Image under leading coefficient map is Noetherian implies finitely generated

Statement
Suppose $$R$$ is a commutative unital ring, and $$I$$ is an ideal in the polynomial ring $$R[x]$$. Let $$l:R[x] \to R$$ be the leading coefficient map. Suppose $$l(I)$$ is a Noetherian ideal. Then $$I$$ is also a finitely generated ideal.

Construction of a generating set
Let $$l_n(I)$$ be the subset of $$l(I)$$ comprising those elements of $$R$$ that occur as leading coefficients of polynomials of degree $$n$$. Then $$l_n(I)$$ is an ideal, and further:

$$l_0(I) \le l_1(I) \le l_2(I) \le \ldots$$

And the ascending union is $$l(I)$$.

Since $$l(I)$$ is finitely generated, there exists a $$n$$ such that $$l_n(I) = l(I)$$ (to see this, take a generating set for $$l(I)$$, and consider a $$n$$ such that $$l_n(I)$$ contains all the generators. Now construct a subset of $$I$$ as follows:


 * For each $$0 \le m \le n$$, consider $$l_m(I)$$. This is a submodule of the Noetherian module $$l(I)$$, hence it is finitely generated. Pick a finite generating set for $$l_m(I)$$, and for each member of this generating set, pick a representative polynomial of degree $$m$$. Call the set of such representative polynomials $$T_m$$.
 * Let $$T = \bigcup_{m=0}^n T_m$$.

Then $$T$$ is a generating set for $$I$$.