Ideals with infinitely many generators As the title may suggest, I am not sure if I understand the ideals with infinitely many generators correctly. Let us take the ring $R=\mathbb{Z}[x_1,x_2,x_3,\dots]$ and consider the ideal $(x_1,x_2,x_3\dots)$ generated by all the indeterminates in the polynomial ring $R$. As a set
\begin{equation}
(x_1,x_2,x_3,\dots)=\{f_1x_1+f_2x_2+f_3x_3+\dots|f_1,f_2,f_3,\dots\in R\}.
\end{equation}
My understanding is that the $+$ in the above notation does not stand for the ring operation (we can't add infinitely many elements) and that $f_1x_1+f_2x_2+f_3x_3+\dots$ is an element of $R$ and so automatically only finitely many $f_i\neq0_R$. Is my understanding correct?

Writing $$(x_1,x_2,x_3,\dots)=\{f_1x_1+f_2x_2+f_3x_3+\dots|f_1,f_2,f_3,\dots\in R\}$$ is rather imprecise, because (as you said), you cannot add infinitely many elements in a ring (unless you have a topology). It would be better to write instead
$$
(x_1,x_2,x_3,\dots) = \left\{\sum_{i\geq 1} f_i x_i\mid f_i\in R\textrm{ and }f_i = 0\textrm{ for all but finitely many }i\right\}.
$$
The sum in the set on the right hand side is now well-defined, and this is the ideal that you want.
