Let \(I\) be the ideal of \(\mathbb Z[x]\) generated by \[ 5 \qquad\text{and}\qquad x^3+x+1. \] Is \(I\) a prime ideal?
Solution
1. There is an isomorphism\[ \ZZ[x]/I \cong \FF_5[x]/(x^3+x+1). \]
Quotienting first by \((5)\) gives \[ \ZZ[x]/(5)\cong\FF_5[x]. \] Under this isomorphism, the image of the second generator of \(I\) is still \(x^3+x+1\). The third isomorphism theorem gives the displayed quotient.
2. The polynomial\[ q(x)=x^3+x+1 \]has no root in \(\FF_5\).
Evaluating at the five elements of \(\FF_5\) gives \[ \begin{aligned} q(0)&=1,\\ q(1)&=3,\\ q(2)&=8+2+1\equiv1,\\ q(3)&=27+3+1\equiv1,\\ q(4)&=64+4+1\equiv4 \end{aligned} \pmod5. \] None is zero.
3. The polynomial \(q\) is irreducible over \(\FF_5\).
A reducible cubic over a field has a linear factor, hence a root in that field. Step 2 shows that \(q\) has no root in \(\FF_5\).
4. The quotient\[ \FF_5[x]/(q) \]is a field.
By step 3, \(q\) is irreducible in the polynomial ring over the field \(\FF_5\). Therefore the principal ideal \((q)\) is maximal, so the quotient is a field.
5. The ideal \(I\) is prime.
By steps 1 and 4, the quotient \(\ZZ[x]/I\) is a field, hence in particular an integral domain. Therefore \[ \boxed{I\text{ is a prime ideal of }\ZZ[x]}. \]
6. Q.E.D.
Step 5 answers the question.