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I have a couple of ideals which I wonder if I correctly classify as maximal/prime ideal.

$I_1 = \langle 2x^2 + 9x -3\rangle$, $I_2 = \langle x - 1\rangle$

$\mathbf 1)$ Is $I_1$ a maximal ideal in $\mathbb{Q}[x]$?

Yes, since $I_1$ is irreducible with $p=3$ using Eisenstein's criterion, thus maximal ideal.

$\mathbf 2)$ Is $I_2$ a prime ideal in $\mathbb{Q}[x]$?

Yes, since $I_2$ is obviously irreducible, and thus a maximal ideal, and every maximal ideal is a prime ideal.

$\mathbf 3)$ Is $I_2$ a maximal ideal in $\mathbb{Z}[x]$?

Yes, $I_2$ is obviously irreducible, and thus a maximal ideal.

$\mathbf{Edit:}$ No, as it is not a field. $$ $$

Am I right in my conclusions?

Appreciate any help.

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    $\begingroup$ Parts 1 and 2 are ok. But irreducibility of a generating polynomial implies maximality only when the ring of coefficients is a field, so 3 is wrong. Can you think of a bigger proper ideal? For examples you might add another generator that is a constant? $\endgroup$ – Jyrki Lahtonen May 31 '14 at 20:32
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There is a theorem that you can use: Given a commutative ring $R$ with identity, $I$ is a maximal ideal in $R$ if and only if $R/I$ is a field. (Similarly, $I$ is a prime ideal if and only if $R/I$ is an integral domain.)

What do elements in $\mathbb{Z}[x]/\langle x - 1\rangle$ look like? Do they form a field?

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  • $\begingroup$ Obviously $\mathbb{Z}$ does not have multiplicative inverse for all elements and is thus not a field. So the extension is not a field. My bad there! Thanks for the help! $\endgroup$ – knordbo May 31 '14 at 21:10
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In a commutative ring $R$ with 1 \begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Downarrow & \\ \Uparrow&&(a) \text{ maximal among principal} & \Longleftarrow & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

In an integral domain $R$ \begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Downarrow & \\ \Uparrow&&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

In a UFD $R$

\begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Updownarrow & \\ \Uparrow &&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

In a PID $R$ \begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ && & & \Updownarrow & \\ \Uparrow &&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ &&\Downarrow & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

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  • $\begingroup$ I have an exercise saying : in a PID give an example of $(p)$ maximal and $p$ not irreducible, but from your answer this can't be true, right?? $\endgroup$ – giannispapav Sep 5 '18 at 5:05
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In 3, your argument is wrong. In an integral domain, $a$ is irreducible iff $(a)$ is maximal among principal ideals, but $\mathbb{Z}[x]$ is not a PID, thus you cannot conclude that $(a)$ is maximal. In fact it is not, because $\mathbb{Z}[x]/\langle x-1\rangle$ is not a field. So for example $\langle 2,x-1\rangle$ is a maximal ideal containing $\langle x-1\rangle$.

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