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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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    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?2012-04-25

2 Answers 2

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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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    Ok, this is good advice, thanks.2012-04-25