings differ by $[x_i,[x_j,x_k]]+[x_j,[x_k,x_i]]+[x_k,[x_i,x_j]]=0$. The Jacobi identity therefore makes the reduced ordered expression independent of the reordering path. Ordered monomials have no remaining relations and form the basis. Their degree-$d$ symbols giveFor generators $m_1,\dots ,m_q$ of the finitely generated module $M$, put $M_d={\sum }_j{F}_{d}R{m}_j$. This is a good filtration, and $\mathrm{gr}M$ is finite over the three-variable polynomial ring. Define the dimension of a filtered module as the Krull dimension of its associated graded module, equivalently the degree of the eventual cumulative polynomial ${\dim }_{\mathbb{C}}{M}_d$, or the Gelfand–Kirillov dimension of a module. Two good filtrations have bounded shifts, so the growth degree is independent of the generators. Sinceits polynomial degree is at most three. Therefore every nonzero finitely generated $R$-module has dimension at most $3$. The zero module may be assigned the usual separate empty-support convention. This bound also follows immediately from its support of a module inside affine three-space.

Take $R=\mathbb{Z}$, $M=\mathbb{Q}/\mathbb{Z}$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\ge 1$,Tensor products commute with filtered colimits, so

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a maximal ideal $\mathfrak{m}$ in the support of $M$. The localized module $M_{\mathfrak{m}}$ is nonzero and finitely generated. By Nakayama lemma,This is a nonzero vector space over the residue field $\kappa (\mathfrak{m})$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak{m}$ of $(M^{\otimes k}{)}_{\mathfrak{m}}$, a contradiction. Thus a nonzero tensor-nilpotent module cannot be finitely generated.

For a ring homomorphism $A\to B$, the module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.