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Am trying to solve the following exercise which appeared in an abstract algebra textbook : Assume that $V$ is an $n$-dimensional vector space over a finite field ${\bf F}_{q}$ which consists of $q$ elements. If $V$ is a finite set theoretic union of $m$ proper linear subspaces ${W_1},\cdots {W_m},$ then it must be the case that $$m\geq{\frac{{q^n}-1}{q-1}}.$$ Prove that there exist ${\frac{{q^n}-1}{q-1}}$ subspaces whose union is $V$.

It is clear that each of the proper subspaces must consist of $q^k$ elements where $k<n$. Not sure how to approach this question!

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    $\begingroup$ This is false as stated. $\endgroup$
    – user208649
    Commented Aug 3, 2020 at 2:07
  • $\begingroup$ Thanks for letting me know. I stated it exactly as it appeared in the book A Course in Galois Theory by D. J. H. Garling. (Exercise 1.18, page 13). After I posted the question, I noticed that it was dropped from a second printing. $\endgroup$
    – student
    Commented Aug 3, 2020 at 3:33
  • $\begingroup$ TokenToucan, the argument below shows one can find $\frac {q^n -1} {q-1}$ subspaces whose union is $V$. However, is the inequality $m\geq {\frac {q^n -1} {q-1}}$ not necessarily true? $\endgroup$
    – student
    Commented Aug 3, 2020 at 4:19
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    $\begingroup$ Yes, it is the inequality which is not true. In fact, it follows relatively quickly from the example that the inequality is false (see comment below). $\endgroup$
    – user208649
    Commented Aug 3, 2020 at 4:40
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    $\begingroup$ For instance $\Bbb F_2^3$ can be covered by three 2D subspaces. $\endgroup$
    – anon
    Commented Aug 3, 2020 at 9:10

1 Answer 1

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Despite the comments, I believe this is true. There are $q^n-1$ nonzero elements $v$, and each such $v$ is in the one-dimensional subspace $W_v=\{fv|f\in F_q\}.$ Each of these subspaces has $q-1$ nonzero elements, so there are $\frac{q^n-1}{q-1}$ distinct $W_v$ whose union is $V$.

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  • $\begingroup$ So this actually partitions $V$ into mutually disjoint subspaces each having cardinality $q-1$. $\endgroup$
    – student
    Commented Aug 3, 2020 at 3:55
  • $\begingroup$ I really like this answer. If $k$ is an infinite field and $V$ is a finite dimensional $k$-vector space, it cannot be the set theoretic union of a finite collection of proper subspaces. The intuitive picture is that no finite collection of lines through the origin of ${\bf R}^2$ can be equal to ${\bf R}^2$ itself-these lines miss the points on the plane between two adjacent lines. This example intuitively underscores that this can happen for finite dimensional vector spaces over finite fields. $\endgroup$
    – student
    Commented Aug 3, 2020 at 4:02
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    $\begingroup$ Please explain the downvote. Is there something I'm overlooking? $\endgroup$
    – saulspatz
    Commented Aug 3, 2020 at 4:26
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    $\begingroup$ @saulspatz The first statement in the question is not true. The particular part you address is true but, in fact, your (correct) example is essentially the reason the inequality can't be true (except in 2 dimensions) because a single two-dimensional subspace can replace some pair of these one-dimensional subspaces (in fact it'll replace more than just two of them). $\endgroup$
    – user208649
    Commented Aug 3, 2020 at 4:33
  • $\begingroup$ I did NOT downvote you saulspatz!! $\endgroup$
    – student
    Commented Aug 3, 2020 at 6:05

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