1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ I do not really understand the notation either. The essential difference between $\oplus$ and $\times$ is that the universal properties that define them are different (most standard texts have these properties). For finite collections of rings and groups (among other things) the constructions that satisfy these two properties are the same, though of course that isn't true for infinite collections of rings and groups. $\endgroup$ Aug 28, 2014 at 16:29

2 Answers 2

1
$\begingroup$

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)

$\endgroup$
1
$\begingroup$

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."

$\endgroup$

You must log in to answer this question.