Let $R =\{f:\{1,2,3,4,\cdots ,10\}\to \mathbb{Z}_2 \}$ be the set of
all $\mathbb{Z}_2$ -valued functions on the set $\{1,2,3……10\}$ of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?
1) $R$ has a unique maximal ideal.
2) Every prime ideal of $R$ is also maximal.
3 ) The number of proper ideals of $R$ is $511$.
4 ) Every element of $R$ is idempotent.
R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless .
Best Answer
You shouldn't have any trouble showing that your ring is isomorphic to $\mathbb Z_2^{10}$ (the product ring of $10$ copies of $\mathbb Z_2$.) The idea is that you send $\phi :\{1,2,\ldots, 10\}\to\mathbb Z_2$ to $(\phi(1),\phi(2),\ldots,\phi(10))\in \mathbb Z_2^{10}$.
Of course $\mathbb Z_2^{10}$ is a boolean ring since it is a product of boolean rings. This takes care of $(4)$.
It is well-known and easy to prove that the prime ideals are maximal in any boolean ring. Of course, it's also well-known and easy to prove that the prime ideals are maximal in finite rings. This takes care of $(2)$.
The ideals of a finite product of rings are easy to describe: they are all possible products of the ideals in the component rings. Since you have $10$ component rings, each of which have $2$ ideals, you have $2^{10}$ possible ideals. The whole ring is the one non-proper ideal you'll have to discard. This takes care of $(3)$.
The maximal ideals of a finite product of rings are also easy to describe: they are all of the form of the ideals described above, of course. It is easy to show that they are exactly those where you have a maximal ideal in one position, and the full ring in all other positions. So that gives you exactly $10$ maximal ideals, one for each position. This takes care of $(1)$.