I have some trouble with the following problem:

Problem. Whether the ideal $I=\langle x^2+1, 5\rangle$ is a maximal ideal of $\mathbb{Z}[x]$ or not?

Solution. For this I check whether the quotient $\mathbb{Z}[x]/I$ is a field or not. So if we take $J=\langle 5\rangle$ then by 2nd isomorphism theorem we have: $$\mathbb{Z}[x]/I \cong \frac{\mathbb{Z}[x]/J}{I/J}\cong \frac{\mathbb{Z}_5[x]}{I/J}.$$

Now we have to calculate the ring $I/J=\langle x^2+1, 5\rangle/\langle 5\rangle.$ I am suspecting it to be isomorphic to $\langle x^2+1\rangle.$ To prove this I try to use 1st isomorphism theorem.

So I have to define a epimorphism from $\langle x^2+1, 5\rangle$ to $\langle x^2+1\rangle$ whose kernel is $\langle 5\rangle.$ The natural intuition should be to define the map $\phi:\langle x^2+1, 5\rangle \to \langle x^2+1\rangle$ by $(x^2+1)f(x)+5g(x) \mapsto (x^2+1)f(x).$ Then it will be a homomorphism, moreover onto. Now clearly, $5\mathbb{Z}[x] \subset \ker \phi$. Again if an element is of the form $(x^2+1)f(x)$ then it doesn't map to $0$. So $\ker \phi=\langle 5\rangle.$

Therefore by First Isomorphism theorem $\langle x^2+1, 5\rangle/\langle 5\rangle \cong \langle x^2+1\rangle.$ Then $$\mathbb{Z}[x]/\langle x^2+1, 5\rangle \cong \mathbb{Z}_5[x]/\langle x^2+1\rangle.$$

Now as the latter ring is not an field (since $x^2+1$ is not an irreducible in $\mathbb{Z}_5[x]$) so the given ideal is not an maximal ideal of $\mathbb{Z}[x]$.

Is this proof is correct? Please tell if I made any mistake. Thank you.

  • $\begingroup$ In general. maximal ideals in $\Bbb Z[x]$ is $(p,f(x))$ where $f(x)$ is irreducible in $\Bbb Z_p$ $\endgroup$ – Mustafa Jun 21 '18 at 14:58
  • $\begingroup$ How to prove this..? $\endgroup$ – Indrajit Ghosh Jun 21 '18 at 14:59
  • $\begingroup$ see ma.utexas.edu/users/voloch/Homework/zx.pdf $\endgroup$ – Mustafa Jun 21 '18 at 15:09
  • $\begingroup$ $x^2+1=(x+2)(x+3)$ in $\mathbb Z_5$. $\endgroup$ – Andres Mejia Jun 21 '18 at 15:31
  • 1
    $\begingroup$ Is my proof right? $\endgroup$ – Indrajit Ghosh Jun 21 '18 at 15:32

Note that $x^2+1=(x+2)(x+3) \in \mathbb Z_5[x]$, so it is not irreducible, and your proof is correct.

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