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Let $R_1$ and $R_2$ be two left (resp. right) Artinian rings. I would like to prove $R_1times R_2$ is also a left (resp. right) Artinian ring.

My proof is the following (only for left):

Let $A_1ge A_2ge A_3gecdots$ be a decreasing chain of left ideals of $R_1times R_2$. Then we can rewrite the chain as $$S_1times T_1ge S_2times T_2gecdots,$$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 ${T_1,T_2ldots}$ have minimal element, say $S_i$ and $T_j$. Let $k=max{i,j}$. Then $S_1times T_1ge S_2times T_2gecdots$ stabilizes at $S_ktimes T_k$.

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 $R_1times R_2$ is of the form $I_1times I_2$ 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 $S_itimes T_isupseteq S_{i+1}times T_{i+1}$ it follows that $S_isupseteq S_{i+1}$ and $T_isupseteq T_{i+1}$.

Since both chains in the component rings stabilize, also the chain in the product stabilizes.