[proofplan] We prove the result by induction on the number of polynomial variables. The base case is exactly [Hilbert's basis theorem](/theorems/2904) for one variable over the [Noetherian ring](/page/Noetherian%20Ring) $R$. For the induction step, we identify the [polynomial ring](/page/Polynomial%20Ring) in $m+1$ variables as an iterated polynomial ring $R[x_1,\ldots,x_m][x_{m+1}]$, then apply the induction hypothesis followed by [Hilbert's basis theorem](/theorems/2907). [/proofplan]

[step:State the one-variable input used in the induction] We use the following external result, the one-variable [Hilbert basis theorem](/theorems/860): if $A$ is a commutative Noetherian ring, then the one-variable polynomial ring $A[t]$ is Noetherian. Its hypotheses are exactly commutativity and Noetherianity of the coefficient ring. This is the only external input in the proof. [/step]

[step:Prove the base case by applying Hilbert's basis theorem to $R$] Define the assertion $P(m)$, for $m \in \mathbb{N}$, to mean that $R[x_1,\ldots,x_m]$ is Noetherian. For $m=1$, the ring $R$ is commutative and Noetherian by hypothesis. Applying Hilbert's basis theorem with $A=R$ and $t=x_1$, we obtain that $R[x_1]$ is Noetherian. Hence $P(1)$ holds. [/step]

[step:Identify the next polynomial ring as a one-variable extension]Let $m \in \mathbb{N}$ with $m \ge 1$, and assume $P(m)$ holds. Define \begin{align*} A_m := R[x_1,\ldots,x_m]. \end{align*} By the induction hypothesis, $A_m$ is Noetherian. Since $R$ is commutative, the polynomial ring $A_m$ is commutative. By the construction of polynomial rings in finitely many variables as iterated polynomial extensions, there is a canonical $R$-algebra isomorphism \begin{align*} A_m[x_{m+1}] \cong R[x_1,\ldots,x_m,x_{m+1}]. \end{align*} Noetherianity is invariant under ring isomorphism: an isomorphism carries ideals bijectively to ideals and preserves finite generation of ideals. Thus it is enough to prove that $A_m[x_{m+1}]$ is Noetherian.[/step]

[guided]Let $m \in \mathbb{N}$ with $m \ge 1$, and suppose the induction hypothesis $P(m)$ is true. This means precisely that the polynomial ring \begin{align*} A_m := R[x_1,\ldots,x_m] \end{align*} is Noetherian. Since $R$ is commutative, the polynomial ring $A_m$ is also commutative. The reason for introducing $A_m$ is that the ring in $m+1$ variables can be viewed as a one-variable polynomial ring over the ring already handled by the induction hypothesis. More explicitly, the variables $x_1,\ldots,x_m$ are regarded as coefficients for a new polynomial variable $x_{m+1}$. The standard iterated-polynomial-ring identification gives a canonical $R$-algebra isomorphism \begin{align*} A_m[x_{m+1}] \cong R[x_1,\ldots,x_m,x_{m+1}]. \end{align*} This identification is compatible with addition, multiplication, and the inclusion of $R$. Therefore the target ring $R[x_1,\ldots,x_m,x_{m+1}]$ is Noetherian if and only if the isomorphic ring $A_m[x_{m+1}]$ is Noetherian. So the induction step reduces exactly to the one-variable case over the coefficient ring $A_m$.[/guided]

[step:Apply Hilbert's basis theorem to complete the induction step] The ring $A_m$ is commutative and Noetherian, as shown in the previous step. Applying Hilbert's basis theorem with $A=A_m$ and $t=x_{m+1}$, the polynomial ring $A_m[x_{m+1}]$ is Noetherian. Using the canonical isomorphism \begin{align*} A_m[x_{m+1}] \cong R[x_1,\ldots,x_m,x_{m+1}], \end{align*} we conclude that $R[x_1,\ldots,x_m,x_{m+1}]$ is Noetherian. Hence $P(m+1)$ holds. [/step]

[step:Conclude the result for every positive number of variables] We have proved $P(1)$ and shown that $P(m)$ implies $P(m+1)$ for every $m \in \mathbb{N}$. By induction on $m$, $P(n)$ holds for every positive integer $n$. Therefore, for every positive integer $n$, the ring $R[x_1,\ldots,x_n]$ is Noetherian. [/step]