Is it true that every commutative ring with unity and finitely many maximal ideals can be written as finite direct product of local rings?

Question

Is it true that every commutative ring with unity and finitely many maximal ideals can be written as direct product of local rings?

Answer 1

The theoretical detail to this problem is that while a ring with finitely many maximal ideals is semilocal, a finite product of local rings is semiperfect.

A semiperfect ring is semilocal, but there is a gap between the two notions, namely that a semiperfect ring is a semilocal ring in which idempotents lift modulo the Jacobson radical.

Now, given an integral domain with two or more maximal ideals, it is easy to produce another integral domain (which isn't a product of any two nonzero rings) which has two maximal ideals. You simply take two maximal ideals $M_1, M_2$, and "localize" at the multiplicative set $S=(R\setminus M_1)\cap (R\setminus M_2)$. The images of $M_1$ and $M_2$ in the localization are guaranteed to be the only two maximal ideals.

Answer 2

How about $$R=\{a/(1+6b):a,b\in\Bbb Z\}?$$ This has two maximal ideals, generated by $2$ and $3$, yet is indecomposable (only idempotents are $\pm1$).