Pilch 5) If NM, and M/N a R* is free, then M = (M/N) - N. Hint: Lise Exercise 6.214) to show that if N... Nabosis of MIN, MEN internal direct sum We will sometimes want to be able to take the direct sum of infinitely many modules, not just finitely many. With Exercise 6.4.741) in mind, this should be done in such a way that if M, R for all 1, then e M is a free module, and a basis is given by leil, where has only one nonzero entry, which is in the ith spot. We begin by generalizing Definition 6.4.2 to allow bases that are infinite. Pilch 5) If NM, and M/N a R* is free, then M = (M/N) - N. Hint: Lise Exercise 6.214) to show that if N... Nabosis of MIN, MEN internal direct sum We will sometimes want to be able to take the direct sum of infinitely many modules, not just finitely many. With Exercise 6.4.741) in mind, this should be done in such a way that if M, R for all 1, then e M is a free module, and a basis is given by leil, where has only one nonzero entry, which is in the ith spot. We begin by generalizing Definition 6.4.2 to allow bases that are infinite