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Hailong Dao
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answered
On the map $\Phi_M: M\otimes_RM^*\xrightarrow{x\otimes y\mapsto \left\{f\mapsto f(x)y\right\}}\text{Hom}_R(M^*,M^*) $
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Thomason's "open letter" to the mathematical community
Why are people voting to close this question? The OP did not ask others to comment on the content of the letter. He simply wants a copy of the letter, and asking the MO community seems like a pretty reasonable move.
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Why is $M$ torsion-free?
@user26857: yes, of course.
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Tensor product by the canonical module preserves Cohen-Macaulayness
Are you sure you want tensor product and not Hom?
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Why is $M$ torsion-free?
@Danimenru: en.wikipedia.org/wiki/…
answered
Why is $M$ torsion-free?
awarded
 Nice Answer
awarded
 Necromancer
awarded
 Revival
answered
How often do you put your research into trash?
answered
Intersecting a ideal generated in degree $\leq a$ with one generated in degree $\leq b$ in a polynomial ring
answered
Ideals in Artinian Gorenstein local ring $(R,\mathfrak m)$ with $\mu(\mathfrak m)=2, \mathfrak m^2\ne 0$ and $\mathfrak m^3=0$
accepted
When is a triangulation of sphere two-colorable?
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When is a triangulation of sphere two-colorable?
@FedorPetrov: that sounds reasonable. Is there a reference, and if not, would you mind giving a full answer? Thanks.
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When is a triangulation of sphere two-colorable?
Yes, but one can formulate the question in higher dimensions.
asked
When is a triangulation of sphere two-colorable?
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When is a maximal Cohen-Macaulay module of finite homological dimension?
The Auslander-Buchsbaum formula would force such $M$ to be projective. en.wikipedia.org/wiki/Auslander%E2%80%93Buchsbaum_formula
awarded
 Nice Answer
comment
When are rings of the form $K[x_1,...,x_n]/(Q)$ principal ideal domains when $Q$ is quadratic?
I don't think they will ever be. The ideal $(x_1,...,x_n)$ will not be principal unless $n=1$, in which case it will not be domain.
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Is Koszul homology of a monomial ideal always generated by the "obvious" things?
@Rellek: Also open, as far as I know.
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