Proof that the solutions are algebraic functions

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mathmari
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Hey! :o

I am looking at the following:

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View attachment 5012 I haven't really understood the proof...

Why do we consider the differential equation $y'=P(x)y$ ? (Wondering)

Why does the sentence: "If $(3)_{\mathfrak{p}}$ has a solution in $\overline{K}_{\mathfrak{p}}(x)$, then $(3)_{\mathfrak{p}}$ has also a solution $y_{\mathfrak{p}}$ in $\overline{K}_{\mathfrak{p}}[x]$." stand? (Wondering)
 
Attachments
  • Printed page of mathematical text introducing a field k of characteristic p > 0, an algebraic number field K, and differential equation (1) with polynomial coefficients a_i(x) in K[x], and its reduction modulo a prime ideal.
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  • Printed text stating Grothendieck's problem: if for almost all prime ideals p, equation (1) has n solutions in the algebraic closure of K_p(x) that are independent over the algebraic closure of K_p(x^p), are all solutions of (1) algebraic functions?
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  • Printed mathematical statement labelled (*): Let K be an algebraic number field of finite degree. If almost all prime ideals of K are of degree one, then K is the rational number field.
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  • Scanned page of a mathematics text showing a differential equation y' = P(x)y with P(x) in K(x), followed by a sentence about solutions in the p-adic closure and a product formula for y_p with factors (x minus a_i) raised to powers.
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  • Printed mathematical text: the equation P(x) mod p = y'_p / y_p = sum over i of c_i / (x - ā_i), then P(x) = sum over i of β_i / (x - α_i), then β_i ≡ (a rational integer) mod p, and y = c product over i of (x - α_i)^β_i, described as an algebraic function.
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mathmari said:
Why do we consider the differential equation $y'=P(x)y$ ?

Since the proof for Grothedieck's problem stands for $n=1$, we consider a differential equation of the form $\alpha_0(x)y'+\alpha_1(x) y=0, \alpha_i(x)\in K[x]$.
So $\alpha_0(x)y'+\alpha_1 (x) y=0 \Rightarrow \alpha_0(x)y'=-\alpha_1(x)y \Rightarrow y'=P(x)y$, where $P(x)=-\frac{\alpha_1(x)}{\alpha_0(x)}\in K(x)$.

Is this correct? (Wondering)
mathmari said:
Why does the sentence: "If $(3)_{\mathfrak{p}}$ has a solution in $\overline{K}_{\mathfrak{p}}(x)$, then $(3)_{\mathfrak{p}}$ has also a solution $y_{\mathfrak{p}}$ in $\overline{K}_{\mathfrak{p}}[x]$." stand? (Wondering)

$\overline{K}$ is the algebraic closure, right?

We have that any polynomial in a field $K[x]$ factors into linear factors in the algebraic closure $\overline{K}[x]$. Does this hold also for the elements of a field of fraction $K(x)$?

Also how do we conclude that $\beta_i \in \mathbb{Q}$ ? (Wondering)