## Hilbert basis theorem

I remember learning this theorem early 2015, but I could not remember its proof at all. Today, I relearned it. It employed a beautiful induction argument to transfer the Noetherianness (in the form of finite generation) from $R$ to $R[x]$ via the leading coefficient.

Hilbert Basis TheoremIf $R$ is a Noetherian ring, then so is $R[x]$.

Proof: Take some ideal $J$ in $R$. Notice that if we partition $J$ by degree, we get from the leading coefficients appearing in each an ascending chain (that has to become constant eventually, say at $k$). Take finite sets $A_n \subset J$ for $m \leq n \leq k$, where $m$ is the smallest possible non-zero degree such that the $I_n$s for the leading coefficient ideals are generated. With this we can for any polynomial $p$ construct a finite combination within $A = \displaystyle\cup_{n=m}^k A_n$ that equates to $p$ leading coefficient wise, and thereby subtraction reduces to a lower degree. Such naturally lends itself induction, with $m$ as the base case. For $m$ any lower degree polynomial is the zero polynomial. Now assume, as the inductive hypothesis that $A$ acts as a finite generating set all polynomials with degree at most $n$. If $n+1 \leq k$, we can cancel out the leading coefficient using our generating set, and then use the inductive hypothesis. If $n+1 > k$, we can by our inductive hypothesis generate with $A$ a degree $n$ polynomial with same leading coefficient (and thereby a degree $n+1$ one multiplying by $x$) and from that apply our inductive hypothesis again, this time on our difference.