Can the Kernel of a Ring Homomorphism Equal 12Z or 13Z?

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SUMMARY

The kernel of a ring homomorphism f: Z -> C cannot equal 12Z or 13Z. This conclusion is established through a proof by contradiction utilizing the First Isomorphism Theorem for rings. Since both 12Z and 13Z are not integral domains and every subring of C must be an integral domain, the assumption that ker f equals either of these values leads to a contradiction, confirming that ker f must be {0}.

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Let f : Z ->C be a homomorphism of rings. Can the kernel of f be equal to 12Z or 13Z?
Ok,the way I'm thinking about it is using a proof by contradiction:asuming ker f=12Z...then by the First Isomorphism Theorem for rings Z/ker f ~im f where I am f is by definition a subring of C.But since I am f=12Z is not an integral domain and every subring in C is an integral domain the I am f will not be a subring oc C which is a contradiction.
The same thing with 13Z,is not equal with the kernel.
 
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Re: Kernerl and homomorphism

think about what happens to f(1).

if f(12) = 0, then f(12) = f(1 + 1 +...+ 1) = f(1) + f(1) +...+ f(1) = 12f(1) = 0.

since C is an integral domain, and 12 ≠ 0, f(1) = 0. but f(1) = 1, since f is a ring homomorphism.
 
Re: Kernerl and homomorphism

ok either ker f=12Z or ker f= {0}
f(n)=0>n*f(1)=0 but f(1)=1since f is a ring homomorphism.So ker f={0} or

Z/ker f is an integral domain since it is a subring of C =>ker f/=12Z. or ker f=/13
I'm I doing this wrong?
 
Re: Kernerl and homomorphism

f is ring homomorphism which means f(1) =1 must be
but if 12Z is the kernel this will drive us to f(1) =0 which contradict with the ring homomorphism
so 12Z,13Z can't be the kernel, ker(f) = {0}
 
Re: Kernerl and homomorphism

Thank you for the confirmation
 

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