5
$\begingroup$

I refer to Dummit 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.
  1. 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.


Edit: 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.

$\endgroup$
8
  • 1
    $\begingroup$ What is $ R\{\bigcup_{i \in I} N_i\}$ supposed to mean? $\endgroup$ – rschwieb Jun 17 '20 at 18:59
  • $\begingroup$ @rschwieb It's in p.351 $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:00
  • 1
    $\begingroup$ For an infinite index set $I$? That's clearly not always going to generate the product. Suppose $I$ is infinite but $R$ is finite. Then the submodule you're describing is countable, but the product is uncountable. $\endgroup$ – rschwieb Jun 17 '20 at 19:01
  • $\begingroup$ @rschwieb Oh right yeah I forgot about the $I$. I weirdly thought you asking only about the $R\{\cdot\}$. Here, $I$ is any nonempty index set. $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:05
  • 1
    $\begingroup$ Also in my comment above I should have specified $I$ countably infinite. $\endgroup$ – rschwieb Jun 17 '20 at 19:08
2
+50
$\begingroup$

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.

$\endgroup$
18
  • $\begingroup$ It sounds like you answer my questions, but I think I'd have to read a little more into your answer. Thanks! $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:07
  • $\begingroup$ Alekos Robotis, do you agree with rschwieb please? $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:16
  • 1
    $\begingroup$ I believe so. Certainly $1.$ is true, and I think the issue he points out in $2.$ is quite valid also. You will not be able to describe the infinite product using sums. A general element of $\prod N_i$ is an infinite sequence $(n_i)_{i\in I}$ such that $n_i\in N_i$ for each $i$. (I have upvoted it, anyway.) $\endgroup$ – Alekos Robotis Jun 17 '20 at 19:31
  • 1
    $\begingroup$ Regarding your first question: this set of infinite sequences can also be turned into an internal notion. If $M$ contains all of the $N_i$, and $M\cong \prod_{i\in I} N_i$ then you might say that this is an internal direct sum (though one might have to be a bit careful here). Regarding the second question: depending how you define the symbol $R\{\bigcup_{i\in I} N_i\}$, I think that it is okay to say $\bigoplus_{i\in I} N_i=R\{\bigcup_{i\in I} N_i\}$. This construction gives you the direct sum, but not the direct product. $\endgroup$ – Alekos Robotis Jun 17 '20 at 20:02
  • 1
    $\begingroup$ @JohnSmithKyon For me, the adjectives "internal/external" don't belong in front of "product", except in this question, which is the first time i"ve entertained the idea to any depth. In my last comment, I meant to compare "the product" with "the (external) sum" $\endgroup$ – rschwieb Jun 17 '20 at 21:30
2
$\begingroup$
  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.

$\endgroup$
8
  • $\begingroup$ Thanks. Re the second question, I don't get it. So am I right? Or what? $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:05
  • $\begingroup$ @JohnSmithKyon I don’t know if any characterization of a product in terms of sub modules. The cardinals try reason I gave shows why your idea won’t suffice. $\endgroup$ – rschwieb Jun 17 '20 at 19:07
  • $\begingroup$ wait so what is 'the (unital $R$-)submodule of $M$ generated by (the union of) all the $N_i$'s' if not $R\{\bigcup_{i \in I} N_i\}$? $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:15
  • $\begingroup$ rschwieb, how should I understand internal direct sum in Chapter 11.5 if not that $\bigoplus_{i \in I} N_i = \sum_{i \in I} N_i = R\{\bigcup_{i \in I} N_i\}$ (with $N_i$'s and $R$ satisfying the relevant conditions) ? $\endgroup$ – John Smith Kyon Jun 17 '20 at 19:54
  • $\begingroup$ @JohnSmithKyon That's a valid description of the internal direct sum but a sum like this will never be able to generate the product out of the same generators ( the union you chose) and addition. $\endgroup$ – rschwieb Jun 17 '20 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.