By R. L. Goodstein

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2: Let R be a commutative ring with 1, and n ≥ 1 be an integer. Mn×n = {(aij) / aij ∈ R; 1 ≤ i, j ≤ n} be the set of all n × n matrices with entries in R where a11 a12 . . a1n a 21 a 22 . . a 2 n . . (aij ) = . . . . a n1 an 2 . . ann Define addition and multiplication in Mn×n as follows: (aij) + (bij) = (aij + bij) n and (aij ) . (bij ) = (cij ) where cij = ∑ aik bkj for all i, j; 1 ≤ i, j ≤ n. It is easily k=1 verified that Mn×n is a ring called the matrix ring of order with entries from R.

Let S = {0, 1 + p1 + p2 + p3 + p4 + p5} be the subset of Z2S3. S is a S-pseudo ideal related to A and S is also a S-pseudo ideal related to Z2. 6: Let Z2 = {0, 1} be the prime field of characteristic 2, G any finite group of order n. Then Z2G has S-pseudo ideals which are ideals of Z2G. Proof: Take Z2 = {0, 1} a field of characteristic two and the group ring Z2G is a S-ring. Let G = {g1, g2, …, gn–1, 1} be the set of all elements of G. S = {0, 1 + g1 + … + gn–1} is a S-pseudo ideal related to Z2 and S is also an ideal of Z2G.

2: Let R be a ring say Z8. Now M = Z8 × Z8 is an abelian group under addition, M is a Z8 – module over Z8. 4: Let M be a module, M is called a simple module if M ≠ (0) and the only submodules of M are (0) and M. 3: Let R be a ring S = R × R is an R-module. (show M = R × {0} and N = {0} × R are not isomorphic as S-modules). PROBLEMS: 1. Let A and B be two submodules of a module M; prove A ∩ B is a submodule of M. 2. M = Z × Z × Z × Z × Z is a module over Z. 1. Find submodules of M. 2. Find two submodules which are isomorphic in M.

### Boolean algebra by R. L. Goodstein

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