# Valid argument ?

1. Mar 19, 2008

### sihag

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 ?

2. Mar 20, 2008

### ircdan

no

f(2z) = 3n, then you said f(2)f(z) = 3n, you can't do this because you don't where z lives and f is a ring homo on 2Z, what if z is in 5Z, etc

you need to use the fact you have an isomorphism, in your proof you are only using the fact you have a homomorphism but those do exist, so what you are trying to do is not going to be enough

edit: also you should be using additive notation here, ie, f(x) + f(y) = f(x + y)

Last edited: Mar 20, 2008
3. Mar 20, 2008

### sihag

let f(2) = 3n , f.s. n in Z
f(2 + 2) = f(4) , and f(2.2) = f(4)

f(2 + 2) = f(2) + f(2) = 6n
f(2.2) = f(2)f(2) = 9n^2

=> 6n = 9n^2
=> n = 0 (contradiction as f is an isomorphism, so must be injective, so only 0 maps to 0)
or, n = 2/3 (contradiction as n belongs to Z)

Q.E.D.

thanks for the help