- #1
sihag
- 29
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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 ?