Submodules of direct sums of simple modules

I'm reading these online notes on representation theory, and I don't fully understand this:

Isn't $V\oplus(\bigoplus_{i\in I}S_i)$ a direct sum by definition, so we'd get $I=\{1,\cdots,n\}$? Can we assume the $S_i$ are pairwise disjoint submodules of $U$ based on how the lemma is phrased (or does that follow necessarily from them each being simple)? Intuitively, I feel the condition should be that $W$ as an external direct sum maps into the internal direct sum situated within $U$ in an obvious way, and we choose $I$ maximal so that this map has no kernel. Is this understanding correct?

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I don't understand your first question in the last parraph above: I is maximal wrt that sum being direct, but somehow you seem to think that the sum of the n submodules $\,S_i\,$ is direct, which is wrong. And no, why could we assume the $\,S_i\,$ have trivial intersection (of course, they can't be disjoint). What they do is to take sum of $\,S_i'\,$s that (1) is direct, but also (2) the sum with $\,V\,T is direct. – DonAntonio Aug 5 '12 at 17:15 Take for example,$S_i$be the subspace spanned by the vector$(1,i)$, for each$i\in\{1,\dots,1000\}$, which are simple$\mathbb R$-submodules of$U=\mathbb R^2$whose sum is$U$. In particular, notice that in this example the subset$I$you can take depends on$V$. – Mariano Suárez-Alvarez Aug 5 '12 at 17:17 @DonAntonio Doesn't the$\oplus$symbol by definition mean "direct sum"? (And yes I meant trivial intersection.) Should I understand$W=V+S_{i_1}+\cdots+S_{i_k}$, and we take$I$maximal so that each of$V,S_{i_1},\cdots,S_{i_k}$has pairwise trivial intersection (which I am guessing is the definition of "direct")? – blue Aug 5 '12 at 17:24 Do not guess the definition of direct sum (you guessed it wrong!): look it up. – Mariano Suárez-Alvarez Aug 5 '12 at 22:27 @seaturtles, Mariano already told you should do, in particular when we deal with such a standard and basic definition. just one note more: no, that the sum$\,V+S_{i_1}+...+S_{i_k}\,$is direct does not mean "each of these has pairwise trivial intersection". The condition is way more interesting and involved when there are three or more summands. – DonAntonio Aug 6 '12 at 2:05 1 Answer Your notes have a typo: the proof should begin "Choose a subset$I$maximal subject to the condition that$V + \sum S_i$is a direct sum." Of course, as you've noted,$V\oplus \bigoplus S_I$is a direct sum by definition-but it might be considerably bigger than$U$, as Mariano's example shows. We can assume that the$S_i$'s pairwise intersections are either$0$or an$S_i.$If the latter case happens,$S_i \cap S_j=S_i,$we know$S_i=S_j$, since otherwise$S_i$would be a proper submodule of$S_j.$Regarding your comments, the definition of direct is not that the summands have pairwise trivial intersections, but that each element of the sum can be written uniquely in terms of the summands. So$(e_1) + (e_2)=(e_1) \oplus (e_2)=V$in some 2-dimensional vector space$V$, but$(e_1)+(e_2)+(e_1+e_2)=V$while$(e_1)\oplus (e_2)\oplus (e_1+e_2)$is three-dimensional and not in$V\$ at all. Note, though, that these three modules have trivial intersection. On the other hand, if a sum is direct, this does imply the summands have trivial intersection: otherwise anything in any intersection can be written in at least two ways. It's just that these properties aren't equivalent.

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Thanks. I believe I understand what's going on after thinking about DonAntonio's comment and reading your answer. Just to clarify though, a sum of submodules is direct precisely when the summands have pairwise trivial intersection, right? –  blue Aug 5 '12 at 17:37
Not quite-see my new paragraph. –  Kevin Carlson Aug 5 '12 at 17:40
I see. Uniquely writeable does imply pairwise trivial intersection (seen via contraposition), but the converse is not true. –  blue Aug 5 '12 at 17:45