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Question 1: Prove there is no surjective ring homomorphism $\mathbb 2\mathbb Z$ -> $\mathbb{Z}_2$

Question 2: Either give an example of a surjective ring homomorphism $\mathbb 2\mathbb Z$ -> $\mathbb{Z}_3$ , or prove none exists.

What I know:

Question 1: For there to be a ring homomorphism I must map f(0) = 0 so that is fixed. Then to show the homomorphism cannot be surjective, I must show there can be no mappings from $\mathbb 2\mathbb Z$ to 1.

Any element of $\mathbb 2\mathbb Z$ must be of the form 2n where n is an integer, so say I map an arbitrary 2n to 1. Then f(2n) = 1 Then f(2n + 2n) = f(2n) + f(2n). f(2n + 2n) = 1 but f(2n) + f(2n) = 1 + 1 = 0 so the mapping is not closed under addition. Then nothing from $\mathbb 2\mathbb Z$ can map to 1.

Is this an acceptable proof or am I thinking something wrong?

Question 2: This seems to be a follow up to the previous question but I'm confused how to approach this. It looks like I should be able to come up with an example but I'm still seeing the issue that I cannot map to 1 from $\mathbb 2\mathbb Z$?

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    $\begingroup$ How do you get $f(2n+2n)=1$? I'm not saying it's wrong, it's just that I don't see what's the reason. If you are fixing your $n$ so that $f(2n)=1$, there is no problem, but I don't see why should $f(4n)$ be $1$. If you're mapping every $2n$ to $1$ then you're doing it wrong. $\endgroup$ – David Molano Apr 20 '17 at 3:54
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    $\begingroup$ @LeviathanTheEsper - I'm not sure looking back at my scratch notes. You're right, I didn't define what f(4n) would be, so I can't say it is mapped to 1... How do I steer my proof back on course? $\endgroup$ – Madoher Apr 20 '17 at 4:00
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    $\begingroup$ Hint: First, prove that if $f(2) = 0$, then $f = 0$. Then use the fact that $2+2 = 2\cdot 2$ to show that $f(2) = 0$. $\endgroup$ – Jason DeVito Apr 20 '17 at 4:08
  • $\begingroup$ @Jason DeVito - Thank you for the hint. So I need to show that this homomorphism can only be the 0 map by extending the reasoning that 2 maps to 0. It's starting to make more sense to me now. $\endgroup$ – Madoher Apr 20 '17 at 4:22
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    $\begingroup$ @Max Note that we are searching epimorphisms from $2\mathbb{Z}$ and you can't use that property, since you can't always guarantee that the homomorphic image of a nonunital ring is nonunital. If we were looking for homomorphisms from $\mathbb{Z}_2$ that would be another thing. $\endgroup$ – David Molano Apr 20 '17 at 15:26

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