Discuss Exercise 9 when \(V\) is a vector space over the field with two elements described in Exercise 5, Section 1.1.
Solution
Let the scalar field be \(\FF_2\), and let \(\alpha,\beta,\gamma\) be linearly independent. Then \[ (\alpha+\beta)+(\beta+\gamma)+(\gamma+\alpha) =2\alpha+2\beta+2\gamma=0 \] because \(2=0\) in \(\FF_2\). The three pairwise sums are therefore linearly dependent.
In fact, any two of them remain linearly independent. For example, if \[ a(\alpha+\beta)+b(\beta+\gamma)=0, \qquad a,b\in\FF_2, \] then \[ a\alpha+(a+b)\beta+b\gamma=0. \] Independence of \(\alpha,\beta,\gamma\) gives \(a=b=0\). The other pairs are identical by symmetry. Thus Exercise 2.3.9 fails precisely at the three-vector level in characteristic two.