Commutative algebra – Show that $ dim (R) = 0. $

Theorem $: $

For a reduced ring $ R $ with only finitely many minimal primes, the following is equivalent.

$ (1) $ $ dim (R) = 0. $

$ (2) $ $ R $ is isomorphic to a direct product of many fields.

Evidence $: $

First, suppose that $ dim (R) = 0. $ Let the first minimal ideals of $ R $ be $ p_1, p_2, cdots, p_n. $ Since $ dim (R) = $ 0 each $ p_i $ is maximum for $ i = 1,2, cdots, n. $ They are therefore co-maximals two by two. We know that $ bigcap limits_ {i = 1} ^ {n} p_i = sqrt {0} = (0), $ since $ R $ is given to be reduced. So, by the Chinese rest theorem, we can say that $$ R cong prod limits_ {i = 1} ^ {n} R / p_i. $$

Conversely, if $ R cong prod limits_ {i = 1} ^ {n} K_i, $ or $ K_i $are fields for $ i = 1,2, cdots, n $ so for all the ideals $ I subseteq R $ the projection of $ i $ in $ K_i $ is either $ (0) $ or $ K_i. $ Therefore, only elements of $ mathcal {Spec} (R) $ are $$ p_i = K_1 times K_2 times cdots times K_ {i-1} times (0) times K_ {i + 1} times cdots times K_n, $$ each of these $ p_i $ are both minimal and maximal.

I have understood the first part of the proof, but I have a hard time understanding the opposite part. How does the projection of the ideals of $ R $ sure $ K_i $is the guarantee that the elements of $ mathcal {Spec} (R) $ are of the form above?

Please help me in this regard. Any help will be greatly appreciated.

Thank you so much.