# “standard” form of a finite morphism

Every etale morphism is locally (passing to affine neighbourhoods and then to their coordinate rings) of the form $A \to (A[x]/(P(x)))_b$ where $P(x)$ has the property that $P'(x)$ is invertible in $(A[x]/(P(x)))_b$.

Is anything like that true for a general finite morphism? finite flat morphism?

(I am trying to understand the proof for the statement for etale morphisms from the stacks project, Algebra 133.16 and I cannot undersnand if the etaleness is crucial to the proof -- of course, one then doesn't need the condition on $P'(x)$ )

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## 1 Answer

No, this is false even for finite field extensions:

Let $K=k(x,y)$ where $k$ is a field of characteristic $p>0$. Let $L=K[x^{1/p}, y^{1/p}]$. Then $L/K$ is finite, flat. But $L$ can't be generated by one element over $K$ (even up to localization).

Edit Add a new example (cf. comments).

Let $A=\mathbb C$, $B=A[x,y]/(x^2, y^2)$. Then $A\to B$ is a finite extension of local rings with trivial residue extension. But it still is not monogeneous.

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indeed, you are right. I guess one has to ask that the residue field extension for a point lying over another be separable (and this is so when the morphism is etale) – Dima Sustretov Jun 5 '12 at 10:39
@DimaSustretov: the answer is still no. – user18119 Jun 6 '12 at 15:54