What is the Theorem regarding the number of polynomial zeros modulo p and H?

Click For Summary

Discussion Overview

The discussion revolves around a theorem in number theory concerning the number of polynomial zeros modulo a prime number \( p \) and an irreducible polynomial \( H \). Participants seek clarification on the theorem's implications and specific terms, particularly regarding congruences and examples.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant requests a simple explanation of the theorem regarding the number of polynomial zeros modulo \( p \) and \( H \).
  • Another participant seeks clarification on the meaning of "congruent modulo \( p \) and \( H \)" after understanding "congruent mod \( p \)".
  • An example is provided involving polynomials \( P \) and \( Q \) with a prime \( p = 3 \), illustrating congruences but lacking clarity on the polynomial \( H \).
  • A participant points out the omission of the definition of \( H \) in the example.
  • One participant defines \( H \) as a non-constant polynomial whose leading coefficient is coprime to \( n \).
  • A later reply summarizes the theorem's conditions and questions the interpretation of zeros and congruences, particularly regarding the division of polynomials and the ring in which it is performed.

Areas of Agreement / Disagreement

The discussion contains multiple competing views and interpretations regarding the theorem, particularly about the definitions and implications of congruences. No consensus has been reached on the specifics of the theorem or the examples provided.

Contextual Notes

Participants express uncertainty regarding the definitions of congruences and the conditions under which the theorem applies, particularly concerning the polynomial \( H \) and the nature of polynomial division.

Kontilera
Messages
176
Reaction score
24
Hello I am currently learning some of the basics of number theory, and struggling to understand this Theorem. Could someone please explain it with maby a simple example? :)

THRM:(Number of polynomial zero mod p and H)
Let p be a prime number and let H be a polynomial that is irruducible modulo p. Furthermore let P be a polynomial that has degree d>=0 modulo p. Then P has at most d polynomial zeros that are pairwise not congruent modulo p and H.
 
Mathematics news on Phys.org
Kontilera said:
Then P has at most d polynomial zeros that are pairwise not congruent modulo p and H.
What does the part in bold mean?
I understand what "congruent mod p" means, but I don't understand what "congruent mod p and H" means.
 
okey, i will explain with an example. say P=X^2 and Q=2X^3+X-2, and p=3, dividing P by H we obtain a rest of +4.
Then dividing Q by H we obtain a rest of -5.
+4 is congurent to +1 mod 3, and -5 is congurent with +1 mod 3.
We now say that the polynomials P is congurent to Q mod(p,H). Because they have the same remainders when dividing by H mod n.
 
Kontilera said:
okey, i will explain with an example. say P=X^2 and Q=2X^3+X-2, and p=3, dividing P by H we obtain a rest of +4.
You haven't said what H is.
Kontilera said:
Then dividing Q by H we obtain a rest of -5.
+4 is congurent to +1 mod 3, and -5 is congurent with +1 mod 3.
We now say that the polynomials P is congurent to Q mod(p,H). Because they have the same remainders when dividing by H mod n.
 
srry!
H is an non-constant polynomial whoose leading coeff is coprime to n.
 
May I summarize:
We have a prime ##p## and polynomials ##P(x), H(x) ∈ ℤ[x]## where ##\deg P = d ## and ##H[x] \mod p ## is irreducible in ##Z_p[x]##.
Then ##d \mod p ≥ 0##. But this is always the case.
Now we have to show that ##P(x)## has at most ##d## zeros ##\{x_1,...,x_d\}## in ##ℤ## or in ##ℤ_p##?
Or did you mean ## \{ x-x_1,...,x-x_d \} ## as "polynomial zeros"?

Those are pairwise incongruent "modulo ##(p,H)##" which you defined as follows:
Two polynomials ##P(x),Q(x) ∈ ℤ[x]## are congruent modulo ##(p,H)## if ##\frac{P}{H} = \frac{Q}{H} \mod n##.
I suppose ##n=p##? Or ##n=d##? And the division of the polynomials is performed in which Ring? Or shall we divide ##\frac{P(x_i)}{H(x_i)}##?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K