Irreducibility of Polynomial


by Square1
Tags: irreducibility, polynomial
Square1
Square1 is offline
#1
Dec5-12, 05:42 PM
P: 115
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]
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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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Square1
Square1 is offline
#2
Dec5-12, 06:36 PM
P: 115
sry. i guess this is homework styled question. I don't see a delete button.


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