Prove that Z15 is isomorphic to Z3 x Z5. Give an explicit formula for an isomorphism and prove that it does indeed provide an isomorphism between these two rings.
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One explicit isomorphism is
f(n) = (n*, n**)
with n* = the residue of n modulo 3
n** = the residue of n modulo 5
Proof this is a homomorphism:
f(n) + f(m) =
(n*,n**) + (m*, m**) = ...by definition of group product
(n* + m*, n** + m**) = ...by nature of modular residues
((n + m)*, (n + m)**) = ... by definition of f
f(n + m)... as desired
Proof of bijection:
This follows from the Chinese remainder theorem, which states that
n = x (mod 3)
n = y (mod 5)
must have a unique solution modulo 3*5 = 15. Since the solution exists, f is onto. Since the solution is unique, f is one-to-one.
Since f is a bijective homomorphism, it is an isomorphism. So, Z15 =~ Z3 x Z5
QED
f(n) = (n*, n**)
with n* = the residue of n modulo 3
n** = the residue of n modulo 5
Proof this is a homomorphism:
f(n) + f(m) =
(n*,n**) + (m*, m**) = ...by definition of group product
(n* + m*, n** + m**) = ...by nature of modular residues
((n + m)*, (n + m)**) = ... by definition of f
f(n + m)... as desired
Proof of bijection:
This follows from the Chinese remainder theorem, which states that
n = x (mod 3)
n = y (mod 5)
must have a unique solution modulo 3*5 = 15. Since the solution exists, f is onto. Since the solution is unique, f is one-to-one.
Since f is a bijective homomorphism, it is an isomorphism. So, Z15 =~ Z3 x Z5
QED