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Find out whether Quotient ring $\mathbb{Z}[x]/(x-3)$ is a field? I know that $x-3$ is irreducible in $\mathbb{Z}[x]$ but further don't know.

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    $\begingroup$ You can study maximality of the ideal generated by $x - 3$. $\endgroup$ – user296602 Jan 3 '16 at 7:28
  • $\begingroup$ Trivially different from math.stackexchange.com/q/1585891/29335 $\endgroup$ – rschwieb Jan 4 '16 at 0:41
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If $\psi :Z[x]\to S$ is a homomorphic surjection to a ring $S$, with $\ker (\psi)=\{(x-3)f :f\in Z[x]\},$ then $S=\{\psi (n):n\in Z\}$ because $\forall f\in F[x]\;\exists g\in Z[x] \;\exists n\in Z\; (f=n+(x-3)g ).$ Now $\psi (n)\ne 0$ for $n\in Z\backslash \{0\}$ otherwise $n\in \ker (\psi).\;$ And $S$ is not a field, else for some $n\in Z$ we will have $\psi(1)=\psi (1)\psi (2)\psi (n)=\psi (2n)$, implying $0=\psi (1-2n)\ne 0.$

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The ring homomorphism $\mathbb Z[x]\rightarrow \mathbb Z$ given by $x\mapsto 3$ is surjective with kernel $(x-3)$. Hence $\mathbb Z[x]/(x-3) \cong \mathbb Z$, which is not a field.

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It is not a field.

You may think that being $x-3$ irreducible, the ideal $(x-3)$ must be maximal and thus $\mathbb{Z}[x]/(x-3)$ would be a field, since:

Given a commutative ring $R$ and $I\subset R$ a maximal ideal, $R/I$ is always a field.

Sadly, this is wrong. $\mathbb{Z}[x]$ is not a PID, and $f$ irreducible does not imply that $(f)$ is maximal.

In fact, $(x-3)\subsetneq (2,x-3)\subsetneq\mathbb{Z}[x]$, therefore $(x-3)$ cannot be a maximal ideal of $\mathbb{Z}[x]$.


Alternatively, you can see that the surjective ring homomorphism $$\varphi:\mathbb{Z}[x]\longrightarrow \mathbb{Z}$$ $$f(x)\mapsto f(3)$$ gives $\ker\varphi = (x-3)$ and $im\varphi = \mathbb{Z}$, thus $\mathbb{Z}[x]/(x-3) \simeq \mathbb{Z}$, which is not a field.

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    $\begingroup$ I believe you mean to say that $\mathbb{Z}[x]$ is not a PID, but instead you said that $\mathbb{Z}$ is not a PID. $\endgroup$ – ASKASK Jan 3 '16 at 7:52
  • $\begingroup$ Yes, thank you. Edited $\endgroup$ – Ottavio Bartenor Jan 3 '16 at 7:54

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