finite product of artinian rings is artinian Let $R_1$ and $R_2$ be two left (resp. right) Artinian rings. I would like to prove $R_1\times R_2$ is also a left (resp. right) Artinian ring.
My proof is the following (only for left):
Let $A_1\ge A_2\ge A_3\ge\cdots$ be a decreasing chain of left ideals of $R_1\times R_2$. Then we can rewrite the chain as $$S_1\times T_1\ge S_2\times T_2\ge\cdots,$$where $S_i$ and $T_i$ are ideals of $R_1$ and $R_2$ respectively.
Then since both $R_1$ and $R_2$ are artinian, the set $\{S_1,S_2,\ldots\}$ and $\{T_1,T_2\ldots\}$ have minimal element, say $S_i$ and $T_j$. Let $k=\max\{i,j\}$. Then $S_1\times T_1\ge S_2\times T_2\ge\cdots$ stabilizes at $S_k\times T_k$.
Could you please help to check if my proof is rigorous? Is there a more obvious (easier) way to show this claim?
Thanks!

Modulo the proof that every ideal of $R_1\times R_2$ is of the form $I_1\times I_2$ for $I_i$ and ideal of $R_1$ and $I_2$ an ideal of $R_2$ (easy and well-known), the proof is mostly good.
However you should also mention that from $S_i\times T_i\supseteq S_{i+1}\times T_{i+1}$ it follows that $S_i\supseteq S_{i+1}$ and $T_i\supseteq T_{i+1}$.
Since both chains in the component rings stabilize, also the chain in the product stabilizes.
