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Is the ideal $(X^2 + 1, X + 2)$ prime/maximal in $\mathbb Z[X]?$

$I = (X^2 + 1, X + 2) = (X + 2, 5)$ since $(X + 2)^2 − 4(X + 2) + 5 = X^2 + 1,$ then $\mathbb Z[X]/I ≃ Z_5[X]/(X + \overline 2)$ where $X + \overline 2$ is irreducible in $\mathbb Z_5[X]$ thus the quotient is a field and I is maximal.

I have two questions:

  1. Why $\mathbb Z[X]/I ≃ Z_5[X]/(X + \overline 2)?$
  2. Why from irreducibility of $X+\overline 2$ follows that quotient field is maximal, what if it wasn't irreducible?
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  1. We have $\mathbb{Z}[x]/(5) \cong (\mathbb{Z}/5 \mathbb{Z})[x]$, hence $\mathbb{Z}[x]/I\cong \mathbb{Z}/5[x]/(x+2)$.
  2. $K[x]/(f)$ is a field if and only if $f$ is irreducible, if and only if $(f)$ is maximal (note that $K[x]$ is a PID for a field $K$).
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  • $\begingroup$ Thanks, i don't understand why we can take quotients in turn, i read that it is because of isomorphism theorem, but i don't know how. $\endgroup$ – Yola Dec 2 '16 at 6:46
  • $\begingroup$ For "taking quotients in turn" see here, they explain it. $\endgroup$ – Dietrich Burde Dec 2 '16 at 9:38

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