In the book of Linear Algebra by Werner Greub, whenever we choose a field for our vector spaces, we always choose an arbitrary field $F$ of characteristic zero, but to understand the importance of the this property, I am wondering what would we lose if the field weren't characteristic zero ?
I mean, right now I'm in the middle of the Chapter 4, and up to now we have used the fact that the field is characteristic zero once in a single proof, so as main theorems and properties, if the field weren't characteristic zero, what we would we lose ?
Note, I'm asking this particular question to understand the importance and the place of this fact in the subject, so if you have any other idea to convey this, I'm also OK with that.
Note: Since this a broad question, it is unlikely that one person will cover all the cases, so I will not accept any answer so that you can always post answers.
Best Answer
Many arguments using the trace of a matrix will no longer be true in general. For example, a matrix $A\in M_n(K)$ over a field of characteristic zero is nilpotent, i.e., satisfies $A^n=0$, if and only if $\operatorname{tr}(A^k)=0$ for all $1\le k\le n$. For fields of prime characteristic $p$ with $p\mid n$ however, this fails. For example, the identity matrix $A=I_n$ then satisfies $\operatorname{tr}(A^k)=0$ for all $1\le k\le n$, but is not nilpotent.
The pathology of linear algebra over fields of characteristic $2$ has been discussed already here.