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Jan
24
comment Merge operation in homological algebra?
nevertheless, if you want to play around with the combinatorics in the example above. Syzygy $\Omega(M)$ of $M$ is: $Q=(1\to 2)$, $\Omega(1)=2$, and $2,\substack{1\\2}$ are projective, so no syzygy. For $Q=1\to 3\leftarrow 2$, $\Omega(1)=\Omega(2)=\Omega(\substack{1 2 \\ 3})=3$, the rest are projectives, so no syzygy.
Jan
24
comment Merge operation in homological algebra?
as I said, I don't understand your idea, just as Tobias commented - we don't see why what you want to do has anything to do with homological algebra. To do homological algebra, one needs to start with some abelian category, which means one want to start with some geometric space or a ring, then consider sheaves or modules (respectively), then one can start doing homological algebra. in particular, you can only talk about syzygy of one particular sheave/module. I don't see why you are obsessed with a translation in hom. alg. (and in particular) syzygy, so I can't really answer your question.
Jan
22
answered Merge operation in homological algebra?
Dec
31
awarded  Tumbleweed
Dec
27
asked Is there any study on noncrossing partitions for cyclic polytopes?
Dec
24
asked Is the notion of strongly graded algebra a Morita invariant?
Dec
22
comment Understanding the Definition of the Tensor Product of Chain Complexes
for explanation of Koszul sign rule, read this: math.umn.edu/~tlawson/papers/signs.pdf
Dec
22
comment Extension of nonisomorphic simple objects
If you have a cycle, then the path algebra becomes infinite dimensional, and representation theory on that algebra may not work with usual quiver representation machinery. For example, $Q$ with one vertex one arrow gives path algebra $\Bbbk[x]$ (the polynomial ring).
Dec
21
comment Extension of nonisomorphic simple objects
finite oriented tree relaced by finite oriented acyclic graph is OK too (if you don't like the word (finite acyclic) quiver). Also, this example is actually the algebra of upper triangular matrix over $\Bbbk$ if one doesn't like quiver representation. The two simple modules correspond to having only entry in $(1,1)$ and $(2,2)$ (with obvious action), then the (unique) non-split extension is associated to the algebra itself.
Dec
9
comment Is there a non-projective submodule of a free module?
I like to think in finite dimensional algebras, and the example I would give is take $R$ the dual number $k[x]/(x^2)$. Obviously $R$ itself is free and projective, and the 1-dim $k$-space $k.x$ spanned by $x$ is a $R$-submodule of $R$, and $k.x\cong R/k.x$ as $R$-module. So we have non-split exact sequence $0\to k \to R\to k\to 0$ which says that $k$ is non-projective submodule of free module $R$.
Dec
3
comment matrix coefficients are representative functions.
when $V$ f.d., say of dim $n$, then $\Phi(g)$ is an $n$-by-$n$ matrix, after choosing an appropriate basis, a matrix coefficient is then a function extracting the $(i,j)$-th component of this matrix (hence the name), and obviously the converse is true. And so the space of matrix coefficients is finite dimensional (at most $n^2$), i.e. matrix coefficients are representative functions.
Dec
2
answered Should Ext-quiver be a full sub-quiver of its AR-quiver for a basic hereditary algebra A over algebraic closed field K?
Nov
24
comment permutation representation of Symmetric group
If you know about Specht module, then it is easy exercise to see that this representation (called natural representation) can be decomposed into $S^{(n)}$ and $S^{(n-1,1)}$. Discretely $S^{(n)}=\mathbb{C}-span\{v_1+\cdots+v_n\}=$ trivial representation, where $v_1,\ldots,v_n$ is the standard basis of the natural rep. $S^{(n-1,1)}$ is the kernel of the projection of the natural representation project onto the trivial representation. The natural rep is the induction of trivial representation as rep of trivial group to $S_n$, so it is a permutation representation by definition.
Nov
3
comment How to compute $Ext_A^{1}(S_1, S_2)$ and $Ext_A^{1}(S_2, S_1)$?
By the way, in general, whether to count $i\to j$ or $j\to i$ depends on how you define Ext-quiver, or which way you work (i.e. left modules or right modules, and direction of multiplying arrows). In your example, we see from how you write down $P(1)$ to determine which direction is the correct one.
Nov
3
answered Compatibility of homomorphisms and quotient maps of abelian groups
Nov
2
comment Compatibility of homomorphisms and quotient maps of abelian groups
How (1) is used is the same (dual) thing of my previous comment: you need $h$ to be well-defined, i.e. any difference in $a$ by an element in $A'$ will give you the same element in the image. Here you will need to use the commutative square of the left.
Nov
2
comment Compatibility of homomorphisms and quotient maps of abelian groups
$g(a'):=f(a')$ is not quite the correct way as the element is in $C$, not in $C'$ by trivial reason. You should think, I have the diagram with the lower left corner taken out, and I have element in $A'$, how to get to some element in $C'$? Well, if you can push the element all the way to lower right corner ($C/C'$) and show that the result element is $[0]=0+C'\in C/C'$, that means you have something lying in $C'$.
Nov
2
comment How to compute $Ext_A^{1}(S_1, S_2)$ and $Ext_A^{1}(S_2, S_1)$?
I will just clarify Mariano and Julian's comment for the OP. The mistake in the calculation is that the ``projective resolution" you quoted works only for module of projective dimension 1. If you actually calculate the kernel of $P_1\to S_1$, you will see it is $S_2$, so the resulting ses is $0\to S_2\to P_1\to S_1\to 0$. In particular, the non-zero element of $Ext_A^1(S_1,S_2)$ corresponds to this non-split extension. If you only want the dimension of the Ext-groups between simples, then all you need to do is to count the number of arrows from one vertex to another -no calculation needed.
Oct
27
awarded  Revival
Oct
24
comment Semisimple modules and the radical
What is your condition on $A$? If $A$ is finite-dimensional algebra, then $\text{rad}V=\text{rad}AV$, and since $A/\text{rad}A$ semisimple, so is $V/\text{rad}V$.