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This is an exercise from Chapter 3 of Golan's linear algebra book.

Problem: Show $\mathbb{Z}$ is not a vector space over a field.

Solution attempt: Suppose there is a such a field and proceed by contradiction. I will write multiplication $FV$, where $F$ is in the field and $V$ is an element of $\mathbb{Z}$.

First we rule out the case where the field has characteristic 2. We would have

$$0=(1_F+1_F)1=1_F1+1_F1=2$$

a contradiction.

Now, consider the case where the field does not have characteristic 2. Then there is an element $2^{-1}_F$ in the field, and

$1=2_F(2^{-1}_F1)=2^{-1}_F1+2^{-1}_F1$

Now $2^{-1}_F1\in\mathbb{Z}$ as it is an element of the vector space, but there is no element $a\in\mathbb{Z}$ with $2a=1$, so we have a contradiction.

Is this correct?

  • 9
    Seems correct..2012-05-31
  • 8
    Seems fine; alternatively, it cannot be a vector space over a field of positive characteristic, because that would require elements of $\mathbb{Z}$ to be of additive order $p$; and it cannot be a vector space over a field of characteristic $0$ because that would require it to be a vector space over $\mathbb{Q}$, which is equivalent to being divisible. This is essentially what you are doing.2012-05-31
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    Thanks for the help. As a side note, I plan on working through Golan's book and posting solutions to some of the more interesting exercises here to keep myself honest and my proof writing skills strong. Would the preferred format be as above, or should I post just the problem in the question and my solution as an answer?2012-05-31
  • 0
    @Potato: My personal preference would be for the latter, since it would let others post with any alternative approaches to solving the problem, instead of just verifying that your solution is correct.2012-05-31
  • 6
    I like the minimality of your approach. Although what Arturo says is indeed correct, proving a field of characteristic $0$ has a subfield isomorphic to $\Bbb{Q}$ is something which "goes too far afield" (excuse the terrible pun), and is not needed, here.2012-05-31
  • 0
    Nice solution! Getting rid of the simplest (characteristic $2$) case as neatly as possible, and then using this directly to support the initial intuition: "fields have fractions." Om nom chomp.2012-05-31
  • 9
    a pedantic (yet important) remark: $\mathbb {Z}$ can be made a vector space over a field if one interprets that as "make the set $\mathbb {Z}$ into a vector space" (just use a bijection with any countable vector field, of which there are plenty). The problem should be stated to say "prove that the additive group $\mathbb {Z}$ is not the additive group part of any vector space".2012-05-31
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    Vector spaces are completely reducible.2012-05-31
  • 1
    @SteveD: Dear Steve, Not as abelian groups (if the field of scalars is char. zero), and the abelian group structure of $\mathbb Z$ is all that one has access to in this question. Regards,2012-06-25

1 Answers 1

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The answer is, "Yes."


If $V$ is a vector space over a field of positive characteristic, then as an abelian group, every element of $V$ has finite order. If $V$ is a vector space over a field of characteristic $0$, then as an abelian group, $V$ is divisible. The abelian group $\mathbb Z$ has neither of these properties.