Abstract algebra: irreducible polynomials

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Homework Help Overview

The discussion revolves around proving the irreducibility of the polynomial f(x) = x^3 - 7x + 11 over the rational numbers Q. Participants explore various methods and criteria for establishing irreducibility in the context of abstract algebra.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the Eisenstein criterion and transformations of the polynomial, as well as attempting to factor the polynomial into smaller components. Questions arise regarding the implications of the product ac = 11 and the nature of rational roots.

Discussion Status

There are multiple lines of reasoning being explored, with some participants suggesting the use of the rational root theorem and questioning the limitations imposed by the coefficients. Guidance has been offered regarding the nature of a and c in relation to their product, but no consensus has been reached on a definitive method for proving irreducibility.

Contextual Notes

Participants note the constraints of working with rational coefficients and the implications of the polynomial's degree. The discussion reflects a mix of attempts and theoretical considerations without a clear resolution.

zero1207
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Homework Statement


Prove that f(x)=x^3-7x+11 is irreducible over Q


Homework Equations





The Attempt at a Solution


I've tried using the eisenstein criterion for the polynomial. It doesn't work as it is written so I created a new polynomial g(x)=f(x+1)=(x+1)^3-7(x+1)+11=x^3+3x^2-4x+5. I did this because g(x) and f(x) are similar and if g(x) is irreducible so is f(x), but the new polynomial I constructed doesn't meet the eisenstein criterion either. Any ideas on where I should turn next?
 
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I've also tried breaking up the polynomial into two smaller ones:
x^3-7x+11=(x+a)(x^2+bx+c)=x^3+(a+b)x^2+(ab+c)x+ac
so from this, I get:
a+b=0; ab+c=-7; ac=11
by rearranging the equations I get b(-b^2+7)=11, but I don't know where I could go from there to show that no solutions for b exist in Q.
 
zero1207 said:
ac=11
Is that a rather severe limitation on a and c?
 
I think I see what you're saying, a and c can't be integers for their product to be 11, but they could both be rationals. I'm trying to prove that the polynomial is irreducible over Q. I do see that when I put b(-b^2+7)=11 into a calculator, my solution is not a rational number. That justifies that it's not reducible to me, but I'm sure there's a more definitive way of showing that without having to say "my calculator says this".
 
If it's reducible over Q then it must have a rational root, since it's a cubic. You might want to use the 'rational root theorem'.
 
zero1207 said:
I think I see what you're saying, a and c can't be integers for their product to be 11,
Actually, they can. (In exactly 4 different ways)

but they could both be rationals.
But they have to be rational integers! (Because you're factoring an integer polynomial, rather than a more general one with nonintegral rational coefficients)
 
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hey zero, how do you like abstract algebra so far? i am debating on whether or not to take it this summer. it's a 6-week course and my friend who tutors with me said it's the biggest B you first take :O
 
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