Question about universal derivation $\Omega_{A/k}$

Question

Let $k$ be a ring, let $A$ be a $k$-algebra. The universal derivation $\Omega_{A/k}$ is the (unique) $k$-module representing the functor of the $k$-derivations of $A$; suppose that $\Omega_{A/k}=0$. If $k$ is an algebraically closed field, we can deduce that, for every maximal ideal $m$, holds $m/m^2=0$. In what settings this is this true? I would say yes for $k$ a generic field, but I don't understand if the arguments we used in the course to get this result are valid when $k$ is just a ring. Thanks in advance for any clarify; also, any reference that could help is welcome

Answer 1

If $k$ is not a field, this may be false. The classic example is the localization $\mathbb{Z}_2$ as a $\mathbb{Z}$-algebra, which has many maximal ideals distinct from their squares ($(p)$ for any odd prime), but $\Omega_{\mathbb{Z}_2/\mathbb{Z}}=0$ by the quotient rule. (See Vakil, The Rising Sea, example 21.2.8.)

For the field case, characteristic zero or $A$ finite-dimensional is sufficient, but probably not quite necessary. If this reminds you of separability, it should: the key obstruction is precisely an inseparable extension. The argument below is inspired by the only-vaguely-related answer to Kähler differentials and ordinary differentials. Eisenbud, Commutative Algebra, proves the local ring case (via the conormal exact sequence) in Corollary 16.13.

Suppose $\dim_k{(A)}<\infty$. Then $A$ is Artinian, and in particular $A$ is a product of field extensions of $k$; say $A=\prod_a{K_a}$. Maximal ideals of $A$ are precisely $0\times\prod_{a\neq b}{K_a}$ for varying $b$, which are their own square, and so $m/m^2$ always vanishes.

Now suppose $\mathrm{char}{(k)}=0$, but $A$ is not a product of fields extending $k$. Consider a "transcendence basis" $S$ of $A$ over $k$: a minimal $S\subseteq A$ that is algebraically independent over $k$ and such that $\dim_{k(S)}(A)<\infty$. For each $s\in S$, there exists a $k$-derivation $\delta$ on $k(S)$ such that $\delta(s)=1$. Since $A$ is algebraic over $k(S)$, $\delta$ extends to $A$. (See Derivations and algebraic extensions in characteristic zero; neither existence nor uniqueness holds in characteristic $p$.) Thus $\Omega_{A/k}\neq0$.

A counterexample for the positive-characteristic scenario is $A=\mathbb{F}_2[\{x_j\}_{j\in\mathbb{N}}]/(\{x_j-x_{j+1}^4\}_{j\in\mathbb{N}})$. To compute $\Omega_{A/\mathbb{F}_2}$, apply Vakil's "Key Fact" 21.2.3, which says that $$\Omega_{A/\mathbb{F}_2}=\left(\bigoplus_{j\in\mathbb{N}}{A\,dx_j}\right)/(\{dx_j\}_{j\in\mathbb{N}})=0$$ Yet $(\{x_j\}_{\in\mathbb{N}})$ is a maximal ideal differing from its square.