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I'm trying to find a maximal ideal in ${\mathbb Z}[x]$ that properly contains the ideal $(x-1)$.

I know the relevant definitions, and that "a proper ideal $M$ in ${\mathbb Z}[x]$ is maximal iff ${\mathbb Z}[x]/M$ is a field."

I think the maximal ideal I require will not be principal, but I can't find it.

Any help would be appreciated.

Thanks.

Maths
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2 Answers2

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All ideals of the form $(p,X-1)$ with $p$ a prime number satisfy your condition. Note that $\mathbb Z[X]/(p,X-1)\simeq \mathbb Z_p$: we have $$\mathbb Z[X]/(p,X-1)\simeq \frac{\mathbb Z[X]/(X-1)}{(p,X-1)/(X-1)}\simeq\mathbb Z/(p).$$ Moreover, you can prove that these are all maximal ideals with the required property.

  • Thanks, that makes sense. So ${\mathbb Z}_p$ is a field since $p$ is prime, then the quotient isomorphic to it must also be a field, and this gives that $(p,x-1)$ is maximal. Can you tell me the isomorphism between the groups? – Maths May 30 '13 at 15:21
  • @Maths Consider the homomorphsim $\phi: \mathbb{Z}[X]\to \mathbb{Z}_p$ given by $p(X) \mapsto p(1) \pmod p.$ Show that this has kernel $(p,X-1)$ and then the first isomorphism theorem finishes it off. – Ragib Zaman May 30 '13 at 15:25
  • Thanks. I used $\phi(q)=q'(1) mod p$, where $q'$ is the polynomial function associated to $q$. Then $ker(\phi)={q in {\mathbb Z}[x] : q'(1)=0 mod p}$. $q'(1)=0$ means that $1$ is a zero of $q(x)$, which by the factor theorem for ${\mathbb Z}[x]$ means that $(x-1)$ is a factor of $q(x)$. Then we have $ker(\phi)=(x-1)$. I'm not sure how to incorporate the prime $p$ into this argument. – Maths May 30 '13 at 16:01
  • Is $u(x)$ in ${\mathbb Z}_p[x]$ and $v(x)$ in ${\mathbb Z}[x]$? Also, does $+ pv(x)$ take $(x-1)u(x)$ from ${\mathbb Z}_p[x]$ to ${\mathbb Z}[x]$? – Maths May 30 '13 at 16:54
  • I was trying to work out how you got the equation $q(x)=(x-1)u(x) + pv(x)$ – Maths May 30 '13 at 17:44
  • Thanks. I wasn't familiar with "lifting", but I just looked it up. – Maths May 30 '13 at 19:48
  • Is there a way to show that the homomorphism $\phi$ has kernel $(p,x-1)$ without considering that $x-1$ is a factor of $q(x)$ in ${\mathbb Z}_p[x]$, because that result uses a different homomorphism (one from ${\mathbb Z}_p[x]$ to ${\mathbb Z}_p$)? – Maths May 30 '13 at 20:04
  • @Ragib Zaman How can I show that the kernel of your homomorphism is $(p,x-1)$? – Maths May 30 '13 at 20:57
  • @Maths Each of $p$ and $x-1$ are in the kernel, so $(p,x-1)\subseteq \ker \phi.$ For the other direction: Suppose $p(X)\in \mathbb{Z}[X]$ is in $\ker \phi$ so then $p(1) =kp$ for integer $k.$ We want to use that information to show that $p(X)\in (p,X-1)$ and the trick to that this is to show that $p(X) + (p,X-1) = (p,X-1).$ – Ragib Zaman May 30 '13 at 21:39
  • If $p(X) = \sum a_i X^i$ then $p(X) + (p,X-1) = \sum a_i X^i + (p,X-1) = \sum a_i (X-1 + 1)^i + (p,X-1) = \sum a_i 1^i + (p,X-1) = p(1) + (p,X-1) = (p,X-1).$ We got the 3rd equality by expanding $( (X-1) +1)^i$ with the binomial theorem and remembering we can put all terms with a factor of $X-1$ into the ideal $(p,X-1).$ – Ragib Zaman May 30 '13 at 21:40
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Hint: the primes containing $\,(x-1)\subset \Bbb Z[x]\,$ are in $1$-$1$ correspondence with the primes in $\,\Bbb Z[x]/(x-1)\cong \Bbb Z,\,$ by a basic property of quotient rings.

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