Abstract

Ruyong Feng pointed out that the proof of Proposition 2.9 contains a mistake and supplied a correct proof. In the second line of the proof, we say that “Therefore ∂(Z) − BZ = DZ for some D ∈ gln(k)”. This is incorrect and the correct statement should be ‘Therefore ∂(Z) − BZ = ZD for some D ∈ gln(k)” . The following is a replacement for the first paragraph of the proof. The remaining parts of the proof are correct. Assume that such a B exists. A calculation shows that σ(∂(Z)−BZ) = A((∂(Z)−BZ). Therefore ∂(Z) − BZ = ZD for some D ∈ gln(k). Since k is differentially closed, there is a U ∈ GLn(k) such that ∂(U) = −DU . We then have ∂(ZU) = ∂(Z)U + Z∂(U) = (BZ + ZD)U − ZDU = B(ZU). For any φ ∈ Autσ∂(R/k), let [φ]Z ∈ GLn(k) denote the matrix such that φ(Z) = Z[φ]Z. We then have

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