Direct sum of noetherian rings If $R_{1}, R_{2}, ..., R_{n}$ is a family of noetherian rings, is $$R = R_{1} \oplus R_{2} \oplus ... \oplus R_{n}$$ a noetherian ring?
From a theorem I have an equivalence that says if a ring $R$ is noetherian then every nonempty set $S$ of left ideals of R has a maximal element. If $I \in S$, then $rI = (r_{1} + r_{2} + ... + r_{n})I = r_{1}I + r_{2}I + ... + r_{n}I\subseteq I$. Would this mean that $I$ is also a left ideal of each $R_{i}$ in the family of noetherian rings? If, so is it correct to conclude that any set $S$ of left ideals of R is also a set of left ideals of $\{ R_{i} \}$, thus having a maximal element and giving that $R$ is noetherian?
I understand what noetherian/artinian rings and modules are, but I'm having some trouble actually working with it.
 A: The picture to have in mind for product rings (a picture which does not work for modules, by the way) is that a right ideal of $\prod_{i=1}^n R_i$ is always of the form $\prod_{i=1}^n T_i$ where each $T_i$ is a right ideal of $R_i$. This only applies to finite products. Containment works like you expect: $\prod_{i=1}^n T_i
\subseteq \prod_{i=1}^n T'_i$ iff $T_i\subseteq T'_i$ for all $i$.
You can use this to solve your problem in a number of ways.
Given an ascending chain of right ideals $_1T\subseteq {_2T}\subseteq \ldots$ (I'm putting the subscripts on the left so they don't look like the subscripts of right ideals of the product ring above), you can find, for each index $i$, an index $i_n$ such that $_1T_i\subseteq {_2T_i}\subseteq \ldots$ has stabilized after $i_n$ steps. Since there are only finitely many $i$, you can just take the maximum $i_n$, and that means all the chains are simultaneously stabilized, and hence the original chain has stabilized.
Alternatively, if $N$ is an $R$ submodule of $M$ and you believe that $M$ is Noetherian iff $M/N$ and $N$ are Noetherian, then you can work inductively by saying that if $R=R_1\times R_2$ where $R_i$ are both Noetherian, you can use the first observation to conclude that the $R$ submodules of $R_1$ are exactly $T_1\times\{0\}$ for each right ideal $T_1\subseteq R_1$, and similarly for $R_2$, so that both of the $R$ modules $R/R_1\cong R_2$ and $R_1$ are Noetherian $R$ modules (since their submodules match those of the $R_i$, which are both Noetherian.)
Your idea of checking by finite generation also works once you note that $T=\prod_{i=1}^n T_i$ is finitely generated as an $R$ module iff each $T_i$ is finitely generated as an $R_i$ module.
