By Toma Albu, Robert Wisbauer (auth.), S. K. Jain, S. Tariq Rizvi (eds.)
"[The publication] exhibits advancements in lots of topics of this very lively box of jewelry and modules and it encompasses a wealth of recent principles, options and effects added via one of the most very important researchers within the field..."
--Mathematica
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O Since B n Be ~ Be, C = Be. Recall that C :::; M is a complement of A :::; M if C is maximal among submodules B :::; M such that An B = o. A Zorn's lemma argument shows that every submodule of a given module has a complement in that module. 4. A module M is supplemented if for any two submodules A and B with A + B = M, B contains a supplement of A in M. The notion of supplement is dual to that of complement. Note that not every submodule of M need have a supplement. Hence, not all modules are supplemented.
The closure of a submodule A ~ M is maximal among essential extensions of A in M. 9. Let MR be a supplemented module and let B B·· be an interior of B in M. +- B·· If Y ~ B··' ~ M. +- y' (2) then Y = B··. BUM Proof. (1): Suppose B·· < U < M and -B + - = - . Then M = B+ •• B·· B·· U. Since M = B· +B··, we have U = B·· +(B·nU), so M = B+{B·nU). e. B· :::; U. Hence M = B· + B·· :::; U so U = M. B B· + Y M B· + Y M (2): Y + Y = Y so Y = y' Hence B· + Y = M. Since B·· is a supplement of B· in M, Y = B··.
The map 4>: is well defined, for if, aeii E (I, J)i and 4>( aeii) = peii + qeTi+l then, as aeiiei+1i+l = 0, we have q = o. 4>: is clearly a K-homomorphism. Since Q is injective as K -module, there exist a E Q such that 4>: (b) = ab for all b E 1;. If 4>i (aeii + beTi+ 1) = peii + qeTi+ 1 then peTi+1 = (peii + qeTi+l)eTi+1 = 4>i(aeTi+l)' so that 4>:(a) = p. Thus P = aa, q = ab. It follows that 4>i(aeii + beTi+l) = aaeii + abeTi+l = aeii(aeii + beTi+l)' Define 4>:+1 : I H1 ---+ Q as follows: if 4>Hl( ae i+li+1 + beT+1H2) = qeii+1' set 4>~+1(a) = q.