Let $(A,\mathfrak{p})\subset(B,\mathfrak{q})$ two local rings, such that $B$ is finitely generated as $A$-module. It's very well known, from Algebraic Geometry, that if $\Omega^1_{B/A}=0$ then the inclusion is unramified, hence in particular we must have $\mathfrak{q}=\mathfrak{p}B$. However I think that at last the equality $\mathfrak{q}=\mathfrak{p}B$ can be proved by using only ring theory, without schemes and such.

Any idea?

  • 8
    $\begingroup$ Apply scalar extension by $A\to A/\mathfrak{p}$ to reduce to the case $A=k$ is a field. Now $B$ is a finite local $k$-algebra. Dropping the locality assumption, $B$ is finite over $k$ with $\Omega^1_{B/k}=0$ and you want to conclude that the local factor rings of $B$ are fields separable over $k$. For this purpose you can extend scalars to $\overline{k}$ so $k$ is alg. closed, and then pass back to the local case. You want $B=k$. If not, it admits as a quotient $B'=k[\epsilon]/(\epsilon)^2$, so $\Omega^1_{B'/k}\ne 0$ by hand, but this is a quotient of $\Omega^1_{B/k}=0$, contradiction. $\endgroup$ – nfdc23 Feb 1 '18 at 19:59

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