MHB Proof that the solutions are algebraic functions

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Functions Proof
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

I am looking at the following:

View attachment 5008

View attachment 5009

View attachment 5010 View attachment 5011

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

  • id.PNG
    id.PNG
    37.7 KB · Views: 111
  • gr.PNG
    gr.PNG
    14 KB · Views: 108
  • tche.PNG
    tche.PNG
    10.2 KB · Views: 115
  • tcheg1.PNG
    tcheg1.PNG
    12 KB · Views: 112
  • tcheg2.PNG
    tcheg2.PNG
    27.1 KB · Views: 118
Physics news on Phys.org
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)
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

Replies
31
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K
Replies
21
Views
1K
  • · Replies 3 ·
Replies
3
Views
742
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K