In ring theory, the Chinese Remainder Theorem is stated as follows.
Let $A_1, \dotsc, A_n$ be ideals in a ring $R$ such that $R^2 + A_i = R$ for all $i$ and $A_i + A_j = R$ for all $i \neq j$. If $b_1, \dotsc, b_n \in R$, then there exists $b \in R$ such that $$b \equiv b_i \pmod{A_i}$$ where $i = 1,\dotsc, n$. Furthermore $b$ is uniquely determined up to congruence modulo the ideal $A_1 \cap \cdots \cap A_n$.
As mentioned in the title, what is $R^2?$ At first, I thought it was the Cartesian product of $R$, but after reading the proof of the above theorem, that's definitely not it.
Thanks!
Edit The way the theorem is written above is taken from Thomas Hungerford's book "Algebra"
In the source you reference, the author takes non-empty subsets $A, B$ of a ring $R$ and defines the subset $AB$ to be the set of all finite sums $$ a_1b_1 + a_2b_2 + \cdots +a_kb_k $$ where $a_i \in A, b_i \in B$. He then declares that $A^n$ will be shorthand for this construction applied to $A$ $n$ times in succession.
He then proves that when $A$ and $B$ are ideals in $R$, so is $AB$.
In particular $R^2$ is then the ideal $RR$ consisting of all finite sums of the form $$ a_1b_1 + a_2b_2 + \cdots +a_kb_k $$ where $a_i , b_i \in R$.
Source: fifth printing of Algebra by Thomas W. Hungerford, Springer-Verlag 1989.