Hilbert's Basis Theorem question

Question

If $F$ is a field and $R = F[t_1, t_2, ... t_k]$ and $Y$ is a set of polynomials in $k$ variables over $F$ then by Hilbert's basis theorem apparently $YR = \sum\limits_{i=1}^m f_i R$ for some $f_1,...,f_m ∈ R$, but I don't see why this is? Any help appreciated, thanks!

Answer 1

Hilbert's Basis Theorem says that if $R$ is Noetherian, then so is $R[t_1]$. By induction, then, it follows that $R[t_1,\cdots,t_n]$ is Noetherian.

Now, as user26857 says in the comments, in a Noetherian ring, every ideal is finitely generated. For completeness, I'll give a proof of that fact:

First, recall that Noetherian means that every ascending chain terminates. I.e, every chain of ideals $I_1 \subset I_2 \subset I_3 \cdots \subset $ must terminate, meaning that $I_n = I_{n+1}$ for all $n \gg 0$.

For an example: Let $R=k[t_1,t_2]$. Then we have the chain of ideals $(0) \subset (t_1) \subset (t_1,t_2)$. It is not possible to squeeze more ideals in this sequence.

Now I'll prove that assuming the ascending chain condition, every ideal is finitely generated. Suppose the opposite, i.e. $I$ requires an infinite number of minimal generators $I=\langle f_1,\cdots,f_k,f_{k+1},\dotsc \rangle$. Then you can make an infinite descending chain ($I_k=\langle f_1,\cdots,f_k \rangle$), contrary to the Noetherian assumption. Thus every ideal in a Noetherian ring is finitely generated.

In fact, the implication goes the other way as well. That is, if every ideal is finitely generated, then the ascending chain condition holds. This is easy: suppose the opposite, i.e. the existence of an infinite ascending chain of ideals $\{ I_k \}_{k=0}^\infty$. Then the ideal $\bigcup_k I_k$ is finitely generated, by say $\{ x_1,\cdots,x_r\}$. Each of these belong to some $I_k$, so choosing $\ell$ to be the maximal such $k$, we see that all $x_r$ belong to $I_\ell$. Hence $I_\ell = \langle x_1,\cdots,x_r \rangle$, so the chain is stationary.