Let \(K\) be a subfield of a field \(F\), and suppose \(f\), \(g\) are polynomials in \(K[x]\). Let \(M_K\) be the ideal generated by \(f\) and \(g\) in \(K[x]\) and \(M_F\) be the ideal they generate in \(F[x]\). Show that \(M_K\) and \(M_F\) have the same monic generator.
Solution
1. Computation of the monic generator in \(K[x]\) via the Euclidean algorithm:
1.1. Since \(K\) is a field, the polynomial ring \(K[x]\) is a principal ideal domain (PID).The ideal \(M_K = \langle f, g \rangle_{K[x]}\) is principal, generated by the unique monic greatest common divisor \(d(x) = \gcd_{K[x]}(f, g)\).
1.2. The polynomial \(d(x)\) is computed by the Euclidean algorithm in \(K[x]\):\[ r_0 = f, \quad r_1 = g, \quad r_{i-1} = q_i r_i + r_{i+1} \quad \text{with } \deg(r_{i+1}) < \deg(r_i), \]terminating with the last non-zero remainder \(r_k(x) \in K[x]\), and \(d(x) = c^{-1} r_k(x)\) where \(c \in K^\times\) is the leading coefficient of \(r_k\).
2. Invariance of the Euclidean algorithm under field extensions:
2.1. By the uniqueness of quotient and remainder in the polynomial division algorithm, performing polynomial division on elements of \(K[x]\) inside \(F[x]\) yields the exact same quotient \(q_i \in K[x] \subseteq F[x]\) and remainder \(r_{i+1} \in K[x] \subseteq F[x]\).
2.2. Therefore the Euclidean algorithm for \(f, g\) in \(F[x]\) produces the exact same sequence of remainders, terminating at the same monic polynomial \(d(x) \in K[x] \subseteq F[x]\).
3. Proof that \(d(x)\) generates \(M_F\) in \(F[x]\):
3.1. By Bézout’s identity in \(K[x]\), there exist polynomials \(a(x), b(x) \in K[x]\) such that:\[ d(x) = a(x) f(x) + b(x) g(x). \]Since \(K[x] \subseteq F[x]\), \(a, b \in F[x]\), so \(d(x) \in M_F = \langle f, g \rangle_{F[x]}\), which implies \(\langle d \rangle_{F[x]} \subseteq M_F\).
3.2. Conversely, because \(d \mid f\) and \(d \mid g\) in \(K[x]\), there exist \(u, v \in K[x] \subseteq F[x]\) such that \(f = u d\) and \(g = v d\).For any element \(h = A f + B g \in M_F\) with \(A, B \in F[x]\):\[ h = A(u d) + B(v d) = (A u + B v) d \in \langle d \rangle_{F[x]}. \]Thus \(M_F \subseteq \langle d \rangle_{F[x]}\).
3.3. Combining both inclusions yields \(M_F = \langle d \rangle_{F[x]}\).Since \(d(x)\) is monic, \(d(x)\) is the unique monic generator of \(M_F\).
4. Conclusion:\(M_K\) and \(M_F\) have the exact same monic generator \(d(x) = \gcd(f, g)\). Q.E.D.