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This is a sort of meta question.

How does one best define the direct sum (of vector spaces or other algebraic structures).

It seems to me that when one writes $$V = M \oplus N$$ One would like for $$M, N \subset V$$

(where subset means subspace) However, if one uses the standard definition of the elements of $V$ as being $$ \{(m, n): m\in M, n\in N\}$$

Then one clearly cannot say that $M\subset V$, because the elements of $M$ and the elements of $V$ clearly are not the same. ($V$ has ordered pairs, $M$ has whatever $M$ has...)

As another example, one would like the direct sum to be associative, but if one uses the standard ordered pair definition,

$$(M\oplus N) \oplus K \neq M \oplus (N\oplus K)$$

As (and this is perhaps a bit pedantic), the elements of the LHS "look like" $((m, n), k)$, whereas the elements of the RHS "look like" $(m, (n, k))$. There would be an extremely easy isomorphism from one space to the other, but the elements of the spaces would be totally different mathematical objects.

However, when I've said "would like," I really mean that this is how I see the direct sum used all the time. And my definition is how I've seen the direct sum defined all the time. So... how can I reconcile this?

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    $\begingroup$ Authors often use the phrase "internal direct sum" when they want to emphasize the two summands as living inside of the space. I agree it's a little confusing/ambiguous; at least in my experience it generally makes little difference in practice, though. $\endgroup$ Mar 16, 2018 at 18:13
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    $\begingroup$ One interpretation I know is that $V=M\oplus N$ iff $M\cap N=\emptyset$ and $\langle M,N\rangle=V$ where $\langle\cdot\rangle$ denotes linear span. $\endgroup$ Mar 16, 2018 at 18:46
  • $\begingroup$ The triples you defined allow a bijective mapping between your associative vector space sum examples, by $((m,n),k)\mapsto(m,(n,k))$. The structure of the vectorspaces is preserved by this, so $(M\oplus N)\oplus K$ and $M\oplus (N\oplus K)$ are isomorph and so essentially the same, so usually mathematicians do not care to differ. In the same way, $M\times\{0\}\subset M\oplus N$ and since $M$ and $M\times\{0\}$ are isomorph, one shortens the notation to $M\subset V$. Wrong if looked at with rigor, but right for any interesting applications. $\endgroup$
    – SK19
    Mar 16, 2018 at 19:48

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The concept you are looking for is the internal direct sum, as opposed to the external direct sum which is the one you use.

Let $M, N$ be two subspaces of $V$. Define the sum of subspaces $M$ and $N$ as

$$M + N = \operatorname{span}\{M \cup N\}$$

Furthermore, if $M \cap N = \{0\}$, then we say that the sum is direct and we denote it as $M \,\dot+\, N$.

There is a characterization of the sum of subspaces which justifies the name:

$$M + N = \{m + n : m \in M, n \in N\}$$

Furthermore, the decomposition of every vector $x \in M+N$ as $$x = \underbrace{m}_{\in M} + \underbrace{n}_{\in N}$$

is unique if and only if the sum is direct (i.e. $M$ and $N$ intersect trivially).

Therefore, when dealing with the internal direct sum, there is a canonical isomorphism $$\phi : M \,\dot+ \,N \to M \oplus N$$ $$m + n \mapsto (m, n)$$ to the external direct sum, hence they are used interchangeably.

Notice that the internal direct sum is associative, as expected.

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It's very common to handle (any fixed or) canonical embedding $j:A\hookrightarrow B$ as a mere substructure, by (implicitly) identifying elements of $A$ with their image in $B$.
Note also that the range $j(A)\subseteq B$ of $j$ is indeed always isomorphic to $A$.

As a sidenote, if one insists on having $M, N\subseteq M\oplus N$, $\, $and$\, M, N$ as sets happen to intersect exactly in $\{0\} $ where $0$ is the zero element both in $M$ and in $N$, then one could also define the direct sum on the set $M\cup N\cup\{(m, n) \mid m, n\ne0\} $.

However, the essence of being the direct sum is a universal property, and as such, it can be captured only up to isomorphism. Namely, direct sum can be generalized in any category, called as coproduct of objects $M, N$:

It consists of an object $V$ and two morphisms $i:M\to V$ and $j:N\to V$ (these will be the canonical embeddings), such that for any similar data given $f:M\to X, \ g:N\to X$, there is a unique morphism $t:V\to X$ making $f=t\circ i$ and $g=t\circ j$.

In the case of vector spaces (or modules over any ring), it happens that $M\times N\cong M\oplus N$, but this is not the general pattern.

Your observation about almost associativity is one of the root causes for the need of coherence axioms of monoidal categories instead of requiring the operation to be strictly associative.

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