1,338 reputation
819
bio website dpmms.cam.ac.uk/~sjw47
location Cambridge
age 36
visits member for 4 years, 10 months
seen yesterday

My main interest is in non-commutative algebra. At present that mostly means working with Iwasawa algebras and related rings.


Jul
22
reviewed No Action Needed Approximation of bounded continuous functions by Lispschitz bounded functions
Jul
22
comment Isomorphism of matrix ring over ore domain
That the maximal right ring of quotients of $Mat_n(R)$ is $Mat_n(q.f(R))$ (in your notation) follows immediately from Corollary 3.1.6 of McConnell and Robson's book `Noncommutative Noetherian rings': ams.org/bookstore-getitem/item=GSM-30
Jul
9
awarded  Pundit
Jul
2
awarded  Curious
Jun
19
comment About the construction of the Universal Enveloping Lie Algebroid
By the way you should probably see section 1.2.5 of math.harvard.edu/~gaitsgde/grad_2009/BB%20-%20Jantzen.pdf if you haven't already.
Jun
19
answered About the construction of the Universal Enveloping Lie Algebroid
May
22
reviewed No Action Needed Implications of non-negativity of coefficients of arbitrary Kazhdan-Lusztig polynomials?
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.