Lemma 10.136.6. Let $S$ be a finitely presented $R$-algebra which has a presentation $S = R[x_1, \ldots , x_ n]/I$ such that $I/I^2$ is free over $S$. Then $S$ has a presentation $S = R[y_1, \ldots , y_ m]/(f_1, \ldots , f_ c)$ such that $(f_1, \ldots , f_ c)/(f_1, \ldots , f_ c)^2$ is free with basis given by the classes of $f_1, \ldots , f_ c$.

Proof. Note that $I$ is a finitely generated ideal by Lemma 10.6.3. Let $f_1, \ldots , f_ c \in I$ be elements which map to a basis of $I/I^2$. By Nakayama's lemma (Lemma 10.20.1) there exists a $g \in 1 + I$ such that

$g \cdot I \subset (f_1, \ldots , f_ c)$

and $I_ g \cong (f_1, \ldots , f_ c)_ g$. Hence we see that

$S \cong R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)[1/g] \cong R[x_1, \ldots , x_ n, x_{n + 1}]/(f_1, \ldots , f_ c, gx_{n + 1} - 1)$

as desired. It follows that $f_1, \ldots , f_ c,gx_{n + 1} - 1$ form a basis for $(f_1, \ldots , f_ c, gx_{n + 1} - 1)/(f_1, \ldots , f_ c, gx_{n + 1} - 1)^2$ for example by applying Lemma 10.134.12. $\square$

## Comments (2)

Comment #3432 by ym on

It's easier to see the isom $S\cong R[x\ldots]/(f\ldots)[1/g]$ if you conclude from nakayama that $I_g = (f\ldots)_g$

Comment #3491 by on

OK, I added the conclusion from Nakyama's lemma. But I kept the other statement as well because it is how I think about it. See change here. Thanks very much.

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• 2 comment(s) on Section 10.136: Syntomic morphisms

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