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In abstract algebra, the direct sum is a construction which combines several modules into a new, larger module. The direct sum of modules is the smallest module which contains the given modules as submodules with no "unnecessary" constraints, making it an example of a coproduct. Contrast with the direct product, which is the dual notion.
A decomposition with local endomorphism rings [5] (cf. #Azumaya's theorem): a direct sum of modules whose endomorphism rings are local rings (a ring is local if for each element x, either x or 1 − x is a unit). Serial decomposition: a direct sum of uniserial modules (a module is uniserial if the lattice of submodules is a finite chain [6]).
The direct sum is also commutative up to isomorphism, i.e. for any algebraic structures and of the same kind. The direct sum of finitely many abelian groups, vector spaces, or modules is canonically isomorphic to the corresponding direct product. This is false, however, for some algebraic objects, like nonabelian groups.
Download as PDF; Printable version; ... Direct sum of modules; Dual module; E. Eilenberg–Mazur swindle; ... Stably free module;
A decomposition of a module is a way to express a module as a direct sum of submodules. dense dense submodule determinant The determinant of a finite free module over a commutative ring is the r-th exterior power of the module when r is the rank of the module. differential A differential graded module or dg-module is a graded module with a ...
Torsionfree modules over a Dedekind domain are determined (up to isomorphism) by rank and Steinitz class (which takes value in the ideal class group), and the decomposition into a direct sum of copies of R (rank one free modules) is replaced by a direct sum into rank one projective modules: the individual summands are not uniquely determined ...
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The direct sum of two nonzero uniform modules always contains two submodules with intersection zero, namely the two original summand modules. If N 1 and N 2 are proper submodules of a uniform module M and neither submodule contains the other, then M / ( N 1 ∩ N 2 ) {\displaystyle M/(N_{1}\cap N_{2})} fails to be uniform, as