[proofplan] We prove both implications. For $(2) \Rightarrow (1)$, since $A$ is a finitely generated algebra over the noetherian ring $A_0$, [Hilbert's Basis Theorem](/theorems/2904) immediately gives noetherianity. For $(1) \Rightarrow (2)$, we first observe that $A_0 \cong A / A_+$ is noetherian as a quotient of a noetherian ring. Then, since $A_+$ is a finitely generated ideal of the noetherian ring $A$, we replace its generators by their homogeneous components to obtain homogeneous generators $x_1, \dots, x_s$ of $A_+$, and prove by induction on degree that $A = A_0[x_1, \dots, x_s]$. [/proofplan]

[step:Prove $(2) \Rightarrow (1)$ via Hilbert's Basis Theorem] Assume $A_0$ is noetherian and $A$ is a finitely generated $A_0$-algebra. By [Hilbert's Basis Theorem](/theorems/2904), every finitely generated algebra over a noetherian ring is noetherian. Since $A_0$ is noetherian and $A$ is a finitely generated $A_0$-algebra, $A$ is noetherian. [/step]

[step:Prove $(1) \Rightarrow (2)$: show $A_0$ is noetherian]Assume $A$ is noetherian. Define the irrelevant ideal $A_+ = \bigoplus_{n=1}^{\infty} A_n$, which is an ideal of $A$. The degree-zero component $A_0$ is isomorphic to the quotient ring $A / A_+$ via the canonical projection $\pi: A \to A / A_+$ that sends $\sum_{n \geq 0} a_n$ to $a_0 + A_+$. This is a surjective ring homomorphism with kernel $A_+$, so $A_0 \cong A / A_+$ by the first isomorphism theorem. Since $A$ is noetherian, every quotient of $A$ is noetherian, so $A_0$ is noetherian.[/step]

[guided]Why is $A_+ \trianglelefteq A$? For any $a = \sum_{n \geq 0} a_n \in A$ and $b = \sum_{n \geq 1} b_n \in A_+$, the product $ab = \sum_k \sum_{i+j=k} a_i b_j$ has no degree-zero component because every summand involves some $b_j$ with $j \geq 1$, so $i + j \geq 1$. Hence $ab \in A_+$. Similarly, $A_+$ is closed under addition. So $A_+$ is an ideal. The projection $\pi: A \to A_0$ defined by $\pi(\sum_{n \geq 0} a_n) = a_0$ is a ring homomorphism: the product of two degree-zero elements has degree zero (since $A_0$ is a subring), and $1_A \in A_0$. The kernel of $\pi$ is $A_+$, so by the first isomorphism theorem, $A_0 \cong A / A_+$. Why is a quotient of a noetherian ring noetherian? The ideals of $A / A_+$ correspond bijectively (via the lattice isomorphism theorem) to the ideals of $A$ containing $A_+$. An ascending chain in $A / A_+$ lifts to an ascending chain in $A$, which stabilises since $A$ is noetherian, so the original chain stabilises.[/guided]

[step:Show $A_+$ is finitely generated by homogeneous elements]Since $A$ is noetherian, every ideal of $A$ is finitely generated. In particular, $A_+$ is finitely generated: there exist $y_1, \dots, y_t \in A_+$ with $A_+ = (y_1, \dots, y_t)$. Each $y_j$ decomposes into its homogeneous components $y_j = y_{j,1} + y_{j,2} + \cdots$ with $y_{j,n} \in A_n$ and only finitely many nonzero terms. Since $A_+$ is a homogeneous ideal (it equals the direct sum $\bigoplus_{n \geq 1} A_n$, so any element of $A_+$ has all its homogeneous components in $A_+$), every nonzero homogeneous component of each $y_j$ also lies in $A_+$. Let $x_1, \dots, x_s$ be the full list of all nonzero homogeneous components of $y_1, \dots, y_t$. Each $x_i \in A_{k_i}$ for some $k_i \geq 1$. Since the $x_i$ include all homogeneous components of the generators $y_j$, and each $y_j$ is a sum of its own homogeneous components, we have $(x_1, \dots, x_s) \supseteq (y_1, \dots, y_t) = A_+$. The reverse inclusion $(x_1, \dots, x_s) \subseteq A_+$ holds because each $x_i \in A_+$. Hence $A_+ = (x_1, \dots, x_s)$.[/step]

