Abstract algebra: irreducible polynomials

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SUMMARY

The polynomial f(x) = x^3 - 7x + 11 is proven to be irreducible over the rational numbers Q. Attempts to apply the Eisenstein criterion were unsuccessful, leading to the construction of a new polynomial g(x) = (x + 1)^3 - 7(x + 1) + 11, which also did not meet the criterion. The Rational Root Theorem was suggested as a definitive method to demonstrate the irreducibility, emphasizing that if f(x) were reducible, it would possess a rational root. The discussion highlights the necessity of rational integers in the factorization of integer polynomials.

PREREQUISITES
  • Understanding of irreducible polynomials in abstract algebra
  • Familiarity with the Eisenstein criterion for irreducibility
  • Knowledge of the Rational Root Theorem
  • Basic polynomial factorization techniques
NEXT STEPS
  • Study the application of the Eisenstein criterion in various polynomial examples
  • Learn about the Rational Root Theorem and its implications for polynomial roots
  • Explore methods for proving irreducibility of polynomials over different fields
  • Investigate advanced topics in abstract algebra, such as Galois theory
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Students and educators in mathematics, particularly those studying abstract algebra, as well as anyone interested in polynomial theory and its applications in higher mathematics.

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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