Skip to main content
9 events
when toggle format what by license comment
Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Feb 13, 2017 at 9:34 comment added js21 @Phil: $A[[T]]$ is never faithfully flat over $A[T]$ (for $A \neq 0$), e.g. $1+T$ generates the unit ideal in $A[[T]]$, but not in $A[T]$.
Feb 11, 2017 at 20:06 comment added M.G. @Phil: thanks for the hint. I will have to think about that.
Feb 11, 2017 at 18:12 comment added Phil Tosteson If you can show that $A[[T]]$ is faithfully flat over $A[T]$ without assuming that $A[T]$ is Noetherian, then you can descend from one to the other.
Feb 11, 2017 at 16:07 comment added M.G. Also, I removed "heavy" as it is somewhat subjective.
Feb 11, 2017 at 16:06 history edited M.G. CC BY-SA 3.0
removed "heavy" as too subjective
Feb 11, 2017 at 15:44 comment added M.G. @nfdc23: No, the point is not the avoid some version of Hilbert's Basis Theorem altogether. I guess the rationale is one of exposition: prove the power series version of HIB as the (seemingly) "more general result" and then give some (short) argument why it subsumes the polynomial version. The only such "argument" I am aware of goes by saying that the proof is nearly the same except we consider the highest non-zero coefficient and so on. On the other hand, we do have a proper argument if we go from $A[T]$ to $A[[T]]$, namely completion.
Feb 11, 2017 at 15:26 comment added nfdc23 How would one be in the situation of having proved $A[\![T]\!]$ is noetherian while regarding the Hilbert Basis Theorem to be a "heavy" tool to be avoided? Very little can be done in the theory of noetherian rings with using the Hilbert Basis Theorem, so the rationale to want to avoid it is unclear. Moreover, the preservation of the noetherian property under completion is a harder fact than the Hilbert Basis Theorem (and its proof uses much more). So it would help to hear more about the rationale.
Feb 11, 2017 at 13:50 history asked M.G. CC BY-SA 3.0