1

I was referring to this post where the author claims that if $R=\oplus_i N_i$ a direct sum of simple left ideals, some of the $N_i$ may repeat.

Is that possible? I mean, by definition of direct sum shouldn't each $N_i$ be necessarily distinct as they have trivial intersection?

Thanks.

Eric Wofsey
  • 342,377
yoyostein
  • 20,428

4 Answers4

13

In this context, "the same" means "isomorphic as $R$-modules", not "equal as subsets of $R$". Of course, they cannot be equal as subsets of $R$ for the reason you point out.

On a related point, the term "direct sum" actually has two different meanings. For simplicity, I will explain them only for binary direct sums. The first meaning is an "internal direct sum": a module $M$ is the internal direct sum of two submodules $N_1,N_2\subseteq M$ iff $N_1\cap N_2=0$ and $N_1+N_2=M$. The second meaning is an "external direct sum": if $N_1$ and $N_2$ are two modules, their external direct sum is the module $N_1\oplus N_2$ (or more generally, any isomorphic module) whose underlying set is cartesian product $N_1\times N_2$ and whose module structure is defined separately on each coordinate. The connection between the two definitions is that whenever $N_1$ and $N_2$ are submodules of $M$, there is a canonical homomorphism from the external direct sum $N_1\oplus N_2$ to $M$, and this homomorphism is an isomorphism iff $M$ is the internal direct sum of $N_1$ and $N_2$. And conversely, the external direct sum $N_1\oplus N_2$ is the internal direct sum of its two submodules $N_1'=N_1\times \{0\}$ and $N_2'=\{0\}\times N_2$ which are isomorphic to $N_1$ and $N_2$.

In an internal direct sum, obviously $N_1$ and $N_2$ must be distinct unless they are both $0$ (since $N_1\cap N_2=0$). However, when defining the external direct sum, it is perfectly fine if $N_1$ and $N_2$ are equal, as discussed in the other answers.

Eric Wofsey
  • 342,377
  • 1
    Note that for infinite direct sums, we don't use the cartesian product, but the subset with only finitely many non-zero values in each tuple. – Paŭlo Ebermann Nov 01 '15 at 11:00
  • @Eric Wofset: Would the canonical homomorphism be the injection $\iota:N_1\oplus N_2 \rightarrow M$? – Dog_69 Dec 15 '17 at 19:31
  • @Dog_69: I don't know what injection you're talking about. The canonical homomorphism need not be injective. It is the map $f:N_1\oplus N_2\to M$ defined by $f(a,b)=a+b$. – Eric Wofsey Dec 15 '17 at 20:00
  • Yes, I was wrong. I was thinking in the injections $\iota_1:N_1\roghtarrow N_1+N_2$ and $\iota_2:N_2\roghtarrow N_1+N_2$. I'm sorry. – Dog_69 Dec 15 '17 at 22:23
3

A good example is $\mathbb{Z}\oplus \mathbb{Z}$. Each component is isomorphic, but $(0,3)$ is different than $(3,0)$. That is what is meant by each summand being unique.

J126
  • 17,707
3

By $N_1 \oplus N_2 \oplus ... N_k$ we actually mean $\bar{N_1} \oplus \bar{N_2} \oplus ... \bar{N_k},$ where $$\bar{N_1} = (x,0,0,...,0)|x \in N_1 \},$$ $$.$$ $$.$$ $$\bar{N_i} = (0,0,...,x,...,0)|x \in N_i \}.$$ We can see that $\bar{N_i} \cong N_i$ and for distinct $i, j$ we always have $\bar{N_i} \cap \bar{N_j} = 0$, even if $N_i = N_j$.

(Edit: By $N_1 \oplus N_2 \oplus ... N_k$ some authors denote $N_1 + N_2 + ... N_k$ but only when the sets happen to be disjoint.)

user3810316
  • 735
Amin
  • 710
1

In general, a direct sum of modules $M$ and $N$ is really the Cartesian product $M\times N$ with the added natural structure of a module. With this structure, you denote the Cartesian product by $M\oplus N$, and as Joe Johnson 126 pointed out, there is no need for $M$ and $N$ to be distinct.

Mankind
  • 13,500