Proceedings of the 44th Symposium on Ring Theory and ...

4. Examples
Example 6. Let M = M(A 5 B7 −1 B 4 ), N = M(A 6 B4 −1 ). By Proposition 1,
l(M ⊗ N) = max { l(M(w) ⊗ M(w ′ )) | w ∈ {A 5 , B7 −1 , B 4 }, w ′ ∈ {A 6 , B4 −1 } }
= l(M(B −1
7 ) ⊗ M(A 6 )) = l(M(B 7 ) ⊗ M(A 6 )) .
Since 7 ̸⊥ 6, by Theorem 4:1 we have, l(M(B 7 ) ⊗ M(A 6 )) = 2 + 7#6 = 9. Hence, <strong>the</strong>
Loewy length <strong>of</strong> M ⊗ N is 9 and, seen as a kD 16 -module, M ⊗ N has a projective direct
summand.
Example 7. By Proposition 1 and <strong>the</strong> inequality (3.1), we have
l(M(A l ) ⊗ M(A m+1 B −1
m )) = max{l(M(A l ) ⊗ M(A m+1 )), l(M(A l ) ⊗ M(B m ))}
= l(M(A l ) ⊗ M(A m+1 )) .
for all l, m ∈ N.
Example 8. While it is clear that l(M(A l B −1
l
, 1)⊗N) l(M(A l B −1
l
)⊗N), <strong>the</strong> difference
between <strong>the</strong> lengths <strong>of</strong> <strong>the</strong> two tensor products may be zero, or arbitrarily large. For
example, if N = M(A m ) and l ̸⊥ m, <strong>the</strong>n l(M(A l B −1
l
, 1) ⊗ N) = l(M(A l B −1
l
) ⊗ N) by
Theorem 4:2. If, on <strong>the</strong> o<strong>the</strong>r hand, l = 2 r and m = 2 s with r > s <strong>the</strong>n
while
l(M(A l B −1
l
, 1) ⊗ M(A m )) = 2 + (l − 1)#m = 2 + 2 r − 1 = 1 + 2 r ,
l(M(A l B −1
l
) ⊗ M(A m )) = l(M(B l ) ⊗ M(A m )) = 1 + l + m = 1 + 2 r + 2 s .
Example 9. Let M = M(a) = M(A 1 ) and N = M(b(ab) l ) = M(B 2l+1 ) for some l ∈ N.
Now 1 ̸⊥ (2l + 1), and 1#(2l + 1) = 2l + 1, so by Theorem 4:1, l(M ⊗ N) = 2l + 3. In this
case, <strong>the</strong> Loewy length actually provides <strong>the</strong> missing piece <strong>of</strong> information to compute <strong>the</strong>
isomorphism type <strong>of</strong> M ⊗ N.
Namely, since k is <strong>the</strong> unique simple module, we have
and similarly,
dim soc(M ⊗ N) = dim Hom kD4q (k, M ⊗ N) = dim Hom kD4q (N ∗ , M)
= dim Hom kD4q (N, M) = 1
dim top(M ⊗ N) = dim Hom kD4q (M ⊗ N, k) = 1 .
Hence M ⊗ N is a module with simple top and simple socle, <strong>of</strong> dimension 4(l + 1), and
Loewy length 2l + 3. A module satisfying <strong>the</strong>se conditions is indecomposable, and must
be isomorphic to M(A 2l+2 B −1
2l+2
, ρ) for some ρ ∈ k {0}. Now if k is <strong>the</strong> prime field, that
is <strong>the</strong> Galois field with two elements, this means that ρ = 1. From this follows that ρ = 1
also in <strong>the</strong> general case, since extension <strong>of</strong> scalars commutes with taking tensor products.
Hence, we have M ⊗ N ≃ M(A 2l+2 B −1
2l+2 , 1).
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