# Abstract-Irreducible plynomials over Q

1. Jan 9, 2010

### WannaBe22

1. The problem statement, all variables and given/known data
Prove that the next polynomials are irreducible over Q:
B. 8x^3 -6x -1

2. Relevant equations
3. The attempt at a solution

Thanks a lot!

Last edited: Jan 9, 2010
2. Jan 9, 2010

### VeeEight

For B, does the polynomial have roots in Q? Use the fact that the polynomial is of degree 3 (if a polynomial of degree 2 or 3 has no roots in F then.......)

3. Jan 9, 2010

### WannaBe22

How can I show the specific plynomial has no roots in Q?
And I can't understand how to continue the theorem you gave...
About the other parts-as you can see, I've managed to solve them...
I only need help in B...

Hope you'll be able to continue helping me...

Thanks a lot

4. Jan 9, 2010

### JSuarez

The solution can be simple, if you covered Gauss' lemma (which states that if the polynomial is irreducible over the integers, then it's also irreducible over the rationals), or quite messy, if you have to prove directly.

5. Jan 9, 2010

### WannaBe22

I've learned thic specific lemma...I know that if a polynomial is irreducible over Z then it's also irreducible over Q...But how can I prove it's irreducible over Z?

6. Jan 9, 2010

### VeeEight

See: 4.2.7. Proposition in http://www.math.niu.edu/~beachy/aaol/polynomials.html

(the proof is pretty simple and involves simply factoring a polynomial of degree 2 or 3 and finding a root in F)

7. Jan 9, 2010

### WannaBe22

Ok...But how can I prove that this polynomia has no root over Q?
Is there no simple way involving Eisenstein's criterion or something?

TNX

8. Jan 9, 2010

### Dick

Use the Rational Roots Theorem.

9. Jan 9, 2010

### WannaBe22

Hmmmm So if k/l is a root of this polynomial then k|(-1) and l | 8
The possibilities are: k=1,-1 , l=+-1,+-2,+-4,+-8

Should I check for each one of the 16 possibilities that it isn't a root?

Thanks

10. Jan 9, 2010

### Dick

Well, I count 8 possibilities, but sure. It doesn't take long to run through them.

11. Jan 9, 2010

### WannaBe22

Yep...8 possibilities-sry...
Well, let's see:
p(1)=1, p(-1)=-3, p(0.5)=-3, p(-0.5)=1, p(0.25)=-19/8, p(-0.25)=3/8,p(1/8)=-111/64
p(-1/8)=-17/64 ... So we have no rational roots to this polynomial and then it's irreducible :)
About the propisition: 4.2.7. Proposition. A polynomial of degree 2 or 3 is irreducible over the field F if and only if it has no roots in F.
If we take a polynomial of degree 2-it must be decomposite into two factors of degree 1 and each and every one of them is a root...When we take a polynomial of degree 3, it can only be reduced into 2-1 factors or 1-1-1 and we must have one factor that defines a root...

Thanks a lot to all of you!

12. Jan 9, 2010

### Dick

That's it. It isn't until you get to degree 4 where a polynomial can have two quadratic factors, neither of which have any roots over Q.

13. Jan 9, 2010

### WannaBe22

Thanks a lot! You've all been very helpful!