Let \(f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb Z[x]\) (where \(a_n\neq 0\)) and let \(R=\mathbb Z[x]/(f)\). Prove that \(R\) is a finitely generated module over \(\mathbb Z\) if and only if \(a_n=\pm 1\).
Solution
(\(\Leftarrow\)) If \(a_n = \pm 1\), then \(R\) is a finitely generated \(\mathbb{Z}\)-module:
1. Since \(a_n = \pm 1\), the polynomial \(f_0(x) = a_n^{-1} f(x) = x^n + a_n a_{n-1} x^{n-1} + \dots + a_n a_0\) is a monic polynomial in \(\mathbb{Z}[x]\) of degree \(n\), with \((f) = (f_0)\).
\(a_n \in \{1, -1\}\) is a unit in \(\mathbb{Z}\), so \((f) = (f_0)\) as ideals in \(\mathbb{Z}[x]\).
2. By the polynomial division algorithm over \(\mathbb{Z}\) for monic polynomials, for every \(g(x) \in \mathbb{Z}[x]\), there exist \(q(x), r(x) \in \mathbb{Z}[x]\) such that:\[ g(x) = q(x)f_0(x) + r(x), \quad \text{with } \deg(r) < n \text{ (or } r = 0\text{)}. \]
division algorithm in \(\mathbb{Z}[x]\) with monic divisor.
3. Passing to the quotient \(R = \mathbb{Z}[x]/(f)\):\[ g(x) + (f) = r(x) + (f) = c_0 \cdot 1 + c_1 \bar{x} + \dots + c_{n-1} \bar{x}^{n-1}, \quad c_i \in \mathbb{Z}. \]
\(q(x)f_0(x) \in (f)\) and \(r(x) = \sum_{i=0}^{n-1} c_i x^i\).
4. Thus \(\{1, \bar{x}, \bar{x}^2, \dots, \bar{x}^{n-1}\}\) spans \(R\) as a \(\mathbb{Z}\)-module, so \(R\) is finitely generated over \(\mathbb{Z}\).
3.
(\(\Rightarrow\)) If \(R\) is a finitely generated \(\mathbb{Z}\)-module, then \(a_n = \pm 1\):
5. Suppose \(R\) is generated as a \(\mathbb{Z}\)-module by finitely many elements \(\{g_1(\bar{x}), \dots, g_m(\bar{x})\}\).
hypothesis.
6. Let \(d = \max_{1 \le i \le m} \deg(g_i) \ge 0\).
maximum degree among the finitely many polynomial generators.
7. Every element of \(R\) can be represented by a polynomial in \(\mathbb{Z}[x]\) of degree at most \(d\).
7.1. Any element \(u \in R\) is a \(\mathbb{Z}\)-linear combination \(u = \sum_{i=1}^m k_i g_i(\bar{x})\) with \(k_i \in \mathbb{Z}\).
\(\{g_i(\bar{x})\}\) is a \(\mathbb{Z}\)-generating set.
7.2. The polynomial \(h(x) = \sum_{i=1}^m k_i g_i(x) \in \mathbb{Z}[x]\) satisfies \(\deg(h) \le \max_i \deg(g_i) = d\).
degree of a sum of polynomials.
7.3. Thus \(u = h(\bar{x})\) with \(\deg(h) \le d\).
1 and 2.
8. For \(k = d + 1\), the element \(\bar{x}^k \in R\) satisfies \(\bar{x}^k = h(\bar{x})\) for some \(h(x) \in \mathbb{Z}[x]\) with \(\deg(h) \le d\).
7 applied to \(u = \bar{x}^k\).
9. The difference \(x^k - h(x)\) belongs to the ideal \((f(x)) = f(x)\mathbb{Z}[x]\).
\(x^k + (f) = h(x) + (f) \iff x^k - h(x) \in (f)\).
10. There exists a polynomial \(q(x) \in \mathbb{Z}[x]\) such that \(x^k - h(x) = q(x)f(x)\).
definition of the principal ideal \((f(x))\) in \(\mathbb{Z}[x]\).
11. Equating leading coefficients:
11.1. Since \(k = d + 1 > d \ge \deg(h)\), the leading term of the left-hand side \(x^k - h(x)\) is \(1 \cdot x^k\).
\(h(x)\) has degree \(\le d < k\), so it cannot cancel the \(x^k\) term.
11.2. The leading term of the right-hand side \(q(x)f(x)\) is \(\operatorname{LC}(q) \cdot a_n \cdot x^{\deg(q) + n}\).
multiplication of polynomials over the integral domain \(\mathbb{Z}\).
11.3. Equating degrees gives \(\deg(q) + n = k\), and equating leading coefficients gives:\[ 1 = \operatorname{LC}(q) \cdot a_n. \]
coefficients of polynomials in \(\mathbb{Z}[x]\).
11.4. Since \(\operatorname{LC}(q), a_n \in \mathbb{Z}\), \(a_n\) is a unit in \(\mathbb{Z}\), which implies \(a_n = \pm 1\).
the only units in the ring of integers \(\mathbb{Z}\) are \(1\) and \(-1\).
12. Conclusion: \(R = \mathbb{Z}[x]/(f)\) is a finitely generated \(\mathbb{Z}\)-module if and only if \(a_n = \pm 1\).
4 and 11.
Q.E.D.