If \(K\) is a subfield of \(L\), show that the ring \(S\) of all polynomials over \(L\) with constant coefficient in \(K\) is Noetherian if and only if \(L\) is finite-dimensional over \(K\).
Solution
Write \[ S=K+xL[x] =\{a_0+a_1x+\cdots+a_nx^n: a_0\in K,\ a_i\in L\text{ for }i\ge1\}. \]
1. If \([L:K]<\infty\), then \(S\) is Noetherian.
Choose a \(K\)-basis \[ 1=b_1,b_2,\ldots,b_r \] of \(L\). Then \[ S=K[x,b_2x,\ldots,b_rx]. \] Indeed, every positive-degree coefficient \(a_d\in L\) can be written as \[ a_d=\sum_{i=1}^r c_{id}b_i, \qquad c_{id}\in K, \] so \[ a_dx^d =c_{1d}x^d+\sum_{i=2}^r c_{id}(b_ix)x^{d-1}. \] Thus \(S\) is a finitely generated \(K\)-algebra. Since \(K\) is Noetherian, Hilbert’s basis theorem implies that \(S\) is Noetherian.
2. If \(S\) is Noetherian, then \([L:K]<\infty\).
Consider the ideal \[ I=xL[x]\subseteq S. \] Since \(S\) is Noetherian, there exist \(f_1,\ldots,f_r\in I\) such that \[ I=(f_1,\ldots,f_r)_S. \] Write \[ f_j=c_jx+\text{terms of degree at least }2, \qquad c_j\in L. \] Let \(a\in L\). Since \(ax\in I\), there exist \(s_j\in S\) with \[ ax=\sum_{j=1}^r s_jf_j. \] Every \(s_j\) has constant term in \(K\). Comparing coefficients of \(x\) on both sides gives \[ a\in Kc_1+\cdots+Kc_r. \] Since \(a\in L\) was arbitrary, \[ L=Kc_1+\cdots+Kc_r. \] Therefore \(L\) is finite-dimensional over \(K\).
3. Hence \(S\) is Noetherian if and only if \(L/K\) is finite.
Combine 1 and 2.