Sentence examples for maximal submodule of from inspiring English sources

Let (N) be a maximal submodule of (M).

(iii) P is called a graded maximal submodule of M if there is no graded submodule K of M such that P ⊂ K ⊂ M.

Let R be a graded ring and M be a graded multiplication module over R. Let P be a graded maximal submodule of M.

Since we know that Im Hom ( C, f ) is a maximal submodule of Im Hom ( C, f ′ ), it follows that Im Hom ( C, f ′ ) / Im Hom ( C, f ′ ) ≃ top Im Hom ( C, f ′ ϕ ) ≃ top Hom ( C, C 0 ).

The map g : R [ t ] → Q j constructed there is a generator of the maximal submodule of Hom ( R [ t ], Q j ) and is used in the proof of Lemma 14.2 in order to show that S Hom ( R [ t ], Q j ) is of the form I ( d ) for some d.

Given a maximal submodule U of Hom ( C, Y ), we may interpret it as a maximal submodule of Y b = Hom ( C, Y ), and we may consider the submodule Λ U of Y generated by U, let u : Λ U → Y be the inclusion map.

Clearly, two distinct graded maximal submodules of a graded module are graded coprime.

The maximal submodules of Y are pairwise non-isomorphic and these are, up to isomorphism, all the strongly regular modules of length | Y | - 1.

In this case, the maximal submodule η C Y ( f : M → Y ) (thus a set of morphisms) uniquely determines the module M. In addition, we will assume that C is a brick, or at least that C is indecomposable and rad End ( C ) annihilates Hom ( C, Y ).

This means that every element of C [ → Y 〉 different from the identity map Y → Y is less or equal to [ g 〉. This means that C [ → Y 〉 has a unique maximal submodule, namely Hom ( C, g ). □.

According to Lemma 14.8, the inclusion map X → Y shows that X is even strongly regular, and of course of length | Y | - 1. Conversely, if X is strongly regular and of length | Y | - 1, Lemma 14.8 yields a monomorphism X → Y. Now assume that two maximal submodules X, X ′ are isomorphic, let f : X → Y and f ′ : X ′ → Y be the inclusion maps.