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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?

ali
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    See the first dupe for a general method (quotient reciprocity), and the 2nd dupe for the specific step you inquire about (which is linked in a comment in the first). – Bill Dubuque May 22 '21 at 21:45

3 Answers3

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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: $$\mathbb Z [x]/(x,2) \cong (\mathbb Z[x]/(x))/((x,2)/(x)) \cong \mathbb Z/(2)$$

J. Darné
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  • sorry this is the first time I'm coming across this so I want to clarify, I would have to use the same map then? – ali May 22 '21 at 19:49
  • What the isomorphism theorem says is that if $I$ and $J$ are two ideals in a ring $A$, with $I \subset J$, then the canonical projection from $A/I$ onto $A/J$ induces an isomorphism $A/J \cong (A/I)/(J/I)$ (because the image $J/I$ of $J$ in $A/I$ is the kernel of the aforementioned projection). So if you have an isomorphism $A/I \cong B$, you need to use this isomorphism to identify $J/I$ with an ideal $K$ of $B$, and then the isomorphism $A/I \cong B$ clearly induces an isomorphism $(A/I)/(J/I) \cong B/K$. – J. Darné May 22 '21 at 21:31
  • But an abstract isomorphism between $J/I$ and some ideal of $B$ is useless for computing the quotient. Does that answer your question ? (With $A = \mathbb Z [x]$, $I = (x)$, $J = (2, x)$, $B= \mathbb Z$, and $K = (2)$, of course) – J. Darné May 22 '21 at 21:35
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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 $\mathbb Z[x]/(x,2)\cong\mathbb Z_2$.

Kenta S
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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 $\mathbb Z/(2)\cong \mathbb Z_4/(2)$.

  • It looks like the OP is trying to convince himself that $\mathbb Z[x]/(x,p)\cong \mathbb Z/(p)$ in the first place. – Troposphere May 22 '21 at 19:32