Exercise 25, § 1

Let I be the semilattice ordered monoid of ideals of a commutative ring A (Ex. 24). For a finite system of congruences $x \equiv a_i (a_i)$ to admit a solution whenever any two of them admit a common solution (that is to say whenever

for every pair of indices i, j), it is necessary and sufficient that each of the laws

in I be distributive over the other ("Chinese remainder theorem"). One may proceed as follows: a) If $(a_1 \cap a_2) + (a_1 \cap a_3) = a_1 \cap (a_2 + a_3)$ and if any two of the three congruences $x \equiv a_i (a_i)$ ($i = 1, 2, 3$) admit a common solution, then the three congruences have a common solution (let $x_{12}$ be a common solution of $x \equiv a_1(a_1)$ and $x \equiv a_2(a_2)$ and let $x_{13}$ be a common solution of $x \equiv a_1(a_1)$ and $x \equiv a_3(a_3)$; show that the congruences $x = x_{12}(a_1 \cap a_2)$ and $x \equiv x_{13}(a_1 \cap a_3)$ have a common solution). b) If any system of three congruences, any two of which have a common solution, admits a common solution, show that the following distributive laws hold

(For the first, note that for all $x \in (a + b) \cap (a + c)$, there exists $y \in b \cap c$ such that $y \equiv x(a)$; for the second, note that for all $x \in a \cap (b + c)$, there exists $y \in a \cap b$ such that $y \equiv x(c)$.) Note that by a) and b), the second distributive law implies the first (cf. Set Theory, III, p. 217, Ex. 16). c) Prove the « Chinese remainder theorem » by induction on the number of congruences under consideration, by an analogous argument to that of a). d) Translation in the case of principal ideal domains.