Reduced ring with only finitely many minimal primes.

Theorem.

For a reduced ring $$R$$ with only finitely many minimal primes show that the following are equivalent.

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

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

Proof.

First of all let us assume that $$\dim (R) = 0.$$ Let the minimal prime ideals of $$R$$ be $$p_1,p_2, \dots , p_n.$$ Since $$\dim (R) = 0$$ each $$p_i$$ is maximal for $$i=1,2, \dots , n.$$ So they are pairwise comaximal. We know that $$\bigcap\limits_{i=1}^{n} p_i = \sqrt {0} = (0),$$ since $$R$$ is given to be reduced. So by Chinese Remainder Theorem we can say that $$R \cong \prod\limits_{i=1}^{n} R/p_i.$$

Conversely, if $$R \cong \prod\limits_{i=1}^{n} K_i,$$ where $$K_i$$ are fields for $$i=1,2, \dots , n$$ then for all ideals $$I \subseteq R$$ the projection of $$I$$ on $$K_i$$ is either $$(0)$$ or $$K_i.$$ Hence the only elements of $$\mathrm{Spec}(R)$$ are $$p_i = K_1 \times K_2 \times \cdots \times K_{i-1} \times (0) \times K_{i+1} \times \cdots \times K_n,$$ and each of such $$p_i$$ are both minimal and maximal.

I have understood the first part of the proof but I find trouble to understand the converse part. How does the projection of ideals of $$R$$ onto $$K_i$$'s guarantee that the elements of $$\mathrm{Spec}(R)$$ are of the above form?

Thank you very much.

EDIT. I think that I have finally understood it. The prime ideals are of the form $$P_1 \times P_2 \times \cdots \times P_n,$$ where $$P_i = (0)$$ or $$K_i$$ for $$i=1,2, \dots,n.$$ But there cannot be more than one $$P_i$$ which are $$(0).$$ For if there is a prime ideal $$P$$ whose $$i$$th and $$j$$th components are $$0$$ then we take elements $$a=(a_1,a_2,\dots, a_{i-1},0,a_{i+1}, \dots ,a_n)$$ and $$b=(b_1,b_2, \dots , b_{j-1},0,b_{j+1} , \dots , b_n)$$ then $$ab \in P$$ though $$a,b \notin P.$$

There is an intermediate step which they glossed over. The projection of any ideal $$I$$ onto $$K_i$$ being either $$(0)$$ or $$K_i$$ (and the existence of elements $$(0,0,\ldots, 0,k,0,\ldots,0)\in R$$) means that all ideals of $$R$$ are of the form $$L_1\times L_2\times\cdots\times L_n$$ where each $$L_i$$ is either $$K_i$$ or $$(0)$$ (independently of one another). We see this because if $$I$$ contains an element where the $$i$$th component is non-zero, then for each $$k\in K_i$$ the ideal contains the element where all the other entries are $$0$$, but the $$i$$th entry is $$k$$. Clearly, $$I$$ also contains any finite sum of such elements.
Such an ideal is prime iff exactly one of the $$L_i$$ is $$(0)$$.
• But there cannot be more than one $L_i$ which are $(0).$ For if there is a prime ideal $P$ whose $i$th and $j$th components are $(0)$ then we can take elements $a=(a_1,a_2,\cdots, a_{i-1},0,a_{i+1}, \cdots ,a_n)$ and $b=(b_1,b_2, \cdots , b_{j-1},0,b_{j+1} , \cdots , b_n)$ then $ab \in P$ though $a,b \notin P.$ – Dbchatto67 Apr 11 at 6:55
• @Dbchatto67 Exactly. Equivalently, dividing out by that ideal gives $K_i\times K_j$, which is not an integral domain. – Arthur Apr 11 at 6:58