I am trying to solve the following problem from Artin:

Every maximal ideal $\mathbb{Z}[x]$ is of the form $(p,f)$ where $p$ is a prime integer and $f$ is a primitive polynomial that is irreducible modulo $p$.

My question is why do we need $f$ to be primitive? I have found this to be true without the assumption that $f$ is primitive.

  • 1
    $\begingroup$ It doesn't say that if $(p,f)$ is maximal, then $p$ is prime and $f$ is primitive. Anyway, why don't you give your example? $\endgroup$ – user2345215 Apr 2 '14 at 13:36
  • $\begingroup$ First, from a logic point of view, if you prove the result without primitive, then it is a weaker statement. Then, we do not "need" $f$ to be primitive, but the result with "primitive" is stronger. Second, if you really have an ideal $(p,f)$ where $p$ is a prime and $f$ is not primitive, then the ideal is not maximal (just add something that divides all coefficients). $\endgroup$ – Jérémy Blanc Apr 2 '14 at 13:38
  • $\begingroup$ @JérémyBlanc $(-2x+2,3)=(x+2,3)$ is maximal, despite $-2x+2$ not being primitive :) But I think we know what you were thinking: $(-2x+2)$ can't be maximal, because $(-2x+2)\subsetneq(2,-x+1)\neq \Bbb Z[x]$ $\endgroup$ – rschwieb Apr 2 '14 at 15:22

It doesn't say that $f$ must be primitive, it says that $f$ can be chosen to be primitive.

For example, $(-2x+2,3)=(x+2,3)$. Notice $-2x+2$ is not primitive in $\Bbb Z[x]$, but $x+2$ is. This ideal is maximal in $\Bbb Z[x]$ whether you write $-2x+2$ or $x+2$.


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