Clarification regarding proof: $ \langle x,2 \rangle$ is a maximal ideal of $\mathbb Z[x]$

Question

The question asked here mentions:

When the polynomial ring Z[x] is quotiented by the ideal (2,x) we get a field as $\mathbb{Z}[x]/(x,2) \cong \mathbb Z/(2) \cong \mathbb{Z}_2$ which is a field.

I am not being able to flesh out my answer and connect the dots.

I know that $\frac{\mathbb Z[x]}{\langle x,2 \rangle }= \frac{\frac{\mathbb Z[x]}{\langle x \rangle }}{\frac{\langle 2,x \rangle}{\langle x \rangle}}$.

From here, I can individually show that: $\frac{\mathbb Z[x]}{\langle x \rangle} \cong \mathbb Z,$

and that, $\frac{\langle 2,x \rangle}{\langle x \rangle} \cong \langle 2 \rangle.$

How do I conclude that, $\frac{\frac{\mathbb Z[x]}{\langle x \rangle}}{\frac{\langle 2,x \rangle}{\langle x \rangle}} \cong \mathbb Z_{2}$?

It seems like I am trying to prove something in this format: $\frac{I}{J} \cong \frac{K}{H}$ because $I \cong K$ and $J \cong H.$ (I believe, this basic concept is essentially what my doubt is)

Am I thinking about this the wrong way?

Answer 1

It's not the maximal ideal, there are others: more generally $(x,p)$ is a maximal ideal and $\mathbb Z[x]/(x,p)\cong\mathbb Z_p$ is a field.

From what you did, the final step is merely that $\mathbb Z/(2)\cong\mathbb Z_2$.

In the linked answer there is a direct proof, which you can generalize, that this ideal is maximal. It is also easy to prove that an ideal is maximal if and only if $\mathcal R/\mathscr I$ is a field.

Your last statement is false: consider $\mathbb Z/(2)\cong \mathbb Z_4/(2)$.

Answer 2

What you write works, but not the end. What you want to use is not "there is an isomorphism between $(x,2)/(x)$ and $(2)$" but "my isomorphism between $\mathbb Z[x]/(x)$ and $\mathbb Z$ identifies $(x,2)/(x)$ and $(2)$". It is this stronger statement that allows you to apply your isomorphism theorem: $$\mathbb Z [x]/(x,2) \cong (\mathbb Z[x]/(x))/((x,2)/(x)) \cong \mathbb Z/(2)$$

Answer 3

Here is an alternative method:

We have the substitution map $\mathbb Z[x]\to\mathbb Z:f(x)\mapsto f(0)$, which is clearly surjective. Composing with the canonical projection $\mathbb Z\to\mathbb Z_2$ gives us the subjective map $\mathbb Z[x]\to\mathbb Z_2:f(x)\mapsto\overline{f(0)}$. The kernel consists of polynomials with even constant terms, which is exactly the ideal $(x,2)$. Thus, the isomorphism theorem tells us that $\mathbb Z[x]/(x,2)\cong\mathbb Z_2$.