The total ring of fractions of a reduced Noetherian ring is a direct product of fields This is question 6.5 in Matsumura's "Commutative ring theory":

How can I prove that the total ring of fractions of a reduced Noetherian ring is a direct product of fields?

 A: Since $A$ is a reduced Noetherian ring, the minimal prime ideals $P_1, \dots, P_n$ are exactly the primes belonging to $(0)$. Pick $S = A \setminus (P_1 \cup \dots \cup P_n)$ the multiplicative set of non-zero-divisors of $A$.
Observe that the only primes of $S^{-1}A$ are $S^{-1}P_1, \dots, S^{-1}P_n$, so they are pairwise coprime. By Chinese Remainder Theorem, the product
$$
f \ \colon \ S^{-1} A \longrightarrow \prod_{i=1}^n S^{-1}A / S^{-1} P_i
$$
of the quotient projections $S^{-1} A \to S^{-1}A / S^{-1} P_i$ is surjective. On the other hand, $f$ is injective because $S^{-1} P_1 \cap \dots \cap S^{-1} P_n = 0$. It is clear that $S^{-1} A / S^{-1}P_i$ is the residue field $k(P_i)$ of $P_i$.
A: The total ring of fraction $Q(A)$ is artinian (its non invertible elements are zero divisors so that all its minimal primes are maximal): it is the product
$$ A=\prod_i A/\mathfrak{m}_i^{k_i} $$
with $\mathfrak{m}_i$ its minimal/maximal prime ideals (see for example Atiyah MacDonald 8.7)
The total ring of fraction $Q(A)$ is still reduced (commutation of $S^{-1}$ and nilradical - Atiyah MacDonald 3.11 - or by hand $(a/s)^k=0\implies ta^k=0\implies (ta)^k=0\implies A/s=0$): all the factor $A/\mathfrak{m}_i^{k_i}$ must be with $k_i=1$.
