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Homework3

# Homework3 - Aaron Goldsmith Homework 3 Section 4.1 Algebra...

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Aaron Goldsmith Homework 3 Section 4.1 Algebra II 4 Let Rx be a unitary cyclic R-module and Define φ : R Rx as r 7→ rx . Then, φ is an R-module homomorphism if R is considered a module over itself: φ ( r + s ) = ( r + s ) x = rx + sx = φ ( r ) + φ ( s ) φ ( rs ) = ( rs ) x = r ( sx ) = ( s ) By a homomorphism theorem, we have Rx = R/ ker φ , and ker φ is a left ideal of R . 9 Let a Ker f Im f with f ( b ) = a . Then, 0 = f ( a ) = f ( f ( b )) = f ( b ) = a and the intersection is trivial. Next, for any a A , we have f ( a - f ( a )) = f ( a ) - f ( f ( a )) = f ( a ) - f ( a ) = 0 and the representation a = ( a - f ( a )) + f ( a ) Ker f + Im f Moreover, this representation is unique for if ( a - f ( a )) + f ( a ) = k + f ( l ) Ker f + Im f then f ( l ) - f ( a ) = a - f ( a ) - k Ker f so f ( l ) - f ( a ) = 0 and f ( l ) = f ( a ). It follows that a - f ( a ) = k . The function ( k, y ) 7→ k + y is then an isomorphism. 11 a Use the ideal test. 1. 0 ∈ O a is nonempty. 2. Given r, s ∈ O a , then ( r + s ) a = ra + sa = 0. Thus, r + s ∈ O a . 3. Given r R , s ∈ O a then ( rs ) a = r ( sa ) = 0 and rs ∈ O a . 1

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b By definition of submodule, 1. O 0 = R and 0 T ( A ) 6 = . 2. If r R, a, b T ( A ), then r ( a - b ) = ra - rb = 0 and a - b T ( A ) 3. if r R, a T ( A ), then 0 = ra T ( A ). c Take A = R = Z 6 . Then 2 , 3 T ( A ) but 2 + 3 = 5 6∈ T ( A ). d Let r R, a T ( A ). Then 0 = f (0) = f ( ra ) = rf ( a ) and f ( a ) T ( B ). e Ker f T Ker f = { 0 } , meaning Ker f T = { 0 } . Let b Im f T . Then, b Im f = Ker
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Homework3 - Aaron Goldsmith Homework 3 Section 4.1 Algebra...

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