Tilted Special Biserial Algberas François Huard and Shiping Liu ...

is a direct summand of K. NotethatM(v1 −1 u) is not projective since uα = p is
non-zero. Therefore M(w) is of projective dimension greater than one. Using a
symmetric argument, one can show that M(w) is of projective dimension greater
than one if a s is the start-point of a binomial relation.
Suppose now that neither a s−1 nor a s is the start-point of a binomial relation.
We may assume that p s u is non-zero. Let v be the path such that q s−1 v is a
string ending in a deep. Then M(v −1 u) is a direct summand of K. Note that
M(v −1 u) is not projective since uα is non-zero. Therefore M(w) is of projective
dimension greater than one. The proof is completed.
We shall now find some necessary conditions for a string module to be of
projective dimension greater than one.
2.5. Lemma. Let A = kQ/I be a special biserial algebra. Let
w = p −1
1 q 1 ···p −1
n
be a string in (Q, I), wheren ≥ 1, thep i and the q j are paths which are nontrivial
for 1