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Use the usual commutative unital-ring convention for Noetherian rings. A ring is Noetherian exactly when every ideal is finitely generated, equivalently when ascending chains of ideals stabilize. Indeed, the union of an ascending chain is an ideal; finitely many generators all lie in a single chain member. Conversely, if an ideal cannot be finitely generated, adjoining a new element at each stage constructs a strictly ascending chain.

First assume $R[X]$ is Noetherian. Evaluation at zero gives a surjective ring homomorphism $R[X]\to R$ with kernel $(X)$. A quotient of a Noetherian ring is Noetherian: lift an ideal to the source, take finitely many generators there, and map those generators down. Thus $R\cong R[X]/(X)$ is Noetherian.

For the converse, assume $R$ is Noetherian and take any ideal $I\subseteq R[X]$. For $d\ge 0$, let $J_d$ be the set of coefficients of $X^d$ in elements of $I$ of degree at most $d$. It is an ideal of $R$, since sums and scalar multiples of such polynomials stay in $I$. Multiplication by $X$ shows $J_d\subseteq J_{d+1}$. The ascending chain stabilizes, say at $N$.

For each $0\le d\le N$, choose finitely many generators $a_{d,j}$ of $J_d$ and corresponding $f_{d,j}\in I$ of degree at most $d$ having coefficient $a_{d,j}$ at $X^d$. Discard zero generators. We claim these finitely many polynomials generate $I$. If a nonzero $f\in I$ has degree $m\le N$, express its leading coefficient as ${\sum }_j{r}_j{a}_{m,j}$ and subtract ${\sum }_j{r}_j{f}_{m,j}$. The degree decreases. If $m>N$, use $J_m=J_N$, express the leading coefficient using the $a_{N,j}$, and subtractAgain the leading term cancels and the degree decreases. Induction on degree, including degree zero, expresses every $f$ in the claimed ideal. Therefore every ideal of $R[X]$ is finitely generated, proving the Hilbert basis theorem andThe argument is the bounded-degree coefficient proof of Hilbert basis theorem; it never divides by a leading coefficient and therefore does not require $R$ to be an integral domain. With the analogous left-ideal convention the same proof applies to left-Noetherian rings with a central indeterminate.

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 $n\ge 0$, 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 $a_{n,1},\dots ,a_{n,r_n}$ 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 $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. 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.