For which pairs of integers \((a,b)\) is the quotient ring \(\ZZ[x]/(x^2+ax+b)\) isomorphic (as a ring) to the direct product of rings \(\ZZ\times\ZZ\)?
Solution
Let \(A=\ZZ[x]/(x^2+ax+b)\) and \(B=\ZZ\times\ZZ\). If \(x^2+ax+b\) is irreducible in the UFD \(\ZZ[x]\), then \((x^2+ax+b)\) is a prime ideal, so \(A\) is a domain. But \(B\) is not a domain. Thus we may assume \(x^2+ax+b=(x-c)(x-d)\) for some \(c,d\in\ZZ\).
Suppose \(p\) is a prime integer dividing \(c-d\). Then \(A\cong B\) implies \(A/pA\cong B/pB\); that is, \(\FF_p[x]/(x-\bar c)^2\cong\FF_p\times\FF_p\), where \(\bar c=\bar d\) is the image of \(c\) in \(\FF_p\). The ring on the left has a nonzero element with square \(0\), namely \(x-\bar c\), whereas the right hand side has no such element. This contradiction shows that \(c-d\) is divisible by no primes, so \(c-d=\pm1\).
Conversely, if \(c-d=\pm1\), then the sum of the ideals \((x-c)\) and \((x-d)\) in \(\ZZ[x]\) is the unit ideal, and their product equals their intersection (since they are generated by non-associate irreducible elements), so the Chinese remainder theorem gives
\[ \frac{\ZZ[x]}{((x-c)(x-d))}\cong\frac{\ZZ[x]}{(x-c)}\times\frac{\ZZ[x]}{(x-d)}. \]
Each factor on the right is isomorphic to \(\ZZ\), because each polynomial in \(\ZZ[x]\) is uniquely expressible as \(q(x)(x-c)+r\) with \(q(x)\in\ZZ[x]\) and \(r\in\ZZ\). Thus \(c-d=\pm1\) implies \(A\cong B\). In other words, the answer is the set of \((a,b)\) such that \(x^2+ax+b\) has the form \((x-n)(x-(n+1))\); that is,
\[ \{(-(2n+1),n(n+1)):n\in\ZZ\}. \]