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| visits | member for | 1 year, 1 month |
| seen | 1 hour ago | |
| stats | profile views | 107 |
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May 17 |
comment |
Specific projective dimension of a module over bound quiver @JulianKuelshammer: yes! thanks for clarification! Actually now I also wonder if algebraically closed changes the situation... |
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May 17 |
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Specific projective dimension of a module over bound quiver @JackSchmidt: thanks, you are correct! |
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May 17 |
revised |
Specific projective dimension of a module over bound quiver added 339 characters in body |
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May 17 |
revised |
Specific projective dimension of a module over bound quiver added 299 characters in body |
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May 17 |
answered | Specific projective dimension of a module over bound quiver |
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May 16 |
answered | How to compute Nakayama functor explicitly? |
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May 15 |
comment |
Question about minimal projective presentations of a module. Have you try with example while you working through the book? You can always try something on the path algebra of $1\to 2$, or even $1\to 2\to 3$; and I mean do draw all the representations (modules) out. e.g. you have simples: $S_1=(k\to 0), S_2=(0\to k)$, projective: $P_1=(k\to k)$, then you see the projective presentation of $S_1$ is just $S_2\to P_1$ by the combinatorics alone, and you can now check this is a minimal projective presentation, then you can play something more to see what a non-minimal presentation would look like, etc. I always play with examples when I learn from this book |
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May 15 |
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Question about minimal projective presentations of a module. If $E_0'$ not isomorphic to $E_1''$, and there is a map $E_0'\to E_1''$, then you get non-trivial cokernel. If $E_0'$ maps bijectively to $E_1''$, then $E_0'$ cannot maps non-trivially to $E_1'$, contradicting the fact that $E_0'$ is projective cover of $ker(E_0'\to E_1')$ (minimality). |
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May 14 |
awarded | Enthusiast |
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May 13 |
comment |
Question about minimal projective presentations of a module. Say $P_X \to X$ is a projective cover, i.e. the map is minimal (or using your notation: superflous), so for any other surjective map $P \to X$ with $P$ projective, $P$ must contain $P_X$ as a direct summand. Of course a submodule of projective is rarely projective, but that's not the reasoning we were using. |
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May 12 |
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Are all of the irreducible representations of (any) symmetric group over $\mathbb{C}$ also irreducible over a finite splitting field. Regardless of the mess in your statements, I sort of understand your problem. The thing you have in mind is called the Specht module, which can be defined over $\mathbb{Z}$, parameterised by partitions of $n$ (bijective to ccl's of $S_n$). These modules are irreducible when extends to $\mathbb{C}$, but most of them are not irreducible anymore when extend to the modular splitting fields (when char$F$ divides $n!$). But they still have a nice property: if you quotient out this module by its radical, it is either irreducible or zero. |
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May 12 |
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Question about minimal projective presentations of a module. Yes, you are correct, shoudl be $Tr(M)$, I have edited it now. For your other question, you know $P_0^t$ and $P_1^t$ are projectives, so they have to be the direct sum of the minimal ones with some other projectives. |
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May 12 |
revised |
Question about minimal projective presentations of a module. added 4 characters in body |
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May 9 |
revised |
Finding the Galois group of $X^8-2$ over Q. fixed notation, and error on sigma_1 |
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May 9 |
suggested | suggested edit on Finding the Galois group of $X^8-2$ over Q. |
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May 8 |
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Expression of basis vectors of permutation modules in different bases. Sorry you are correct, it's Fulton's... |
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May 7 |
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Endomorphisms of Simple A-modules where A is a Complex algebra Your $M$ is not an $A$-module if $n\neq 2$; even so, your $\phi$ is not compatible with $A$-action on $M$, i.e. it is not an $A$-module map. |
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May 7 |
revised |
Path Algebra for Categories deleted 74 characters in body |
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May 7 |
answered | Path Algebra for Categories |
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May 6 |
awarded | Caucus |
