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I have no idea how to solve such an exercise:

Given is a field K, $ n \in \mathbb{N}, 1_K:=(1,1,..1) \in K^n $ $e_1,..,e_n$ is a standard basis of $K^n$. One show that: $1_K-e_1,..,1_K-e_n$ are linearly dependent, if the characteristic(K)>0, and $characteristic(K)\mid(n-1)$

If I am getting it correctly, $1_K-e_1,..,1_K-e_n$ must be all 0 and that does not make any sence to me, since it is obviously linearly dependent to each other. I am reading at wikipedia, that If char(K)=0 than the field must be infinite? I do not get this either.. Thank you

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First, $1-e_i$ are most certainly not $0$, unless $n=1$. To see that a field of characteristic $0$ is infinite, just see that $0,1,1+1,1+1+1,\ldots$ are distinct in such a field.

For the actual hint: write a matrix corresponding to this set of vectors (it will be all $1$s except the diagonal, which is all zeroes). By simple row operations, you can make it into a triangular form, where it becomes apparent that the determinant is $(-1)^{n-1}\cdot (n-1)$ (so in fact these vectors are linearly dependent if and only if characteristic of $K$ divides $n-1$).

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Hint: What happens if you add up all the $1_K - e_i$?

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$1_K-e_i$ are most definitely not 0. I think that's where your problem is.

Also yes it is true that if Char($K$) = 0 then the field must be infinite. In fact it needs to contain $\mathbb{Q}$ in a very natural way as it contains $\mathbb{Z}$ (do you see why?) and hence must contain its field of fraction.

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  • $\begingroup$ Thank you guys. Now I see the question clearly. And I got the point $\endgroup$ – Eu2718 Jun 17 '14 at 15:15

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