# What is “prime factorisation” of polynomials?

I have the following question:

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

For starters, what does prime factorization mean? If we look at $x^3 - 1$ as an example, I can see that this can't be factorized 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 factorization of $(x^4 -1 ) = (x^2 + 1)(x - 1)(x + 1)$?

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

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.

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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 $(x-1)(x^2+x+1)$.

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