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Let $A_1,A_2,\dots,A_n$ be a family of (right) ideals. $A_1+A_2+\cdots +A_n$ is the sum of these ideals (it is the smallest ideal containing the $A_i$'s). Another way to denote this sum is $\sum_{i=1}^n A_i$. It can be proven that the sum is the set $$ \{a_1+a_2+\cdots + a_n \;|\; a_i \in A_i, i=1,\dots n \}. $$ The following is a definition:

A sum $A=\sum_{i=1}^n A_i$ of (right, left) ideals in a ring $R$ is called a direct sum if each element $a\in A$ is uniquely expressible in the form $\sum_{i=1}^n a_i$ with $a_i \in A_i$. If the sum is a direct sum we write it as $A=A_1 \oplus A_2\oplus \cdots \oplus A_n=\bigoplus\sum_{i=1}^n A_i$.

My question is: in the above definition, is the sum sign (I mean the capital sigma) superfluous? Why don't we simply write $A$ as $\bigoplus_{i=1}^n A_i$ ?

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I think it's written without the $\Sigma$. Where did you happen to see the $\Sigma$? –  user21436 Feb 14 '12 at 15:05
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I've never seen $\bigoplus\sum$, and I read quite a bit of algebra. –  Dylan Moreland Feb 14 '12 at 15:06
    
Yeah, that looks like a typo. –  Thomas Andrews Feb 14 '12 at 15:23
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In a freshman's course maybe someone does this to distinguish between the concept of an internal and an external direct sum. –  Julian Kuelshammer Feb 14 '12 at 15:38
    
It is in Bhattacharya's "Basic Abstract Algebra (2ed)" at page 196. I need the counterexample at page 210 (a nil ideal is not necessarily nilpotent), that can be found in Hungerford's "Algebra" too (but with different notation) where one can read "the subring $\sum \mathbb{Z}_{p^n}$ of $\prod \mathbb{Z}_{p^n}$"... –  Oo3 Feb 14 '12 at 17:31

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Then I suspect that the $\oplus$ is just an assertion that the $\sum$ is a direct sum.

Work by analogy with vector spaces. Assume $V$ is a vector space over a field $k$, and assume $W_1,...,W_n$ are sub-spaces of $V$.

Then we can define the sum $\sum W_i$, which is the smallest subspace of $V$ containing all the $W_i$. You can also define the vector space, $\oplus_i W_i$. This is an "exterior" sum, because its members are not members of $V$, but rather tuples $(w_1,...,w_n)$ with $w_i\in W_i$.

Now, there is a simple linear map, $\phi: \oplus W_i \rightarrow \sum W_i$ defined as $\phi(w_1,...,w_n) = \sum_i w_i$.

This map is not necessarily 1-1. Indeed, if $W_1\subset W_2$ are finite dimensional, then the dimension of $W_1\oplus W_2$ is the sum of the dimensions of $W_1$ and $W_2$, while the dimensions of $W_1+W_2=W_2$ is the dimension of $W_2$.

When this map is $1-1$, then the vector spaces $\oplus W_i$ and $\sum W_i$ are "isomorphic."

In that case, the statement $W=\oplus \sum_{i=1}^n W_i$ is a statement in two parts:

  1. $W=\sum_{i=1}^n W_i$ - that is, it is the smallest subspace of $V$ that contains all the $W_i$.
  2. The map $\oplus_{i=1}^n W_i \rightarrow \sum_{i=1}^n W_i$ is 1-1.

So this is one of those cases where the language is confusing, because it the $\oplus$ symbol here is an assertion about the nature of the $\sum$.

There is a way in which this analogy is precise, in that there is a notion of (left/right) $R$-modules which is analogous to the notion of a $k$-vector space, and (left/right) ideals of $R$ are sub-modules of $R$ when $R$ is considered as a (left/right) $R$-module.

Then a set of ideals has a direct sum as $R$-modules, but that sum is not a sub-module of $R$. However, as with vector spaces, there is a map from the direct sum to the sum inside $R$, and what $\oplus\sum$ is saying is that we are taking the sum inside $R$, but we are asserting that the map from the direct sum to the sum inside $R$ is 1-1.

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