I want to show that $\left<2,x\right>$ is a maximal ideal of $\Bbb Z[x]$.

My game plan is to use the 3rd isomorphism theorem to somehow get that $Z[x]/\left<2,x\right>$ isomorphic to $Z_2$ (since every this would mean that $Z[x]/\left<2,x\right>$ is a field and hence $\left<2,x\right>$ would be a maximal ideal) but I'm not entirely sure how to arrive at that conclusion.

My first thoughts are that since $\left<x\right>$ is an ideal of $Z[x]$ and $\left<x\right>\subset\left<2,x\right>$ we have that $Z[x]/\left<2,x\right> \cong Z[x]/\left<x\right>/\left<2,x\right>/\left<x\right>$ by the 3rd isomorphism theorem, but I'm not sure if this is the right direction. Any hints or tips would be great.

  • $\begingroup$ That probably works. I'd personally go with something more direct, using this idea: In any commutative ring $R$, an ideal $I \subseteq R$ is maximal if and only if for all $f \in R$ with $f \notin I$, there exists $g \in I$ such that $f + g = 1$. (In other words, an ideal is maximal if and only if adding in anything more gives the ideal $(1) = R$.) But that's mostly a matter of taste, since both methods amount to the same thing. $\endgroup$ – Daniel Hast Nov 18 '13 at 3:23

This seems like a possible direction. As a next step, you can say that: $$ \mathbb{Z}[x]/(2,x) \simeq \left( \mathbb{Z}[x]/(x)\right) /\left((2,x)/(x) \right) \simeq \mathbb{Z}/(2) = \mathbb{Z}_2,$$ and you're done. The second isomorphism needs some justification, perhaps. This would involve either waiving your hands around it, or doing it by hand.

You can also refer to the second isomorphism theorem. Because $ \mathbb{Z}[x] = \mathbb{Z} + (2,x)$, we have that: $$ \mathbb{Z}[x]/(2,x) = \left( \mathbb{Z} + (2,x) \right) /(2,x) \simeq \mathbb{Z}/\left( \mathbb{Z} \cap (2,x) \right) \simeq \mathbb{Z}/(2) = \mathbb{Z}_2.$$

  • $\begingroup$ That's the thing, I'm not sure how to make that 2nd isomorphism work. Should I just do it from straight definition? $\endgroup$ – user23793 Nov 18 '13 at 2:36
  • $\begingroup$ @user23793 Ah, OK. See if the edited version suits you better. $\endgroup$ – Jakub Konieczny Nov 18 '13 at 3:29
  • $\begingroup$ I see! This problem has been bugging me for a couple of days. Thank you for the assistance. $\endgroup$ – user23793 Nov 18 '13 at 3:45
  • $\begingroup$ @user It means sum of (sub)rings. If $A,B \subset R$ are subrings of a ring $R$ then $A+B = \{a+b \ : \ a \in A,\ b \in B\}$, and it's a subring if $B$ is an ideal. $\endgroup$ – Jakub Konieczny Nov 18 '13 at 12:05

You can look at something constructive, if you may. Note that $P\in\Bbb Z[X]$ is in $(2,X)$ if and only if its constant coefficient is divisible by $2$, that is, the elements of $(2,X)$ are of the form $2a_0+XP$ with $a_0\in\Bbb Z$ and $P\in\Bbb Z[X]$. Then consider the surjective homomorphism $\Bbb Z[X]\to \Bbb Z/(2)$ given by $P\mapsto { P(0)}+(2)$, and recall $\Bbb Z/(2)$ is a field.

In general, we have that $$\frac{R[x]}{(I,x)}\simeq \frac{R}{I}$$

  • $\begingroup$ I'm sorry, but the notation here "$\widehat{P(0)}$" means what exactly? I like that you said this because I thought about this but didn't think it would work. $\endgroup$ – user23793 Nov 18 '13 at 4:56
  • $\begingroup$ @user23793 I mean the class of $P(0)$ (an integer) in $\Bbb Z/(2)$. Edited to a better notation. $\endgroup$ – Pedro Tamaroff Nov 18 '13 at 4:59
  • $\begingroup$ Oh, I think I see it now. Are you thinking of using the 1st iso theorem? If I did the work correctly, the kernel of this mapping $\left<2,x\right>$, correct? $\endgroup$ – user23793 Nov 18 '13 at 5:06
  • $\begingroup$ @user23793 Yes, that is it. $\endgroup$ – Pedro Tamaroff Nov 18 '13 at 5:06

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