Abstract algebra: irreducible polynomials

In summary, the student is trying to find a rational root for the irreducible polynomial f(x)=x^3-7x+11, but is having difficulty because the coefficients a and c cannot both be integers.
  • #1
zero1207
6
0

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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  • #2
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.
 
  • #3
zero1207 said:
ac=11
Is that a rather severe limitation on a and c?
 
  • #4
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".
 
  • #5
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'.
 
  • #6
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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  • #7
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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1. What is the definition of an irreducible polynomial in abstract algebra?

An irreducible polynomial is a polynomial with coefficients in a given field that cannot be factored into two non-constant polynomials. In other words, it cannot be broken down into simpler components, making it a building block for creating more complex polynomials.

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

One method is to check if the polynomial has any roots in the given field. If it does not have any roots, then it is irreducible. Another method is to use the Eisenstein's criterion, which states that if a polynomial's coefficients satisfy certain conditions, then the polynomial is irreducible.

3. What is the importance of irreducible polynomials in abstract algebra?

Irreducible polynomials play a crucial role in the study of field extensions and polynomial rings. They provide a way to construct new fields by adjoining roots of irreducible polynomials, and they also have applications in coding theory and cryptography.

4. Can an irreducible polynomial have multiple degrees in abstract algebra?

Yes, an irreducible polynomial can have multiple degrees. For example, a polynomial of degree 2 can be irreducible in a field where the characteristic is not 2, and a polynomial of degree 3 can be irreducible in a field where the characteristic is not 3. However, in fields where the characteristic is the same as the degree of the polynomial, every polynomial of that degree is irreducible.

5. Is every polynomial in abstract algebra reducible or irreducible?

No, not every polynomial is either reducible or irreducible. Some polynomials can be factored into simpler polynomials, while others cannot. For example, a polynomial with a single term is irreducible, while a polynomial with two terms can be reducible if it can be factored into two polynomials with lower degrees. However, there are polynomials that cannot be factored into simpler polynomials and are therefore irreducible.

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