Find an ideal $I$ in $A$ so that $A/I$ is a finite field with $25$ elements.

Question

Let $A = \frac {\Bbb Z[X]} {\left ( X^4+X^2+1 \right )}.$ Find an ideal $I$ in $A$ such that $A/I$ is a finite field with $25$ elements.

I have seen that the polynomial $X^4+X^2+1$ is reducible in $\Bbb Z[X]$ since $$X^4+X^2+1 = \left ( X^2+X+1 \right ) \left ( X^2-X+1 \right ).$$ So by Chinese remainder theorem in ring theory we have $$\frac {\Bbb Z[X]} {\left ( X^4+X^2+1 \right )} \cong \frac {\Bbb Z[X]} {\left ( X^2+X+1 \right )} \times \frac {\Bbb Z[X]} {\left ( X^2-X+1 \right )} \cong \Bbb Z[\omega] \times \Bbb Z[\omega].$$

How do I prove the above result using the fact that $$\frac {\Bbb Z[X]} {\left ( X^4+X^2+1 \right )} \cong \Bbb Z[\omega] \times \Bbb Z[\omega].$$

Any help will be highly appreciated. Thank you very much.

Answer 1

We want to try to use your idea.

Third isomorphism theorem for rings: Let $I$ and $J$ be ideals of the ring $R$, with $I \subseteq J$. Then $J/I$ is an ideal of $R/I$, and $R/J \cong (R/I)/(J/I)$.

Here, $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, that is, the ideal generated by $x^2+x+1$ contains the ideal generated by $x^4+x^2+1$.

Hence, $(x^2+x+1,5)/(x^4+x^2+1)=P$ is an ideal of $A$.

Then $A/P \cong \Bbb{F}_{25}$.

Hope, this will help. Looking for the improvement.