[guided]Why can we replace the generators by homogeneous ones? The ideal $A_+$ is *homogeneous*: it is defined as the direct sum of all positive-degree components, so if $f \in A_+$, then every homogeneous component of $f$ also lies in $A_+$. Given generators $y_1, \dots, y_t$ of $A_+$, each $y_j$ is a finite sum of homogeneous elements, all of which lie in $A_+$. Since $y_j$ is the sum of its own homogeneous components and each component lies in $(x_1, \dots, x_s)$ (they are among the $x_i$), we have $y_j \in (x_1, \dots, x_s)$. Hence $(y_1, \dots, y_t) \subseteq (x_1, \dots, x_s)$, giving $A_+ \subseteq (x_1, \dots, x_s)$. The reverse inclusion is immediate since each $x_i \in A_+$.[/guided]

[step:Prove by induction on degree that $A_0[x_1, \dots, x_s]= A$] Let $A' = A_0[x_1, \dots, x_s]$ denote the $A_0$-subalgebra of $A$ generated by $x_1, \dots, x_s$. We prove $A_n \subseteq A'$ for all $n \geq 0$ by induction on $n$. **Base case ($n = 0$).** By definition, $A_0 \subseteq A'$. **Inductive step.** Let $n \geq 1$ and assume $A_j \subseteq A'$ for all $0 \leq j < n$. Let $y \in A_n$. Since $n \geq 1$, the element $y \in A_+$. Because $A_+ = (x_1, \dots, x_s)$, there exist $r_1, \dots, r_s \in A$ with \begin{align*} y = \sum_{i=1}^{s} r_i x_i. \end{align*} Each $r_i$ has a homogeneous decomposition $r_i = \sum_{j \geq 0} r_{i,j}$ with $r_{i,j} \in A_j$. Since $x_i \in A_{k_i}$, the product $r_{i,j} x_i$ lies in $A_{j + k_i}$. Comparing the degree-$n$ components on both sides (using the fact that $A = \bigoplus_{n \geq 0} A_n$ is an internal direct sum, so the homogeneous decomposition is unique), we obtain \begin{align*} y = \sum_{i=1}^{s} r_{i, \, n - k_i} \, x_i, \end{align*} where $r_{i, \, n-k_i} \in A_{n - k_i}$ and we set $r_{i, \, n-k_i} = 0$ when $n - k_i < 0$. Since $k_i \geq 1$ for each $i$, we have $n - k_i < n$. By the inductive hypothesis, each $r_{i, \, n-k_i} \in A'$. Since $x_i \in A'$ by definition, each product $r_{i, \, n-k_i} \, x_i \in A'$, and so $y \in A'$. By induction, $A_n \subseteq A'$ for every $n \geq 0$, so $A = \bigoplus_{n \geq 0} A_n \subseteq A'$. Since $A' \subseteq A$ by construction, we conclude $A = A' = A_0[x_1, \dots, x_s]$. Hence $A$ is a finitely generated $A_0$-algebra.[/step]

[guided]The goal is to show that the $A_0$-subalgebra $A' = A_0[x_1, \dots, x_s]$ generated by the homogeneous elements $x_1, \dots, x_s$ equals all of $A$. We use induction on the grading degree. For the base case, $A_0 \subseteq A'$ holds by definition (every $A_0$-subalgebra contains $A_0$). For the inductive step, take any homogeneous element $y \in A_n$ with $n \geq 1$. Since $y \in A_+$ and $A_+ = (x_1, \dots, x_s)$, we can write $y = \sum_{i=1}^{s} r_i x_i$ for some $r_i \in A$. The key technique is to **extract the degree-$n$ component** from both sides of this equation. Because the grading makes $A$ an internal direct sum $A = \bigoplus_{m \geq 0} A_m$, the equation $y = \sum_i r_i x_i$ can be decomposed degree by degree. On the left side, $y$ is purely in degree $n$. On the right side, the degree-$n$ contribution from $r_i x_i$ comes from the component $r_{i, \, n-k_i} \in A_{n-k_i}$ (since $x_i \in A_{k_i}$, and degrees add under multiplication). Therefore \begin{align*} y = \sum_{i=1}^{s} r_{i, \, n - k_i} \, x_i. \end{align*} Why does the induction close? Because each $k_i \geq 1$, so $n - k_i \leq n - 1 < n$. By the inductive hypothesis, $A_j \subseteq A'$ for all $j < n$, so each coefficient $r_{i, \, n-k_i}$ lies in $A'$. Since the generators $x_i$ also lie in $A'$, the product $r_{i, \, n-k_i} x_i \in A'$, and summing gives $y \in A'$. By induction, every homogeneous component of $A$ lies in $A'$, so $A \subseteq A'$. Since $A' \subseteq A$ by construction, $A = A'$ and $A$ is finitely generated as an $A_0$-algebra.[/guided]