We represent the kernel of K by ker K, and the image by Im KkerK={m∈M:K(m)=0N}PROOFf(0)=0N⟹0m∈kerK=∅Let m, n ∈kerK⟹ K(m) = 0N, K(n) = 0N .Then K(m)=K(n)=0N⟹K(m)−K(n)=0N⟹K(m−n)=0N(Since K is a homomorphism)⟹m−n∈kerKAlso for r ∈ R, a ∈ kerK, then K(a)=0N. By the definition of R-module homomorphismK(ra)=rK(a)=r0N=0Ni.eK(ra)=0N. So ra∈kerK. And kerK is a module
ImK={K(x):x∈M}Let n1,n2∈K(m). Then there exists m1,m2∈M such that K(m1)=n1,K(m2)=n2n1+n2=K(m1)+K(m2)=K(m1+m2)(Since K is a homomorphism)Therefore n1+n2∈ImKNow let r ∈ and n∈ImK. Then there exists m such that K(m) = n and rn = rK(m) =f(rm).But rn∈N, since N is an R-module. Therefore K(rm)∈ImK
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