Why Do Prime Numbers Play a Role in Proving the Irreducibility of Polynomials?

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In summary, the polynomial f = (x^3) + 2(x^2) + 1, f belongs to Q[x] will be shown to be irreducible by contradiction. The solution introduces a prime number p that divides the coefficient s in the polynomial, and then shows that if this were the case, it would lead to a contradiction. Additionally, the steps taken to show that r/s is not a root, therefore proving the polynomial is not irreducible in Q[x], are explained.
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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.
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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
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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...
 
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sry. i guess this is homework styled question. I don't see a delete button.
 

1. What is irreducibility of polynomial?

The irreducibility of a polynomial refers to whether or not it can be factored into smaller polynomials with integer coefficients. A polynomial is irreducible if it cannot be factored into smaller polynomials without using complex numbers.

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

One way to determine if a polynomial is irreducible is by using the rational root theorem, which states that any rational root of a polynomial must be a factor of the leading coefficient divided by the constant term. If no rational roots are found, the polynomial is irreducible.

3. What are the implications of a polynomial being irreducible?

If a polynomial is irreducible, it means that it cannot be simplified any further and can only be factored using complex numbers. This has implications in fields such as algebraic geometry and number theory.

4. Can a polynomial with integer coefficients be irreducible over the complex numbers?

Yes, a polynomial with integer coefficients can be irreducible over the complex numbers. An example of this is the polynomial x^2 + 1, which cannot be factored using only rational numbers but can be factored using complex numbers as (x + i)(x - i).

5. How does the degree of a polynomial relate to its irreducibility?

The degree of a polynomial is the highest power of the variable present in the polynomial. In general, polynomials with higher degree are more likely to be irreducible. However, this is not always the case as there are exceptions, such as the polynomial x^4 + 4, which is irreducible even though it has a relatively low degree.

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