1,288 reputation
718
bio website dpmms.cam.ac.uk/~sjw47
location Cambridge
age 36
visits member for 4 years, 6 months
seen yesterday

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