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Mathematics
Linear and Abstract Algebra
Generating modules and sub modules Blyth Theorem 2.3
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[QUOTE="fresh_42, post: 6003281, member: 572553"] The essential point of all such definitions is the finiteness of sums. All coefficients, resp. summands up to finitely many have to be zero. It can be different sets ##J \in \mathbb{P}^*(I)## from case to case, but in any sum there are still only finitely many. As to your question: Theorem ##2.3## starts: Let ##(M_i)_{i\in I}## be a family of ##R-##submodules of an module ##M##. But as submodules, ##\lambda m = m'## is basically the same, i.e. it isn't necessary to carry the ##\lambda## all along the road, if the statement only requires ##m'##. The more, as we additionally have to bother the index sets: ##m=\sum_{i\in J_1}\lambda_i m_i\; , \;m'=\sum_{j \in J_2}\lambda'_j m_j\; , \;m+m'=\sum_{k \in J_1\cup J_2}(\lambda_k m_k+\lambda'_km'_k)## where ##\lambda_k=0## for ##k\in J_2-J_1## and ##\lambda'_k=0## for ##k\in J_1-J_2## ... etc. I understand, that it makes sense to drop all this unnecessary stuff and write ##m+m'=\sum_{k=1}^nm_k##. If we talk about linear combinations, the ##\lambda## are necessary, because they are what makes it linear. If we talk about elements of modules, then ##\lambda_k m_k=m'_k\in M_k## both carry the same information, namely being elements of ##M_k##. [/QUOTE]
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Mathematics
Linear and Abstract Algebra
Generating modules and sub modules Blyth Theorem 2.3
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