= Solution
\b[Hilbert's basis theorem:] if $R$ is a commutative <Noetherian ring> with identity, then the <polynomial ring> $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.
Here is a <bounded-degree coefficient proof of Hilbert basis theorem>. Let $I$ be any <ideal> of $R[x]$. For $n\ge0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the <polynomials> of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an <ideal> of $R$. Multiplication of a <polynomial> by $x$ shows $J_n\subseteq J_{n+1}$. By the <ascending chain condition>, there is $N$ such that $J_N=J_{N+1}=\cdots$.
Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\ldots,a_{n,r_n}$ and corresponding <polynomials> $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen <polynomials>. We prove they generate $I$ by induction on degree.
Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a=\sum_jc_ja_{d,j}$ and subtract $\sum_jc_jf_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a=\sum_jc_ja_{N,j}$, and subtract $\sum_jc_jx^{d-N}f_{N,j}$. In both cases the leading coefficient cancels, leaving a <polynomial> in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.
Thus every <ideal> of $R[x]$ is finitely generated, proving the <Hilbert basis theorem>. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has <zero divisors>.
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