2

enter image description here

Here, shouldn't the $x_i$s map to $x_i + \mathfrak{q}^2$, rather than to $x_i +\mathfrak{q}$? If not, then everything would be mapped to degree $1$ elements.

This is another one: enter image description here

Here, where is the fact that $k$ maps isomorphically onto $A/\mathfrak{m}$ used? If $f_s$ has coefficients in $\mathfrak{m}$, they can't be invertible and hence can only be zero as all the other elements of $k$ are invertible($k$ is a field).

luxerhia
  • 3,538
Jehu314
  • 913
  • 5
  • 14
  • 4
    The error you found in the proof of Proposition 11.20 is mentioned at https://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald (with the same correction). – darij grinberg Aug 16 '18 at 17:00
  • 2
    I think you're right about Corollary 11.21. Atiyah and Macdonald just didn't think of the invertibility argument, it seems. – darij grinberg Aug 31 '18 at 23:13
  • I think you meant 'eveything would be mapped to degree 0 elements' instead of 'degree 1'. – luxerhia Oct 27 '20 at 09:08

0 Answers0