Definition of modulo of Kahler differentials $\Omega_{S/R}$ when a ring homomorphism $\phi:R\rightarrow S$ exists is the $S$-free module generated by elements of $R$ modulo the relations $D(s-s')-Ds-Ds',D(rs)-rDs,D(ss')-sDs'-s'Ds$. But how can you compute a modulo of Kahler differentials sxplicitly. Say one picks up the identity homomorphism $R\rightarrow R$, then what will happen?
How to compute module of Kahler differentials
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abstract-algebra
algebraic-geometry
commutative-algebra
modules
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1For $R\to R$ identity, clearly, you have $\Omega^1_{R/R}=0$, by the description given in the answer by Bernard. – 2017-01-09
1 Answers
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If $S$ is a finitely presented $R$-algebra: $\;S\simeq R[X_1,\dots,X_n]/(f_1,\dots,f_r)$, then $$\Omega_{S/R}\simeq \frac{S\, \mathrm dX_1\oplus\dots\oplus S\, \mathrm dX_n}{S\,\mathrm df_1+\dots+S\,\mathrm df_r}.$$
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1How do we get this result? Can you introduce a very elementary reference? Thanks. – 2017-01-08
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1I'm not sure it's *very* elementary, but you have a nice chapter in Matsumura, *Commutative Ring Theory*, ch. 9: Derivations (Cambridge University Press). – 2017-01-08