TITLE: If the field of a vector space weren't characteristic zero, then what would change in the theory?
QUESTION [57 upvotes]: 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.

REPLY [2 votes]: Nowhere is the difference between characteristic $0$ and characteristic $p>0$ as stark as in the representation theory of finite groups.
Let me try to illustrate this with a baby example:
Let $V$ be a vector space together with a linear map $f\colon V\to V$ such that $f^p=1$ (where $1$ denotes the identity on $V$). This is precisely a representation of the cyclic group $\mathbb Z/p\mathbb Z$ but you can safely ignore that if you don't know about groups.
Assume we have a subspace $W\subset V$ which is stable under $f$, in the sense that $f(w)\in W$ for all $w\in W$. Then we could ask, if we can find a complement $X$ of $W$ in $V$ which is also stable under $f$. In other words, we want to know, whether we can "break up" $V=X\oplus W$ such that $f$ "acts separately on the two parts", i.e. we can write $f(x,w)=(f(x),f(w))$.
It turns out that you can always do this in characteristic zero! (you can look this up under "Maschke
's theorem"; don't let the name scare you, it is not a difficult proof)
But consider for instance the puny $2$-dimensional vector space $V= F^2$, where $F=\mathbb F_p$ is the field with $p$ elements. Consider $f$ given by the matrix $\left(\begin{matrix}1& 1\\ 0 &1\end{matrix}\right)$. Clearly $f^p=1$ (because $p\cdot 1=1+1+\dots+1=0$ in $F$). Let $W$ be subspace of $W$ spanned by the first basis vector $e_1=\left(\begin{matrix}1\\ 0\end{matrix}\right)$; clearly $W$ is $f$-stable because $f(e_1)=e_1$. But a complement of $W$ would have to be spanned by a vector of the form $x=\left(\begin{matrix}a\\ b\end{matrix}\right)$ with $b\neq 0$; hence $f(x)=\left(\begin{matrix}a+b\\ b\end{matrix}\right)$ is not a multiple of $x$. So $W$ has no $f$-stable complement.
It turns out that this seemingly small issue has enormous ripples. For instance, the representation theory of the symmetric group is extremely well understood in characteristic zero (you can find everything you need to know in any introductory textbook); on the other hand, understanding the representation theory of the symmetric group in positive characteristic  is one of the big open problems of modern mathematics.