Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following question:

Find the prime factorisation in $\mathbb{Z}[x]$ of $x^3 - 1, x^4 - 1, x^6 - 1$ and $x^{12} - 1$. You will need to check the irreduciblity in $\mathbb{Z}[x]$, of three quadratic polynomials and of one quartic. In the case of the quartic, you will need to check that it has no integer zeros and does not factorise as a product of two quadratics with integer coefficients.

For starters, what does prime factorisation mean? If we look at $x^3 - 1$ as an example, I can see that this can't be factorised any further, so it is irreducible in $\mathbb{Z}[x]$. Does that mean there is no prime factors for $x^3 - 1$?

If we then look at $x^4 -1 = (x^2 - 1)(x^2 + 1)$. Here, we see that it is irreducible and so would it simply be that the prime factorisation of $(x^4 -1 ) = (x^2 + 1)(x - 1)(x + 1)$?

share|improve this question
1  
Why the downvote? –  Git Gud Apr 14 '13 at 19:58
1  
$x^3-1=(x-1)(x^2+x+1)$ –  Git Gud Apr 14 '13 at 19:58

2 Answers 2

up vote 4 down vote accepted

For quadratics and cubics, irreducibility is easy to test, for if there are no rational roots, the polynomial must be irreducible. And the polynomials $x^2+x+1$ and $x^2-x+1$ don't even have real roots. In more complicated cases, the Rational Roots Theorem could be helpful.

Let's deal with $x^{12}-1$. This immediately factors as $(x^6-1)(x^6+1)$. We leave further decomposition of $x^6-1$ to you. For $x^6+1$, note that it is equal to $(x^2+1)(x^4-x^2+1)$.

We would like to show that $x^4-x^2+1$ is irreducible. It certainly cannot be written as a linear polynomial with coefficients in $\mathbb{Z}$ times a cubic, since $x^4-x^2+1=0$ has no rational roots. Can we express $x^4-x^2+1$ as a product of quadratics with integer coefficients?

If we can, without loss of generality we would have that the decomposition has shape $(x^2+ax+b)(x^2+cx+d)$. Expand. Since $x^4-x^2+1$ has no $x^3$ term, we must have $a+c=0$, so $c=-a$.

Also, we must have $b=1$, $d=1$ and $b=-1$, $d=-1$. This gives the two possibilities (i) $(x^2+ax+1)(x^2-ax+1)$ and (ii) $(x^2+ax-1)(x^2-ax-1)$.

We check there is no factorization of type (i). For the coefficient of $x^2$ in the product $(x^2+ax+1)(x^2-ax+1)$ is $2-a^2$. This cannot be $-1$ for integer $a$.

Similarly, one can show (ii) can't work.

Remark: We want to express each polynomial as a product of irreducibles. The factorization $x^4-1=(x-1)(x+1)(x^2+1)$ is such a factorization. It is not quite unique. For example we also have $x^4-1=(-x+1)(x+1)(-x^2-1)$, and two other obvious variants. But the decomposition is essentially unique, so giving one decomposition is enough.

share|improve this answer
    
How did you know that x^4-x^2+1 has no rational zeroes right off the bat? Did you quickly check + or -1? By the way Andre, could you please answer my 2 questions? Thanks a lot! math.stackexchange.com/questions/361703/… math.stackexchange.com/questions/361691/… –  Ovi Apr 14 '13 at 23:24
    
When the leading coefficient of a polynomial is 1, the Rational Zeroes Theorem tells us that any rational zeroes must be integer divisors of the constant term (so, yes, here they'd have to be +1 or -1). –  RecklessReckoner Apr 15 '13 at 1:44
    
Instead of comparing coefficiants like that, could I write it in the from $A^2 - A + 1$ where $x^2 = A$ and use the quadratic formula to show that this has no real roots? –  Kaish Apr 15 '13 at 9:31
    
I don't see how that idea can be made to work. For example, $x^4+4=(x^2-2x+2)(x^2+2x+2)$, yet putting $A=x^2$ we get $A^2+4$, which has no real roots. –  André Nicolas Apr 15 '13 at 9:36

The prime factorization is the factorization into irreducible polynomials. For example, since $x^3-1$ can be factorized as $(x-1)(x^2+x+1)$ and both these are irreducible, the prime factorization of $x^3-1$ in $\mathbb Z[x]$ of $x^3-1$ is $(x-1)(x^2+x+1)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.