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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?

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9  
Seems correct.. –  lhf May 31 '12 at 1:05
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. –  Arturo Magidin May 31 '12 at 1:12
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. –  David Wheeler May 31 '12 at 1:41
8  
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". –  Ittay Weiss May 31 '12 at 9:05
2  
Vector spaces are completely reducible. –  user641 May 31 '12 at 11:10

1 Answer 1

up vote 5 down vote accepted

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.

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