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Consider the ideal $I$ defined by $$I : = \left \{ f(x) \in \Bbb Z_{11}[X]\ :\ f(2) = 0 \right \}$$ in $\Bbb Z_{11}[X].$ Is $I$ a maximal ideal in $\Bbb Z_{11} [X]$?

My attempt $:$ What I think is that $I = \langle X-2 \rangle$ and $X-2$ is irreducible in $\Bbb Z_{11} [X] .$ So $I$ is a non-zero prime ideal in the PID $\Bbb Z_{11} [X]$ and hence it has to be maximal. Am I doing any mistake?

Please help me in this regard. Thank you very much for your valuable time.

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    $\begingroup$ I assume by $\mathbb Z_{11}$ you mean $\mathbb Z/11$. $\endgroup$ Apr 12 '20 at 7:05
  • $\begingroup$ @Torsten Schoeneberg indeed it is. $\endgroup$ Apr 12 '20 at 7:59
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Define evaluation homomorphism $e$ from $\Bbb{Z}_{11}[X]$ on to $\Bbb{Z}_{11}$ by $e(g)=g(2), \forall\ g\in \Bbb{Z}_{11}[X]$. Now check that $e$ is onto and $\ker(e)=I$ and make use of Fundamental Theorem of Homomorphisms. Can you take it from here?

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This looks good to me. In fact $I= \langle x-2 \rangle$, since for every $f \in \langle x-2 \rangle$ we have: $f=(x-2)g$, where $g \in \mathbb{Z}/11\mathbb{Z}[X]$ and therefore $f(2)=0$. So we have $\langle x-2 \rangle \subset I$.

The other direction follows by the same argument that you gave: $\langle x-2 \rangle$ is a maximal ideal in $\mathbb{Z}/11\mathbb{Z}[X]$ and clearly $I\neq \langle 1 \rangle$.

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If $R$ is commutative ring with $1$, then $$\langle r \rangle \ \text{is prime ideal} , r\in R-\{0,1\} \Rightarrow r \text{ is irreducible} \ \text{...(#1)} $$ And $$r \text{ is irreducible }\Rightarrow \langle r \rangle \text{ is maximal among principal ideals } $$

Converse of (#1) is true only when $R$ is UFD, for example consider $$R=\mathbb Z[\sqrt {-5}] \text{ which is not UFD} \ , \ r=2 \text{ which is irreducible }$$

Observe that $2 \mid 6 = (1 + \sqrt{-5})(1 - \sqrt{-5}) $ but $2 \nmid (1 + \sqrt{-5}), (1 - \sqrt{-5})$. Hence , $2$ is not prime.


$Z_{11}$ is field $\Rightarrow \ R=Z_{11}[X]$ is ED , and hence $R$ is PID and UFD.

So we can conclude that $\langle X-2 \rangle$ is prime and maximal.


Another way to see this is:

$\mathbb Z_{11}[X]/\langle X-2 \rangle \cong \mathbb Z_{11}$

By defining homomorphism from $\mathbb Z_{11}[X]$ to $\mathbb Z_{11}$, $f(x) \mapsto f(2)$

From here also we can conclude that $\langle X-2 \rangle$ is maximal, since $\mathbb Z_{11}[X]/\langle X-2 \rangle $ is field.

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