If $I = \langle 2\rangle$, why is $I$ not a maximal ideal of $\mathbb Z$, even though $I$ is a maximal ideal of $\mathbb Z$?

Let $I = \langle 2\rangle$. Prove $I[x]$ is not a maximal ideal of $\mathbb Z[x]$ even though $I$ is a maximal ideal of $\mathbb Z$.

My professor mentioned that I should try adding something to it to prove that $I[x]$ is not the maximal ideal. I want to assume he meant to add $ \langle 2 \rangle$ to $I[x]$?

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Hint $\ $ The point of the hint is that if $(2)$ isn’t maximal then it will be properly contained in some ideal $\rm\:M\ne (1),\:$ so we can enlarge $(2),$ while keeping it $\ne(1),\:$ by choosing any $\rm\:f\in M,\ f\not\in(2).\:$ Then $\rm\:(2) \subsetneq (2,f) \subset M\subsetneq (1).\:$ Clearly $\rm\:f\:$ must be a nonunit not divisible by $2,\:$ and conversely any such $\rm\:f\:$ works. Can you find one?

$\langle 2 \rangle$ is strictly contained in $\langle 2,x \rangle$.

$\mathbb Z[x]/I[x] \simeq (\mathbb Z/I)[x] = \mathbb F_2[x]$, which is a domain but not field.

Note: it’s not the maximal ideal but rather a maximal ideal, since for every prime $p$, $\langle p \rangle$ is maximal in $\mathbb Z$. To see this, note that $\langle p \rangle = p \mathbb Z \subsetneq \mathbb Z$ and if $\langle p \rangle \subsetneq I$ then there is an $i \in I$ such that $i$ is not a multiple of $p$. Since $p$ is prime you have $\gcd (p,i) = 1$ which means that there exists $k,n \in \mathbb Z$ such that $ki + np = 1 \in I$ and hence $I = \mathbb Z$.

Now to show that $I[x]$ is not maximal in $\mathbb Z[x]$ use Belgi’s answer. You need to find a proper ideal $J$ of $\mathbb Z$ such that $I[x] \subsetneq J \subsetneq \mathbb Z[x]$.

Since $I = \langle 2 \rangle$, adding $\langle 2 \rangle $ to $I[x]$ yields $I[x]$ again, so your idea doesn’t work in this case.

Maybe this helps: what is $I[x]$? It’s all polynomials with even coefficients. To get an ideal $J$ that properly contains $I[x]$ you can adjoin another element. The element you adjoin cannot be in $\mathbb Z$ only since if it’s even, you end up with $I[x]$ again and if it’s odd you end up with $\mathbb Z$. So what you adjoin has to contain $x$.

$\langle x,2 \rangle = \{ x g(x) + 2 h(x) ; g(x),h(x) \in Z[x] \}$ and

$I[x] = 2\mathbb{Z}[x] = \{a_{0} + a_1x + \dots \dots + a_nx^n; a_{i} \in 2\mathbb{Z} \}$

Note that $3x^2 \in \langle x,2 \rangle $ by taking $g(x)$ as $3x$ and $h(x)$ as $0$. But $3x^2 $is not in the other set.