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## Main Question or Discussion Point

I was proving 2Z is non-isomorphic to 3Z

I tried it by contradiction, of course.

If possible there exists a ring homomorphism f : 2Z ---> 3Z

Then,

f (x.y) = f(x).f(y) must hold

x,y belong to 2Z

So x = 2a, f.s. a belonging to Z

y = 2b, f.s. b belonging to Z

so x.y = 2z, f.s. z belonging to Z

Also, f(x) belongs to 3Z, and so does f(y)

=> f(x).f(y) = 3n , f.s. n belonging to Z

So,

f(2z) = 3n

f(2).f(z) = 3n, (since f is a ring homomorphism)

which is a meaningless statement, since f is not defined for all z belonging to Z, but only for elements of the form 2z.

Is it correct ? Someone got something neater ?

I tried it by contradiction, of course.

If possible there exists a ring homomorphism f : 2Z ---> 3Z

Then,

f (x.y) = f(x).f(y) must hold

x,y belong to 2Z

So x = 2a, f.s. a belonging to Z

y = 2b, f.s. b belonging to Z

so x.y = 2z, f.s. z belonging to Z

Also, f(x) belongs to 3Z, and so does f(y)

=> f(x).f(y) = 3n , f.s. n belonging to Z

So,

f(2z) = 3n

f(2).f(z) = 3n, (since f is a ring homomorphism)

which is a meaningless statement, since f is not defined for all z belonging to Z, but only for elements of the form 2z.

Is it correct ? Someone got something neater ?