bio | website | dpmms.cam.ac.uk/~sjw47 |
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location | Cambridge | |
age | 36 | |
visits | member for | 4 years, 6 months |
seen | yesterday | |
stats | profile views | 1,233 |
My main interest is in non-commutative algebra. At present that mostly means working with Iwasawa algebras and related rings.
Jan 27 |
accepted | Are admissible open subsets of an affinoid space of countable type? |
Jan 27 |
comment |
Are admissible open subsets of an affinoid space of countable type?
Ok. I see now. Thanks. |
Jan 24 |
comment |
Are admissible open subsets of an affinoid space of countable type?
I'm struggling to process this slightly. Do you have a reference for the first sentence? |
Jan 20 |
asked | Are admissible open subsets of an affinoid space of countable type? |
Dec 8 |
comment |
An invariant number of modules over Auslander Gorenstein modules
If my interpretation above is correct then I think you are asking about the quotient category $\mathcal{M}^\mu$ in the notation of that paper. |
Dec 8 |
comment |
An invariant number of modules over Auslander Gorenstein modules
I'm not completely sure what you are asking. Does being $\mu$-critical mean that $\mathrm{Hom}_R(M,R)\neq 0$ (ie canonical dimension $\mu$) but for every proper quotient $M/N$ of $M$, $\mathrm{Hom}_R(M/N,R)\neq 0$? In any case this paper math.washington.edu/~smith/Research/asz6.pdf of Ajitabh, Smith and Zhang is likely to be useful. |
Nov 18 |
comment |
When does the homological dimension of a tensor product equal the sum of dimensions?
A good reference for the above claim: degruyter.com/view/j/jgth.2000.3.issue-4/jgth.2000.034/… |
Oct 30 |
comment |
When to pick a basis?
If you can prove that (1) the trace of the identity map on $V$ is $\dim V$ (and this is an integer), (2) the trace of the zero map is $0$ and (3) trace is additive in the sense that if $T_1$ acts on $V_1$ and $T_2$ acts on $V_2$ then $\mathrm{tr} (T_1\oplus T_2)=\mathrm{tr} T_1+\mathrm{tr} T_2$ then what you ask for is straightforward since if $E$ is a projector on $V$ then it decomposes canonically as $I_{\ker E}\oplus 0_{\mathrm{Im} E}$. I would imagine that for any sensible definition of trace (1), (2) and (3) should be straightforward. |
Oct 24 |
comment |
Are all (possibly infinite dimensional) irreducible representations of a commutative algebra one-dimensional?
Quillen removed the cardinality assumption in ams.org/journals/proc/1969-021-01/S0002-9939-1969-0238892-4/…. |
Oct 14 |
awarded | Yearling |
Oct 13 |
awarded | Constituent |
Oct 1 |
awarded | Caucus |
Sep 19 |
comment |
$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?
I'm not even sure that it is easy to extend my strategy to $G=\mathbb{Z}^n$ for $n>1$, although it may be easier than I think. Classifying right ideals in $\mathbb{C}G$ for $G$ a discrete Heisenberg group is probably hard. To get an idea why, see arxiv.org/abs/math/0102190. The first sentence of section 7 of your reference suggests that $Tor^1_{\mathbb{C}G}(\mathbb{C}G/f\mathbb{C}G,l^1(G))=0$ for all non-zero $f\in \mathbb{C}G$ whenever $G$ is torsionfree polycyclic. However, it isn't clear to me how hard this result was to prove already in that generality. |
Sep 17 |
awarded | Revival |
Sep 17 |
comment |
$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?
It occurs to me that in the final paragraph one might as well assume that $f$ is irreducible in $R$. Since (by the fundamental theorem of algebra) such an $f$ is a unit times $(x-\lambda)$ for $\lambda\in \mathbb{C}$ non-zero it should be very easy to complete the case $G=\mathbb{Z}$ by hand. |
Sep 17 |
revised |
$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?
moved some stray commas. |
Sep 17 |
answered | $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module? |
Aug 24 |
revised |
D-modules on rigid analytic spaces
added 172 characters in body |
Jul 12 |
reviewed | No Action Needed Classifying Equivariant Maps Between Fin-Dim Irreducible Modules |
Jun 27 |
awarded | Custodian |