Hilbert's Basis Theorem

Theorem
If $$A \ $$ is a Noetherian ring with identity, then so is the polynomial ring $$A[x] \ $$.

Proof
Let $$I\in A[x] \ $$ be an ideal in $$A[x] \ $$ and define $$I_a = \left\{{a\in A:\exists f \in I \ \text{s.t.} \ f(x)=ax^n+\dots }\right\} \ $$. wait does this definition make sense?


 * Lemma: $$I_a \ $$ is an ideal in $$A \ $$ and $$I_k\subset I_{k+1} \ $$.
 * Proof:

Since $$A \ $$ is Noetherian, this chain stabilizes for some $$N\in\mathbb{N} \ $$. Also, $$I_i \ $$ has a finite number of generators $$\left\{{a_{i1}, \dots, a_{i \ m(i)} }\right\} \in I_i \ $$. For each $$a_{ik} \ $$ choose some polynomial $$f_{ik}(x)=a_{ik}x^i+\dots\in I \ $$.


 * to finish: show the  $$f_ik \ $$ generate $$I \ $$ with $$i=0,\dots,N, k=1, \dots, m(i) \ $$. Since $$I \ $$ was any ideal, the fact that it is finitely generated implies $$A[x] \ $$ is Noetherian.