This question concerns the polynomial ring \(R=\mathbb Z[x,y]\) and the ideal \(I=(5,x^2+2)\) in \(R\).
Prove that \(I\) is a prime ideal of \(R\) and that \(R/I\) is a PID.
Give an explicit example of a maximal ideal of \(R\) which contains \(I\). (Give a set of generators for such an ideal.)
Show that there are infinitely many distinct maximal ideals in \(R\) which contain \(I\).
<1>2. The polynomial \(x^2+2\in\mathbb F_5[x]\) is irreducible.
<1>3. Hence \[ R/I\cong \mathbb F_{25}[y]. \] In particular, \(I\) is prime and \(R/I\) is a PID.
<1>4. The ideal \[ \mathfrak m=(5,x^2+2,y) \] is a maximal ideal of \(R\) containing \(I\).
<1>5. The ring \(\mathbb F_{25}[y]\) has infinitely many monic irreducible polynomials.
<1>6. There are infinitely many distinct maximal ideals of \(R\) containing \(I\).
Solution
1. There is an isomorphism\[ R/I\cong \mathbb F_5[x,y]/(x^2+2). \]
The squares in \(\mathbb F_5\) are \[ 0,1,4. \] Thus \(-2\equiv3\pmod5\) is not a square, so \(x^2+2\) has no root in \(\mathbb F_5\). Since it has degree \(2\), it is irreducible.
By 2, \[ \mathbb F_5[x]/(x^2+2) \] is a field with \(5^2=25\) elements. Therefore \[ \mathbb F_5[x,y]/(x^2+2) \cong \bigl(\mathbb F_5[x]/(x^2+2)\bigr)[y] \cong \mathbb F_{25}[y]. \] A polynomial ring in one variable over a field is a PID and an integral domain. Therefore \(R/I\) is a domain, so \(I\) is prime.
We have \[ R/\mathfrak m \cong \mathbb F_5[x]/(x^2+2) \cong \mathbb F_{25}, \] which is a field. Hence \(\mathfrak m\) is maximal.
Suppose instead that the monic irreducible polynomials were exactly \[ q_1(y),\dots,q_r(y). \] Consider \[ Q(y)=q_1(y)\cdots q_r(y)+1. \] This is a nonconstant polynomial, so it has an irreducible factor \(q(y)\). But no \(q_i\) divides \(Q\), because \[ Q\equiv1\pmod{q_i}. \] Thus \(q\) is a monic irreducible not on the list, a contradiction.
By the correspondence theorem, maximal ideals of \(R\) containing \(I\) are in bijection with maximal ideals of \[ R/I\cong\mathbb F_{25}[y]. \] Since this is a PID, each monic irreducible \(q(y)\in\mathbb F_{25}[y]\) generates a maximal ideal \((q)\), and distinct monic irreducibles give distinct maximal ideals. By 5 there are infinitely many of them. Their inverse images under the quotient map \(R\to R/I\) are therefore infinitely many distinct maximal ideals of \(R\) containing \(I\).