Isn't $I$ a maximal ideal in $\Bbb Z_{11} [X]$?

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.

• I assume by $\mathbb Z_{11}$ you mean $\mathbb Z/11$. Apr 12 '20 at 7:05
• @Torsten Schoeneberg indeed it is. Apr 12 '20 at 7:59

3 Answers

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?

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 $$I\neq \langle 1 \rangle$$.

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.