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I don't realize what to do. Could you give me some hints or point to solution?

I tried to construct surjective homomorphism with $Ker(\phi)=(x, 3)$ by intuition to use First Isomorphism Theorem for Rings, but couldn't find such.

Roma
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  • Do you mean $\mathbb{Z}[x]/(x,3)$ ? – MAS May 25 '20 at 10:36
  • @M.A.SARKAR , I didn't understand a meaning of the question. – Roma May 25 '20 at 10:39
  • I mean are you sure about $\mathbb{Z}x$ ? I think it should be the quotient ring $\mathbb{Z}[x]/(x,3)$ by the ideal $(x,3)$ . – MAS May 25 '20 at 10:41
  • @M.A.SARKAR , oh didn't notice this typo. Yeah, you are right, I ment $\mathbb{Z}[x]/(x,3)$, thank you. Already edited. – Roma May 25 '20 at 10:45

2 Answers2

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Hint: You know such a homomorphism should send $x$ to $0$ (because $x$ is in the kernel), and $1$ to $1$ (because $1$ is not in the kernel, and there is no other way to make $\phi(1)^2=\phi(1)$). There aren't many choices left after that.

Arthur
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Well, you can use the third isomorphism theorem as follows: $$ \mathbb{Z}[x]/(x,3) \cong (\mathbb{Z}[x]/(x))/((x,3)/(x)) \cong \mathbb{Z}/(3) \cong \mathbb{Z}_3.$$ Here $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ and $(x,3)/(x) \cong (3)$.

MAS
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