bio  website  wwwmath.unimuenster.de/u/… 

location  Münster, Germany  
age  27  
visits  member for  4 years, 9 months 
seen  yesterday  
stats  profile views  34,024 
I am interested in the interactions between algebraic geometry and category theory. More specifically, I "model" algebraic geometry on cocomplete symmetric monoidal categories.
Here is a link to my PhD thesis. Comments are welcome.
Email: [my last name] [at] unimuenster.de
20h

awarded  Nice Question 
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Classification of rings satisfying $a^4=a$
I've just seen your edit. Thank you, I will look at it! 
2d

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Open covering of the Hilbert functor of points
On the other hand, I don't want to disturb authors will silly questions. 
2d

accepted  Open covering of the Hilbert functor of points 
2d

answered  Open covering of the Hilbert functor of points 
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Classification of rings satisfying $a^4=a$
In his book "Modules Over Commutative Regular Rings", Pierce studies as an example rings satisfying $a^n=a$ and their corresponding structure sheaves. But does the book also offer a classification? 
2d

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Classification of rings satisfying $a^4=a$
Hm, $\mathcal{X}$ is equivalent to the category of boolean rings with an action of $C_2$. Is there a direct algebraic way to see that this is $\mathcal{R}$? 
Sep 27 
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Classification of rings satisfying $a^4=a$
Yes, I've studied this result and its proof in detail. Probably the shortest and quickest proof ($\approx$ one page) is given in "Elementary proofs of a theorem of Wedderburn and a theorem of Jacobson" by Nagahara and Tominga. 
Sep 27 
revised 
Classification of rings satisfying $a^4=a$
added 347 characters in body; edited tags 
Sep 27 
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Classification of rings satisfying $a^4=a$
Thanks, but that seems to me what I already know. I am interested in a more explicit classification (or even better, an equivalence of categories). 
Sep 27 
revised 
Classification of rings satisfying $a^4=a$
added 490 characters in body 
Sep 27 
asked  Classification of rings satisfying $a^4=a$ 
Sep 15 
awarded  Popular Question 
Sep 13 
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When does sheaf cohomology commute with arbitrary direct sums?
Direct sums are filtered colimits of finite direct sums. And finite direct sums are no problem. 
Sep 10 
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Canonical presentation of promodules over prorings
The maps $M_{j+1} \to M_i$ are surjective. It follows easily that each projection $\widehat{M} \to M_j$ is surjective (construct inverse images recursively). Hence, also $\alpha_j$ is surjective. 
Sep 10 
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Canonical presentation of promodules over prorings
From $K_{ji} = M_{j+i} I_{ij}$ for all $i$ we only get $\widehat{K_j} = \varprojlim_i \, (M_{j+i} \cdot I_{ji})$, but why does this equal $\varprojlim_i \, M_{j+i} \cdot \varprojlim_i \, I_{ji} = \widehat{M} \widehat{I_j}$? Again, $\subseteq$ is unclear. 
Sep 10 
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Canonical presentation of promodules over prorings
Thank you. But $\widehat{K_j} = \widehat{I_j} \widehat{M}$ is unclear to me (and clearly this is a direct reformulation of the claim). How do you prove $\subseteq$? 
Sep 8 
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Canonical presentation of promodules over prorings
Sure, the isomorphisms belong to the data. A morphism $M \to N$ is a family of morphisms $M_i \to N_i$ compatible with the isomorphisms. Actually all this follows from the definition $\mathcal{M} := \varprojlim_i \, \mathsf{Mod}(A_i)$. 
Sep 8 
asked  Canonical presentation of promodules over prorings 
Aug 30 
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A short proof for $\dim(R[T])=\dim(R)+1$
R. Gilmer is the author, but of which paper? 