Nilpotence and conjugacy in $M(p,\mathbb F_p)$ I have to solve the following problem: 

Characterize matrices $X\in M(p,\mathbb F_p)$ (note that $p$ is the dimension and the characteristic of the field) such that there exists $Y$ with the property $YX-XY=X$.

The text is not clear, since it doesn't explain whether $Y$ must be in $M(p,\mathbb F_p)$ or just in $M(p,\overline{\mathbb F_p})$. 
In the case $K=\mathbb C$ (instead of $\mathbb F_p$) and the dimension is $n$, a solution makes use of the fact that, if $char\ K=0$, $X$ is nilpotent if and only if $tr(X^i)=0$ for all $i>0$, and it turns out that such $Y\in M(p,\mathbb C)$ exists if and only if $X$ is nilpotent. In fact, if $X$ is nilpotent one finds the correct $Y$ for its Jordan form $J$, where $X=SJS^{-1}$:
$$\tilde Y=diag(n-1,n-2,...,0)$$
$$\tilde YJ-J\tilde Y=J$$
 and the statement follows easily by considering $Y=S\tilde YS^{-1}$; conversely, if such $Y$ exists all $tr(X^i)$ are $0$, then $X$ is nilpotent.
But in characteristic $p$ that fact isn't true (it is true, for example, if $0\in Sp(X)$; on the contrary, if $X=Id_p$, $tr(X^i)=0$ for all $i$ but $X$ is not nilpotent). Worse, even though $\tilde Y\in M(p,\mathbb F_p)$, $Y$ could be in $M(p,\overline{\mathbb F_p})\setminus M(p,\mathbb F_p)$ (or am I wrong?).
Thank you in advance.
 A: Remark that $X\in M_p(F_p)$ and the equation in the unknown $Y$  is linear; consequently, if there exists a solution $Y\in M_p(\overline{F_p})$, then there exists a solution in $M_p(F_p)$.
Proposition 1. Necessarily, $X$ is either nilpotent or invertible.
Proof. By recurrence, we show that, for every positive integer $k$,$YX^k-X^kY=kX^k$. We deduce that, for $1\leq k<p$, $tr(X^k)=0$. Using the Newton's formulae, we obtain that the characteristic polynomial of $X$ is in the form $x^p-\alpha$ where $\alpha\in F_p$ and we are done. 
Proposition 2. Every nilpotent matrix $X$ works.
Proof. Note that the Jordan form of $X$ is over $F_p$. Then it suffices to consider the case when  $X$ is a nilpotent Jordan block of dimension $k\leq p$. Then a solution is $Y=diag(k-1,k-2,\cdots,0)$.
Proposition 3. Among the invertible matrices $X$, work only those that are similar to a matrix of the form $\alpha I_p+J_p$ where $\alpha\in F_p\setminus 0$ and $J_p$ is the nilpotent Jordan block.
Proof. Since $X$ is invertible, $X^p-\alpha I_p=(X-\alpha I_p)^p=0$ and $X$ is in the form $\alpha I_p+N$ where $N$ is nilpotent and $\alpha\in F_p\setminus 0$. Then $YX-XY=YN-NY=\alpha I_p+N$. By recurrence, $YN^k-N^kY=k(\alpha N^{k-1}+N^k)$. Suppose that there is $k\leq p-1$ s.t. $N^k=0,N^{k-1}\not=0$; then $k\alpha N^{k-1}=0$, a contradiction. Then we may assume that $N=J_p$, the nilpotent Jordan block of dimension $p$.
A solution $Y$ is the $2$-band matrix defined, for every admissible $i$, by $Y_{ii}=p-i,Y_{i+1,i}=(p-i)\alpha$.
