By Wu X.

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Extra resources for 8-ranks of Class Groups of Some Imaginary Quadratic Number Fields

Example text

Hence it leaves irreducibles invariant, since they are determined up to isomorphism by their character according to the theorem. 3 Let M ∈ Suppose: m -mod and N ∈ N indn+m M; (i) indm+n mn M nm N m+n ind (ii) M N appears in resm+n mn M mn Then indm+n mn M n -mod be irreducible modules. N with multiplicity one. N is irreducible. Proof Suppose for a contradiction that K = indm+n N is reducible. mn M Then we can find a proper irreducible submodule S, and set Q = K/S. By Frobenius reciprocity, M N appears in resm+n m n Q with non-zero multiplicity.

1(i). 3 Let M ∈ n -mod and N ∈ n+1 -mod be irreducible modules, and a ∈ F . Then f˜a M N if and only if e˜ a N M. 4). 1(i). 4 Let M N ∈ n -mod be irreducible. Then f˜a M f˜a N if and only if M N . Similarly, providing a M a N > 0, e˜ a M e˜ a N if and only if M N . 1 The map ch K n -mod →K n -mod is injective. Crystal operators 50 Proof We need to show that the characters of the irreducible modules in n -mod are linearly independent in K RepI n . Proceed by induction on n, the case n = 0 being trivial.

Now t = hd = 0, as required. d (iii) We remind that h denotes the map with h t = n → th . Given (i), it is straightforward to check that f is a homomorphism of -bimodules. To see that it is an isomorphism, it suffices by dimension n to show that it is injective. Suppose t lies in the kernel, then f t h = ht = 0 for all h ∈ n . Hence t = 0 by (ii). Now we extend this result to n. 2 Define a linear map fw = for each f ∈ n n →d by f d−1 w if w ∈ dS , otherwise, 0 w ∈ Sn . Then: -bimodules; (i) is a homomorphism of (ii) the map f n is an isomorphism of → Hom n d n h→h -bimodules.