Rationale for expressing as a direct sum and a direct product

Question

In "Ireland and Rosen" page 35, it says if $R_1, R_2, ..., R_n$ are rings, then

$R_1 \oplus R_2 \oplus \dots \oplus R_n = S$ is the direct sum of the $R_i$.

Later in a proposition it says if $S = R_1 \oplus R_2 \oplus \dots \oplus R_n$,

then the group of units $U(S) = U(R_1) \times \dots \times U(R_n)$.

I would appreciate help understanding why in the first instance $S$ is expressed as a direct sum, and the group of units (over the same set of $R_i$) is expressed as a direct product. Thanks

Answer 1

It is standard to use $\oplus$ for the biproduct of modules. (aside: for infinite products of modules, $\oplus$ is interpreted as the coproduct)

If the modules $M$ and $N$ have an algebra structure, then they induce a canonical algebra structure on their direct sum $M \oplus N$.

It is standard, but strange, notation to write $R \oplus S$ for the algebra constructed by forgetting the algebra structure on $R$ and $S$, taking the direct sum of the modules, and then putting the canonical algebra structure back on the direct sum. I really don't understand why this notation is used, since it is rather misleading, as it is very much not the coproduct of the rings. (although it is their product)

Answer 2

Every element of $S$ can be written additively as $r_1+\cdots+r_n$ where $r_i\in R_i$.

Every element of $U(S)$ can be written multiplicatively as $u_1\cdots u_n$ where $u_i\in U(R_i)$.

We view the $R_i$s as subrings of $R$, so the sums/products are "internal."