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I'm having quite a bit of trouble figuring out why $\mathbb{Z}[x]/\left<2,x\right>$ is isomorphic to $\mathbb{Z}_2$. So far I have figured out there is an onto map $\zeta: \mathbb{Z}\rightarrow\mathbb{Z}[x]$ given by $\zeta(n) = n$ (as a polynomial with degree 0), and that there's another onto map $\phi: \mathbb{Z}[x]\rightarrow\mathbb{Z}[x]/\left<2,x\right>$ given by $\phi(n) = n + (x) + (2)$, so then $\mathbb{Z}/\ker\phi$ is isomorphic to $\mathbb{Z}[x]/\left<2,x\right>$ by the isomorphism theorem.

I believe $\ker\phi =\{n\in\mathbb{Z}:n = e \} = \{n\in\mathbb{Z}: n + (2) + (x) = 0 + (2) + (x)\} = (2) + (x)$, but I could be wrong about that. In any event, this is as far as I can go.

Does what I've worked out even help? If not, what's a better approach to take?

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  • $\begingroup$ The inclusion $\Bbb Z\to\Bbb Z[x]$ is injective, not onto. $\endgroup$ – blue Apr 23 '14 at 1:42
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The third isomorphism theorem is the easiest thing to use.

$\Bbb Z[x]/\langle x,2\rangle\cong (\Bbb Z[x]/\langle x\rangle)/(\langle x,2\rangle/\langle x \rangle)\cong \Bbb Z / \langle 2\rangle$

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  • $\begingroup$ for the theorem you use A to be <x,2> and B=<x> so Z[x]/<x,2> is isomorphic to Z[x]/<x,2> that's what the theorem says how did you found Z/<2> ? $\endgroup$ – Manolis Lyviakis Jun 13 '14 at 14:21
  • $\begingroup$ @ManolisLyviakis "Z[x]/<x,2> is isomorphic to Z[x]/<x,2>" is a tautology! The first $\cong$ is exactly what you are quoting in your comment, and the second $\cong$ is just expressing that $\Bbb Z[x]/\langle x\rangle\cong \Bbb Z$ and $\langle x,2\rangle/\langle x\rangle\cong \langle 2\rangle$. $\endgroup$ – rschwieb Jun 13 '14 at 14:35
  • $\begingroup$ and you got the last isomorphism from the same theorem ? $\endgroup$ – Manolis Lyviakis Jun 13 '14 at 16:03
  • $\begingroup$ @ManolisLyviakis No: you could find the second isomorphism with the first isomorphism theorem, if you like. $\endgroup$ – rschwieb Jun 13 '14 at 16:03
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One can easily argue $\{0,1\}$ is a system of coset representatives.

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