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Suppose $\Delta(n,k)$ is the algebra of upper triangular $n$ by $n$ matrices over a field $k$. Furthermore, let $M$ is an artinian module over $\Delta(n,k)$, and let $$ \cdots\to C_n\stackrel{d_n}{\to}C_{n-1}\to\cdots\to C_1\stackrel{d_1}{\to}C_0\stackrel{\epsilon}{\to} M\to 0 $$ be a projective resolution of $M$, so $\epsilon$ is a homomorphism and $\epsilon d_1=0$.

Edited: Based on the comments I received, I'm trying to revise the question to something less abstract. Although I see now that the projective resolution need not be finite, is it always possible to truncate it to yield a projective resolution of length two? Is there some justification for this, or have I misunderstood what I've been told?

Thanks.

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Dear Koji, An Artinian module is a successive extension of simple modules, so if your statement is true for an Artinian module, it will be true in particular for simple modules. Conversely, if it is true for two Artinian modules, it will be true for any extension of the two of them, so in fact it suffices to verify it for simple modules. Now, what are the simple modules over the upper triangular matrices? Once you identify them, you can try to compute their projective resolutions by hand. (I think this will be more profitable than trying to find abstract arguments.) Regards, –  Matt E Apr 23 '12 at 2:35
    
The resolution does not have to be finite, however you can truncate it at some point and get a finite projective resolution. For example, take $M=A$, $C_n=A$ for all $n$, $d_i=0$ if $i$ is odd and $d_i=id$ if $i$ is even. This is an infinite resolution of $A$ but you can truncate it at $C_1$ and get the (finite) resolution $A\to A$. –  Quimey Apr 23 '12 at 2:56
    
Thanks for your comment @MattE. I believe the simple modules are just those where each upper triangular matrix acts by multiplication on its $i$-th diagonal entry, correct? These are one dimensional, so does it somehow follow that the projective resolutions are necessarily of length two? –  Koji Hamada Apr 25 '12 at 20:20

2 Answers 2

One way to sort of sidestep the construction of the quiver and all is to show that all left ideals in your algebra are projective, in other words, that the algebra is hereditary.

From this, it is easy to see that resolutions are of length $1$ (starting from zero!)

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Your algebra is isomorphic to the path algebra of a quiver of type $A_{n-1}$. Quiver algebras are hereditary: submodules of projective modules are projective. Thus minimal projective resolutions of non-projective modules have length two.

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Thanks mt_. Is there an explanation that doesn't resort to the path algebra of a quiver? I haven't heard of those before. –  Koji Hamada Apr 25 '12 at 20:09
    
This property of having projective resolutions of length at most two is very special: essentially it happens only for path algebras. If you don't want to use the path algebra, you will need to compute explicitly projective resolutions of simple modules. To begin doing this you need the structure of the projective indecomposables. By the time you've worked all this out, you probably could have learned enough of the basics of quiver algebras to understand my answer... –  mt_ Apr 25 '12 at 20:48
    
Ok, this is good advice, thanks. –  Koji Hamada Apr 25 '12 at 21:53

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