I am looking at the following part of a paper:
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When we reduce the differential equation $(1)$ modulo the prime $p$ we do the following: $$\alpha_i \equiv \tilde{\alpha}_i \pmod p$$ So $$(1)_p : \tilde{\alpha}_0 (x) y^{(n)}+\tilde{\alpha}_1 (x) y^{(n-1)}+\dots +\tilde{\alpha}_n (x) y=0$$
right?
What does it mean that we reduce the differential equation $(1)$ modulo the prime ideal $\mathfrak p$ ?
Isn't the prime ideal a subset of a ring?
Apparently, the operation that is performed is this: the coefficients of the polynomials $a_i(x)$ belong to the ring of algebraic integers $A$ of an algebraic number field $K$. For any prime ideal $\mathfrak p$ of $A$, we can consider the quotient ring $A/\mathfrak p$, which is an integral domain and has a field of fractions, which is called the residue (class) field. Alternatively, we can consider the localisation $A_{\mathfrak p}$ of $A$ at $\mathfrak p$, which is a local ring with maximal ideal $\mathfrak pA_{\mathfrak p}$, and its residue field $\,A_{\mathfrak p}/\mathfrak pA_{\mathfrak p}$.
In short, we reduce each coefficient of the $a_i(x)$ modulo the prime ideal $\mathfrak p$, in the same way as we reduce a polynomial ideal with integer coefficients modulo a prime number.