I’m using the definitions: $\underline{a}=(a_i)_{i\in I} \in A^I $ (where $A$ is a ring),
$$A^{(I)}:=\{ \underline{a} \in A^I \mid |\mathrm{supp} (\underline{a})| < \infty \} \subset A^I,$$
where $\mathrm{supp}(\underline{a}):=\{i\in I \mid a_i \neq 0 \} \subset I$.
I’d like to understand why the ideal $A^{(\mathbb{N})}\subset A^\mathbb{N}$ isn’t finitely generated and therefore $A^\mathbb{N}$ isn’t noetherian.
This seems counterintuitive to me right now from the definitions.
Let $\underline a_1,\dots,\underline a_n\in A^{(\mathbb N)}$. Write $\underline a_i=(a_{i1},\dots)$ with $a_{ij}\in A$. There is $m\ge1$ such that $a_{ij}=0$ for all $i=1,\dots,n$ and $j>m$. So we can write $\underline a_i=(a_{i1},\dots,a_{im},0,0,\dots)$. Since the multiplication in $A^{\mathbb N}$ is component-wise, a linear combination $\sum_{i=1}^n\underline b_i\underline a_i$ with $\underline b_i\in A^{\mathbb N}$ has the same property: all its components from $m+1$ on are zero. This shows that $A^{(\mathbb N)}$ is not a finitely generated ideal, since an element like $(0,\dots,0,1,0,\dots)$ with $1$ on the $m+1$ position, which obviously belongs to $A^{(\mathbb N)}$, can't be written as a linear combination of the supposed generators.