Ring of $\mathbb{Z}_2$-valued functions

Solution 1:

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 $(1)$.

Solution 2:

4) is true because every element of $\mathbb{Z}_2=\{0,1\}$ is idempotent, so $\forall \Phi \in R, \forall k\in \{1,...,10\}, \Phi^2(k)= \Phi(k) * \Phi(k) = \Phi(k)$.