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I would like to prove that given an irreducible polynomial $g(X)\in \Bbb Z[X]$, then the ideal $\langle g(X) \rangle \trianglelefteq \Bbb Z[X]$ is not maximal.

One can think to prove $$\langle g(X) \rangle \subsetneq \langle p,g(X)\rangle \subsetneq \Bbb Z[X]$$ and maybe to use the ring isomorphism $$\frac{\Bbb Z[X]}{\langle p,g(X) \rangle}\cong\frac{\Bbb Z_p[X]}{\langle \overline g(X) \rangle }$$ where $p$ is a prime number and $\overline g(X)$ is $g(X)$ with coefficients in $\Bbb Z_p$. But how could we proceed?

I face difficulty to prove $\langle p,g(X)\rangle \subsetneq \Bbb Z[X]$.

I know that this may be an easy question, but I have stuck. Also, any other ideas are welcome!

Chris
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3 Answers3

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Hint: $\langle g\rangle$ cannot contain any elements with degree lower than the degree of $g$ (apart from the zero polynomial).

Arthur
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  • Thank you for your answer. Ok, we know that $$\langle p , g(X) \rangle:={a(X)\cdot p + b(X) \cdot g(X):a(X),b(X)\in \Bbb Z[X]}.$$ But how does this help? Maybe with $1\in \langle p,g(X)\rangle \iff \langle p,g(X)\rangle \neq \Bbb Z[X]$? – Chris May 28 '19 at 14:28
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The ideal $\langle p,g(x)\rangle$ cannot contain the residues $1,\ldots,p-1$ if the degree of $g(x)$ is $\geq 1$.

Wuestenfux
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  • Thank you for your answer. Why does this happen? Also, if $g(X):=q$ prime number, is it true that $\langle p,q\rangle =\langle \gcd(p,q) \rangle = \langle 1 \rangle =\Bbb Z[X]$? – Chris May 28 '19 at 14:45
  • And in this case $\Bbb Z[X]/\langle q \rangle \cong \Bbb Z_q[X]$, which is not a field, so our ideal is not maximal. Could you please explain why if $\deg g(X) \geq 1$, then $1,2,\dots,p-1 \notin \langle p,g(X) \rangle $? (and so $\langle p,g(X) \rangle \neq \Bbb Z[X])$ – Chris May 28 '19 at 14:56
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    Degree zero means that $g(x)$ is a constant. – Wuestenfux May 28 '19 at 15:01
  • It's ok if $\deg g(X)=0$. But I stuck on the case you said that $\deg g(X) \geq 1$. – Chris May 28 '19 at 15:04
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    Just to be at the safe side. – Wuestenfux May 28 '19 at 15:08
  • Please check my answer :) – Chris May 28 '19 at 15:53
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Taking into consideraction Arthur's and Wuestenfux's answers, I ll try to write down the full answer.

At first, if $\deg g(X)=0\implies g(X):=m\in \Bbb Z$, then $\Bbb Z[X]/\langle m \rangle \cong \Bbb Z_m[X]$, which is not a field, so $\langle g(X)\rangle $ is not maximal.

Now, we will assume that $\deg g(X)\geq 1$.

  • $\langle g(X)\rangle \subsetneq \langle p ,g(X) \rangle$, because if so, $p\in \langle g(X)\rangle \implies p=q(X)g(X),\ q(X) \in \Bbb Z[X \implies \deg (p)=\deg q(X)+\deg g(X) \implies 0=d_1+d_2,$ where $d_1,d_2\in \Bbb N$ and $d_2\geq 1$, contradiction.
  • $\langle p ,g(X) \rangle \subsetneq \Bbb Z[X]$: We will show that $$1\notin \langle p ,g(X) \rangle \iff \langle p ,g(X) \rangle \neq \Bbb Z[X].$$ We suppose, contrary, that $1\in \langle p ,g(X) \rangle$. Then, \begin{alignat*}{2} 1\in \langle p ,g(X) \rangle \iff 1=a(X)p+b(X)g(X), \tag{*} \end{alignat*} for some $a(X),b(X)\in \Bbb Z[X]$. We consider the following cases:

(1) If $a(X),b(X)=0$, we have immediately contradiction, through (*).

(2) If $a(X)\neq 0,\ b(X)=0$, then $1=b(X)g(X) \implies \deg g(X)=0$, contradiction.

(3) If $b(X)\neq 0,\ a(X)=0$, then $1=a(X)p \implies p|1$, contradiction.

(4) If $a(X)\neq 0,\ b(X)\neq 0$, then if we take degrees in (*), we will have $$0=\max\{ \deg(a(X)p),\underbrace{\deg (b(X)g(X))}_{\geq 1} \},$$ so again we have a contradiction.

Chris
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