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If $\{V_i\}_{i\in I}$ is a family of vector spaces over $F$ with basis $B_i$ for each $V_i$, then there is a vector space $\prod_i V_i$ over $F$, called the direct product of $V_i$'s; its definition involves a certain universal property in terms from projections from it onto $V_i$'s (see this wiki)

Since every vector space has a basis, $\prod_i V_i$ has so.

Q. Can we obtain basis a of $\prod_i V_i$ from given basis $B_i$ of each $V_i$?

I am not too familiar with Category theory; please explain in as elementary fashion as you can, so that this will be also accessible to undergraduates; I want to explain this in my Linear Algebra course to undergraduates, and my aim is to introduce maximum number of advanced concepts of other areas of mathematics from starting point in Linear Algebra.

Eric Wofsey
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p Groups
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  • This is not an answer. But I see it as a clue. The ring $\mathbb{Z}$ is free as a $\mathbb{Z}$-module, but the countable product of copies of $\mathbb{Z}$ is not. Then such algorithm, if it exists, has to use the operation of division of the field. – Gustavo Sep 22 '17 at 13:04

2 Answers2

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No, there isn't. It's difficult to prove a negative like this (since the question isn't precisely formulated), but let me try to provide some evidence. If there were a canonical way to give a basis for $\prod V_i$ given bases $B_i$ for each $V_i$, then in the case that the $V_i$ are all the same and the $B_i$ are all the same, the basis chosen should be invariant under permuting the factors. But this is impossible, at least in some cases.

For instance, let us take $F=\mathbb{F}_2$, $I=\mathbb{N}$, and $V_n=\mathbb{F}_2$ for all $n$. Then we are asked to give a basis $B$ for the vector space $V=\mathbb{F}_2^\mathbb{N}$ which is invariant under permuting the coordinates. Let us think of elements of $V$ as subsets of $\mathbb{N}$. Note that if $B$ contains any infinite coinfinite set, it contains every infinite coinfinite set, and any one can be sent to any other by a permutation of the coordinates. The collection of all infinite coinfinite sets is not linearly independent, so $B$ cannot contain any of them.

So every element of $B$ is either finite or cofinite. But the collection of finite-or-cofinite subsets of $\mathbb{N}$ is a proper vector subspace of $V$, and so the span of $B$ is contained in this subspace. So $B$ cannot span all of $V$, and cannot be a basis.

Eric Wofsey
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Assuming index set $I$ is infinite, the answer is roughly no, as the axiom of choice is needed to describe the basis for an infinite direct product of vector spaces.

Consider a minimal example, the countable infinite product $\prod F_2 = F_2 \times F_2 \times \ldots $, where $F_2$ is the field of characteristic two. Every element of this infinite product can be realized as a binary sequence. A basis is such that every binary sequence can be written as a finite sum of the basis elements. This amounts to finding basis elements that describe all possible "tail" behavior of binary sequences. You can show, by contradiction, that such a basis cannot be countable.

Christopher A. Wong
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  • sorry; a little confusion in question. My question is that can we obtain basis of direct product, from given basis of components; not necessarily taking union; but by any other way known. – p Groups Dec 27 '16 at 09:24
  • you may be right, but, I will think on it a little and say my reply! – p Groups Dec 27 '16 at 09:31
  • Unfortunately the concept of "obtaining" a basis is not well defined as far as I know. You're right that the uncountability of the basis may not be convincing for what you're asking. I'll think about whether there's a better way for me to interpret and answer your question. – Christopher A. Wong Dec 27 '16 at 19:19
  • +1: Thank you Chrostpher for understanding the problem; I am happy to hear more on the question. Sure, I will try to write question more precisely as you noted. – p Groups Dec 28 '16 at 06:03
  • @Christopher A. Wong maybe he wants an answer if there is a Turing machine that can do the job. – Gustavo Sep 22 '17 at 12:50