Let \(f \in \mathbb{Z}[X] \setminus \mathbb{Z}\) such that \(\gcd(f, f') = 1\). Let \(S\) be the set of non-zero divisors in the quotient ring \(\mathbb{Z}[X]/(f)\).
Show that the ring \(S^{-1}(\mathbb{Z}[X]/(f))\) is isomorphic to a direct sum of fields.
Specify the fields in (a) above if \(f = X^5 - 1\).
Is the hypothesis \(\gcd(f, f') = 1\) necessary in order for the conclusion in (a) above to hold true? Justify.
Solution
1. The hypothesis \(\gcd(f,f')=1\) implies that \(f\) is primitive and squarefree over \(\mathbb Q\).
If the content of \(f\) had a nonunit divisor \(d\in\mathbb Z\), then \(d\) would divide every coefficient of both \(f\) and \(f'\), contradicting \(\gcd(f,f')=1\). Thus \(f\) is primitive.
A polynomial over a field has a repeated irreducible factor exactly when it has a nonconstant common divisor with its derivative. Hence the image of \(f\) in \(\mathbb Q[X]\) is squarefree.
2. Write the factorization over \(\mathbb Q\) as\[ f=c f_1\cdots f_r, \]where \(c\in\mathbb Q^\times\) and the \(f_i\in\mathbb Q[X]\) are distinct monic irreducibles. Then\[ \mathbb Q[X]/(f)\cong\prod_{i=1}^r K_i, \qquad K_i:=\mathbb Q[X]/(f_i). \]
The ideals \((f_i)\) are pairwise comaximal in the PID \(\mathbb Q[X]\). The Chinese remainder theorem therefore gives \[ \mathbb Q[X]/(f) \cong \prod_{i=1}^r\mathbb Q[X]/(f_i). \] Each quotient is a field because \(f_i\) is irreducible.
3. Every nonzero integer is a non-zero divisor in\[ R:=\mathbb Z[X]/(f). \]
Suppose \(0\ne n\in\mathbb Z\) and \(ng\in(f)\) in \(\mathbb Z[X]\). Then \(f\) divides \(ng\) in \(\mathbb Q[X]\), hence divides \(g\) there because \(n\) is a unit in \(\mathbb Q[X]\). Since \(f\) is primitive by 1, Gauss’s lemma implies that \(f\) divides \(g\) in \(\mathbb Z[X]\). Thus the class of \(g\) in \(R\) is zero.
4. Localizing \(R\) at the nonzero integers gives\[ (\mathbb Z\setminus\{0\})^{-1}R \cong \mathbb Q[X]/(f). \]
Localization commutes with quotient, so \[ (\mathbb Z\setminus\{0\})^{-1}(\mathbb Z[X]/(f)) \cong \mathbb Q[X]/(f). \] By 3, this localization inverts only non-zero divisors.
5. The total quotient ring \(S^{-1}R\) is isomorphic to the product of fields in
2.
4. Under the decomposition\[ \mathbb Q[X]/(f)\cong\prod_i K_i, \]a non-zero divisor is exactly a tuple with every coordinate nonzero, and such a tuple is already a unit. Therefore localizing further at the images of elements of \(S\) changes nothing. Hence\[ S^{-1}R \cong \mathbb Q[X]/(f) \cong \prod_{i=1}^rK_i. \]This proves part (a).
6. If \(f=X^5-1\), then\[ S^{-1}(\mathbb Z[X]/(X^5-1)) \cong \mathbb Q\times\mathbb Q(\zeta_5). \]
Over \(\mathbb Q\), \[ X^5-1=(X-1)\Phi_5(X), \qquad \Phi_5(X)=X^4+X^3+X^2+X+1. \] The fifth cyclotomic polynomial \(\Phi_5\) is irreducible over \(\mathbb Q\). Therefore 2 gives \[ \mathbb Q[X]/(X^5-1) \cong \mathbb Q[X]/(X-1)\times\mathbb Q[X]/(\Phi_5) \cong \mathbb Q\times\mathbb Q(\zeta_5). \] This proves part (b).
7. The hypothesis \(\gcd(f,f')=1\) is not necessary for the conclusion in part (a).
Take \[ f=2X. \] Then \(f'=2\), so \(\gcd(f,f')=2\ne1\). Nevertheless \[ R=\mathbb Z[X]/(2X) \] is reduced because \[ (2X)=(2)\cap(X) \] in \(\mathbb Z[X]\), an intersection of prime ideals. A reduced Noetherian ring with minimal primes \(\mathfrak p_1,\dots,\mathfrak p_s\) has total quotient ring equal to the product of the fraction fields of the domains \(R/\mathfrak p_i\). Here the two minimal primes are the images of \((2)\) and \((X)\), so \[ S^{-1}R \cong \operatorname{Frac}(R/(2))\times\operatorname{Frac}(R/(X)) \cong \mathbb F_2(X)\times\mathbb Q, \] a product of fields. Thus the stated gcd hypothesis is sufficient but not necessary.