What is internal direct sum or internal direct product in Dummit and Foote?

Question

I refer to Dummit and Foote Chapter 10.3 specifically pages 351,353,354,356 and 357.

1. Does Exercise 10.3.21 on pages 357 (By the way, there's some errata here. Condition (iii) should be $i_1,...,i_k$) define a notion of internal direct sum (of unital $R$-submodules of a unital $R$-module over a unital, but not necessarily commutative, ring $R$)?

• I think this is an internal direct sum for an infinite or a finite index set that generalises the notion of internal direct sum for a finite index set given in page 354.

2. Do we have a notion of 'internal direct product'?

• For the finite case, I believe this is the '$N_1 + ... + N_k$' part of Proposition 10.5 in page 353.

• For the finite or infinite case, I believe this is the 'the (unital $R$-)submodule of $M$ generated by (the union of) all the $N_i$'s' part of Condition (i) of Exercise 10.3.21 because '$N_1 + ... + N_k$' in Proposition 10.5 is actually equal to (see page 351) the (unital $R$-) 'submodule of $M$ generated by (the union of) all the $N_i$'s' such that Condition (i) generalises the '(1)' in Proposition 10.5.

• Therefore: I think of internal direct product of $N_i$'s of $M$ as $\sum_{i \in I} N_i = R\{\bigcup_{i \in I} N_i\}$, which like external direct product and external direct sum, is always defined. And then I think of internal direct sum as not always defined but, whenever defined, as equal to internal direct product.

• Possibly relevant: 'Semidirect product'. This wikipedia page: https://en.wikipedia.org/wiki/Direct_sum_of_groups#Generalization_to_sums_over_infinite_sets

Context: I'm trying to understand the direct sum parts of graded rings and graded ideals in later in Chapter 11.5. I'm hoping these can be internal instead of just external. I ask more here.

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Edit 1: Thank you for the upvotes or views. I feel like all the hours I spent trying to understand this seemingly minor thing was really worth it.

Edit 2: For (not necessarily Abelian) groups: Internal direct product/sum in groups: Is join and independent equivalent to unique expression?

Answer 1

Let me know if this answers your question. Let's let $R$ denote a unital ring. There are two similar but slightly distinct notions of internal and external direct sum, which I think is at the core of the question here. First and foremost, let $M$ denote an $R-$module, and let $N_1,\ldots, N_k$ denote submodules of $M$. In particular, as sets $N_i\subseteq M$ for each $i$. We say that $M$ is an internal direct sum of the $N_i$, denoted by $$ M=\bigoplus_{i=1}^k N_i$$ if every element of $M$ can be written uniquely as a sum of elements in the $N_i$. That is, for every $m\in M$, there exists a unique tuple $(n_1,\ldots, n_k)$ such that $m=\sum n_i$. It is equivalent to require that $N_1+\cdots+N_k=M$ and $N_i\cap N_j=\varnothing$ for $i\ne j$.

There is a slightly different notion of (external) direct sum where we take a collection of $R-$modules $N_1',\ldots, N_r'$ and we say that $M$ is an (external) direct sum of the $N_i'$ if there exists an isomorphism $\phi:M\to \bigoplus_{i=1}^k N_i'$. I.e. $$ \boxed{M\cong \bigoplus_{i=1}^k N_i'}$$ There is a bit of a distinction here, because we need to define this operation $\oplus$ for modules that do not both belong to a bigger module a priori. This is defined by the familiar rule $$ A\oplus B=\{(a,b): a\in A,b\in B\}$$ subject to the obvious $R-$module structure. So, being an external direct sum can be translated into the terminology of the internal direct sum as follows: $M$ is the external direct sum of $\{N_i'\}_{i=1}^k$ $$ \phi:M\xrightarrow{\sim} \bigoplus_{i=1}^k N_i'$$ if and only if there exist $N_i\subseteq M$ with $\phi(N_i)=N_i'$ for $i=1,\ldots, k$ and in fact $M$ is the internal direct sum of the $N_i$. That is, the $N_i'$ define $N_i=\phi^{-1}(N_i')$ so that $M$ is an internal direct sum of the $N_i$.

example: We should interpret what $\mathbb{R}^2=\Bbb{R}\oplus \Bbb{R}$ means. It means that there is a pair of subspaces $L_1,L_2$ of $\mathbb{R}^2$, each isomorphic to $\mathbb{R}$ so that $\mathbb{R}^2$ is their direct sum. In particular we can take $L_1$ to be the $x-$axis and $L_2$ to be the $y-$axis. These choices are far from unique.

Anyway, as you might know: this carries over almost verbatim to the case of an infinite indexing set, except that for $I$ a general indexing set, $\bigoplus_{i\in I}N_i$ consists of the finite sums of elements in the various $N_i$. So, you can re-define these notions in that case as an exercise.

If you are really interested in direct products, i.e. $M=\prod_{i=1}^k N_i$, then you should notice that for $R-$modules, finite products are isomorphic to finite coproducts (direct sums). I.e. $$ \prod_{i=1}^k N_i\cong \bigoplus_{i=1}^k N_i$$ and so the discussion carries over verbatim. In the case of infinite products, we get distinct notions: $$ \prod_{i\in I} N_i\not\cong \bigoplus_{i\in I} N_i$$ but you can still define the analogous notion of "internal" direct product using the same strategy.

Answer 2

1. Yes, it does.
2. Well, the factors of a product are clearly submodules of the product, but the issue is that since addition is finitary, it can never be additively generate the whole product.

I have never seen the notion of an "internal direct product" entertained, but there could be something to be said about characterizing it.

Proposition 10.5 proves that for finite sets, the direct sum and direct product coincide.

If it is helpful, here is my version of explaining how internal/external sums are related. Maybe it will help you see why there is a finitary constraint on sums, and not on products.