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.


  • 5
    $\begingroup$ Hint: Maximal ideals containing $x-1$ correspond to maximal ideals in $\mathbb{Z}[x]/(x-1)$. But this is isomorphic to $\mathbb{Z}$. $\endgroup$ – Martin Brandenburg May 30 '13 at 14:46

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.


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.

  • $\begingroup$ 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? $\endgroup$ – Maths May 30 '13 at 15:21
  • $\begingroup$ @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. $\endgroup$ – Ragib Zaman May 30 '13 at 15:25
  • $\begingroup$ 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. $\endgroup$ – Maths May 30 '13 at 16:01
  • $\begingroup$ 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 ${\mathbb Z}[x]$? $\endgroup$ – Maths May 30 '13 at 16:54
  • $\begingroup$ I was trying to work out how you got the equation $q(x)=(x-1)u(x) + pv(x)$ $\endgroup$ – Maths May 30 '13 at 17:44

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