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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?

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

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 $a,b \notin P.$

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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 $(0)$.

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  • $\begingroup$ 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 $a,b \notin P.$ $\endgroup$ – Dbchatto67 Apr 11 at 6:55
  • $\begingroup$ @Dbchatto67 Exactly. Equivalently, dividing out by that ideal gives $K_i\times K_j$, which is not an integral domain. $\endgroup$ – Arthur Apr 11 at 6:58
  • $\begingroup$ Yeah. You are right! $\endgroup$ – Dbchatto67 Apr 11 at 7:02

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