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Mathematics
Linear and Abstract Algebra
Directly Finite and Directly Infinite R-Modules - Bland S2.2
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[QUOTE="fresh_42, post: 5485933, member: 572553"] Hi Peter! You can settle an isomorphism ##⊕_{ℕ} ℤ ≅ (⊕_{2ℕ} ℤ) ⊕ (⊕_{2ℕ+1} ℤ)## by the following map: If ##a_n## denotes an element ##a## at the ##n##-th position, i.e. an element of ## ⊕_{ℕ} ℤ##, then we map it to ##(a_{2k},0)## if ##n=2k## is even and to ##(0,a_{2k+1})## if ##n=2k+1## is odd. Again the index only shows the position of ##a## and I wrote the elements of ##(⊕_{2ℕ} ℤ) ⊕ (⊕_{2ℕ+1} ℤ)## as pairs. It doesn't matter, you could write it as well as a sum. I simply felt that ##a_{2k}+ 0## might be confusing. So it remains to show that, e.g. ##⊕_{ℕ} ℤ ≅ ⊕_{2ℕ} ℤ##. This could be done by mapping ##a_n## to ##a_{2n}##. [/QUOTE]
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Directly Finite and Directly Infinite R-Modules - Bland S2.2
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