What role do prime numbers play in proving the irreducibility of polynomials?

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In summary, the conversation is discussing a polynomial f = (x^3) + 2(x^2) + 1, where f belongs to Q[x]. It is shown that the polynomial is irreducible by contradiction. The next part of the solution introduces a prime number p that divides s and uses this to show that (r,s) = 1. The next part of the solution considers two cases for the rational root r/s, and shows that neither case works, implying that the root must be irrational. The conversation also touches on the use of the rational root theorem and how it relates to the given example.
  • #1
Square1
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Hi. There is a polynomial f = (x^3) + 2(x^2) + 1, f belongs to Q[x]. It will be shown that the polynomial is irreducible by contradiction. If it is reducible, (degree here is three) it must have a root in Q, of the form r/s where (r,s) = 1. Plugging in r/s for variable x will resolve to
r^3 + 2(r^2)s + (s^3) = 0


I don't understand the next part of the solution. Why introduce this prime number that divides s.
--
Suppose a prime number p divides s.
This implies p divides 2(r^2)s + (s^2)
Or, this implies also the above equals 2(r^2)pa + (p^3)(a^3) = p(2(r^2)a + (p^2)(a^3)) which implies p divides (r^3) which also means p divides r which is contradiction because (r,s) = 1
---
So I follow the above steps, but what does prime number p dividing s or r have anything to do with this?


Next part of solution.
--> If s = 1, (r^3) + 2(r^2) + 1 = 0 means r((r^2) + 2r) = -1 implies r = 1 or -1, but 1 and -1 are not roots, seen by evaluating.
Same argument for s = -1.
This means that r/s is not a root, so not irreducible in Q[x]
---
So I'm lost about this step too. Only thing that I think is that if you let s = 1 or -1, it's as if you are trying to find something about a root in Z...

Homework Statement

 
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  • #2
Square1 said:
Hi. There is a polynomial f = (x^3) + 2(x^2) + 1, f belongs to Q[x]. It will be shown that the polynomial is irreducible by contradiction. If it is reducible, (degree here is three) it must have a root in Q, of the form r/s where (r,s) = 1. Plugging in r/s for variable x will resolve to
r^3 + 2(r^2)s + (s^3) = 0


I don't understand the next part of the solution. Why introduce this prime number that divides s.
--
Suppose a prime number p divides s.
This implies p divides 2(r^2)s + (s^2)
Or, this implies also the above equals 2(r^2)pa + (p^3)(a^3) = p(2(r^2)a + (p^2)(a^3)) which implies p divides (r^3) which also means p divides r which is contradiction because (r,s) = 1
---
So I follow the above steps, but what does prime number p dividing s or r have anything to do with this?


Next part of solution.
--> If s = 1, (r^3) + 2(r^2) + 1 = 0 means r((r^2) + 2r) = -1 implies r = 1 or -1, but 1 and -1 are not roots, seen by evaluating.
Same argument for s = -1.
This means that r/s is not a root, so not irreducible in Q[x]
---
So I'm lost about this step too. Only thing that I think is that if you let s = 1 or -1, it's as if you are trying to find something about a root in Z...

Homework Statement


You can easily show it's irreducible by using the rational roots theorem. http://en.wikipedia.org/wiki/Rational_root_theorem What they doing here is not using the theorem but working through the proof of the theorem for this specific polynomial. Either s has a prime factor which contradicts (r,s)=1 or s=1 or -1. That doesn't work either.
 
  • #3
This is basically a proof by cases for a reduced rational number r/s

Case 1, s is not a unit in Z (i.e. s is not 1 or -1): Then by prime factorization in Z, there is a prime p which divides s. Yada yada. This contradicts r and s coprime. It sounds like you understand the yada yada part. So Case 1 doesn't work.

Case 2, s is a unit in Z (i.e. s is either 1 or -1): Then r must also be a unit in Z (do you understand why this must be true?). However neither unit works. So we can't be in Case 2 either.

Case 1 and Case 2 exhaust all possible cases for rational roots. Neither works. So the root(s) must be irrational.
 
  • #4
I should have mentioned, the example was given as a warm up before learning the rational root test, so I can't make use of it.

When I have r^3 + 2(r^2)s + (s^3) = 0, if I just show that it also equals
s(2(r^2) + (s^2)) = -(r^3), isn't this enough to show that s divides r so (s,r) != 1 . So now I eliminated rational numbers with GCD = 1 , ie all of them?
 
  • #5
Square1 said:
I should have mentioned, the example was given as a warm up before learning the rational root test, so I can't make use of it.

When I have r^3 + 2(r^2)s + (s^3) = 0, if I just show that it also equals
s(2(r^2) + (s^2)) = -(r^3), isn't this enough to show that s divides r so (s,r) != 1 . So now I eliminated rational numbers with GCD = 1 , ie all of them?

No, you can't say s*n=r^3 implies that s divides r. 4*2=2^3 but 4 doesn't divide 2. That's exactly why they picked the prime. If p*n=r^3 and p is prime then it's correct to say p divides r.
 
  • #6
omg'z ...

I see. thank you thank you thank you.
 

Related to What role do prime numbers play in proving the irreducibility of polynomials?

1. What is the concept of irreducibility in polynomials?

Irreducibility in polynomials refers to the property of a polynomial that cannot be factored into polynomials of lower degree with coefficients in the same field. In other words, an irreducible polynomial cannot be broken down into simpler polynomials.

2. How do you determine if a polynomial is irreducible?

To determine if a polynomial is irreducible, one approach is to use the Rational Root Theorem, which states that if a polynomial with integer coefficients has a rational root, then that root must be a divisor of the constant term. If all possible rational roots are tested and none are found, the polynomial is irreducible.

3. Can a polynomial be both irreducible and reducible?

No, a polynomial cannot be both irreducible and reducible. A polynomial is either irreducible or reducible, but not both. If a polynomial can be factored into polynomials of lower degree, then it is reducible. If it cannot be factored, then it is irreducible.

4. What is the significance of irreducibility in polynomials?

Irreducibility is an important concept in algebra and number theory. It allows us to understand the structure and behavior of polynomials in a simpler way. Irreducible polynomials also have important applications in cryptography and coding theory.

5. Are there any techniques to prove the irreducibility of a polynomial?

Yes, there are various techniques to prove the irreducibility of a polynomial, such as Eisenstein's criterion, the degree test, and the mod p test. These methods use different properties and criteria to determine if a polynomial is irreducible or not.

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