The Grothendieck problem for differential equations (Grothendieck-Katz conjecture) is the following:
$$\alpha_n(x)y^{(n)}(x)+\dots +a_1 (x)y'(x)+a_0(x)y(x)=0, a_i \in \mathbb{Z}[x]\ \ \ \ (*)$$
We suppose that for almost each prime $p$, $(*)$, modulo $p$, has $n$ linearly independent solutions (powerseries in $\mathbb{F}_p((x))$, with field of constants $\mathbb{F}_p((x^p))$). Then $(*)$ has $n$ linearly independent solutions (powerseries in $\mathbb{C}((x))$ with field of constants $\mathbb{C}((x))$) and all are algebraic.
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Could you explain to me the above problem when we have for example the differential equation $xy'-ky=0, k\in \mathbb{Z}$ ?
We have to find a prime such that for all primes $p$ greater or equal than that one, it stands that modulo $p$ the differential equation has in this case one solution in $\mathbb{F}_p((x))$.
Then the differential equation has one algebraic solution in $\mathbb{C}((x))$.
Is this correct?
Suppose you have a Laurent series $y=\sum_{-N}^\infty a_n x^n\in\mathbb{F}_p((x))$ which satisfies $xy'-ky=0$. Then $na_n-ka_n=0$ in $\mathbb{F}_p$ for all $n$, so $a_n=0$ unless $n\equiv k$ mod $p$. That is, the set of solutions consists exactly of series of the form $y=x^kf$, where $f\in \mathbb{F}_p((x^p))$. So for all $p$, the space of solutions is one-dimensional over $\mathbb{F}_p((x^p))$.
Over $\mathbb{C}$, we have the same story: any solution $y=\sum a_n x^n$ must satisfy $na_n-ka_n=0$. This time, the only way that can happen is if $a_n=0$ for all $n\neq k$, so the only solutions are $cx^k$ for $c\in\mathbb{C}$. So, as expected, we have a one-dimensional space of solutions, and the solutions are algebraic.