Let \(F\) be a field and let \(R\) be a subring of \(F[x]\) with \(F\subset R\). Prove that there exist finitely many polynomials \(f_1,\ldots,f_n\in F[x]\) such that \[ R=F[f_1,\ldots,f_n]. \]
Solution
1. If \(R=F\), the conclusion holds.
For example, \[ R=F[0]. \] Thus one may assume below that \(R\) contains a nonconstant polynomial.
2. Let\[ D=\{\deg h:0\neq h\in R\}\subseteq\NN. \]Then \(D\) is an additive submonoid of \(\NN\) containing a positive integer.
Since \(F\subset R\), nonzero constants show that \(0\in D\). If \(h_1,h_2\in R\) are nonzero, then \(F[x]\) is a domain and \[ \deg(h_1h_2) = \deg h_1+\deg h_2. \] Thus \(D\) is closed under addition. By step 1, the case under consideration contains a nonconstant polynomial, so \(D\) contains a positive integer.
3. The additive monoid \(D\) is finitely generated.
Let \[ d=\min(D\setminus\{0\}). \] For each residue class \(r\pmod d\) that occurs among elements of \(D\), choose the least such element \(s_r\in D\). There are at most \(d\) choices.
If \(s\in D\) and \(s\equiv r\pmod d\), then by minimality \[ s\geq s_r, \] and \[ s-s_r=qd \] for some integer \(q\geq0\). Hence \[ s=s_r+qd. \] Therefore \(D\) is generated by the finite set consisting of \(d\) together with the nonzero \(s_r\).
4. Choose positive generators\[ d_1,\ldots,d_s \]for \(D\), and choose \(f_i\in R\) with\[ \deg f_i=d_i. \]Put\[ R_0=F[f_1,\ldots,f_s]. \]Then\[ R_0\subseteq R. \]
The generators \(d_i\) exist by step 3, and by definition of \(D\) each is the degree of some nonzero polynomial \(f_i\in R\). Since \(R\) is a subring containing both \(F\) and all the \(f_i\), it contains the \(F\)-subalgebra they generate.
5. Every element of \(R\) belongs to \(R_0\).
Proceed by induction on degree. Every element of degree \(0\) is a constant in \(F\), hence lies in \(R_0\).
Now let \(0\neq h\in R\) have degree \(m>0\), and assume every element of \(R\) of degree less than \(m\) lies in \(R_0\). Since the \(d_i\) generate \(D\) and \(m\in D\), write \[ m=\sum_{i=1}^s e_i d_i \] with \(e_i\in\NN\). Then \[ P=\prod_{i=1}^s f_i^{e_i}\in R_0 \] has degree \(m\).
Let \(c_h,c_P\in F^\times\) be the leading coefficients of \(h\) and \(P\), and put \[ \lambda=\frac{c_h}{c_P}\in F. \] The leading terms of \(h\) and \(\lambda P\) agree, so either \[ h-\lambda P=0 \] or \[ \deg(h-\lambda P)<m. \] Moreover \(h-\lambda P\in R\). In the first case \(h=\lambda P\in R_0\); in the second, the induction hypothesis gives \(h-\lambda P\in R_0\), and hence again \[ h=(h-\lambda P)+\lambda P\in R_0. \] Thus the induction closes.
6. Consequently,\[ \boxed{R=F[f_1,\ldots,f_s]}. \]
Step 4 gives \(R_0\subseteq R\), while step 5 gives \(R\subseteq R_0\). Hence equality holds.
7. Q.E.D.
Steps 1 and 6 cover all cases.