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    The de Rham differential is not $\OO_X$-linear

    FE-DERHAMNONLIN

    The de Rham differential \(d \colon \OO_X \to \Omega^1_{X/k}\) satisfies the Leibniz rule \(d(fg) = f\, dg + g\, df\), which differs from the \(\OO_X\)-linearity condition \(d(fg) = f\, dg\) by the term \(g\, df\). On \(X = \AA^1_k = \Spec k[x]\), take \(f = x\) and \(g = 1\): \(d(x \cdot 1) = dx\), while \(x\, d(1) = 0\), and \(dx \neq 0\) since \(\Omega^1_{k[x]/k}\) is free on \(dx\). So \(d\) is \(k\)-linear but not \(\OO_X\)-linear, and the de Rham complex \(\Omega^\bullet_{X/k}\) is a complex of sheaves of \(k\)-vector spaces whose terms are \(\OO_X\)-modules, not a complex of \(\OO_X\)-modules.

    This is why algebraic de Rham cohomology is the hypercohomology \(\mathbb{H}^\bullet(X, \Omega^\bullet_{X/k})\) of a complex of \(k\)-sheaves, and why its Hodge-to-de Rham spectral sequence starts from the \(\OO_X\)-module cohomology groups \(H^q(X, \Omega^p_{X/k})\) of the individual terms.

    Show that the differential in the de Rham complex is not $\OO_X$-linear.

    Dependencies

    • D-4GCH6The sheaf of differentials, and its quasicoherenceuses

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