Basic Fundamentals...?

Number Theory Level 2

When $x^9 - x$ is factorized as completely as possible into polynomials and monomials with integral coefficients, how many factors are there?

3 8 5 9 7 2

2 solutions

Mark Hennings
Aug 9, 2019

We can certainly write $x^9 - x \; = \; x(x-1)(x+1)(x^2+1)(x^4+1)$ We need to know that $x^2+1$ and $x^4+1$ are irreducible over $\mathbb{Z}[x]$ . The first is easy, but it is simplest to use Eisenstein's Irreducibility Criterion to handle $x^4 + 1$ , and $x^2+1$ can be dealt with the the same manner. Since $(x+1)^2 + 1 \; = \ x^2 + 2x + 2 \hspace{2cm} (x+1)^4 + 1 \; = \; x^4 + 4x^3 + 6x^2 + 4x + 2$ and since $2$ divides $2,4,6$ , while $2$ does not divide $1$ and $4=2^2$ does not divide $2$ , we deduce that $x^2 + 1$ and $x^4 + 1$ are both irreducible over $\mathbb{Z}[x]$ . This makes the answer $\boxed{5}$ .

Callum Cassidy-Nolan
Aug 11, 2019

Recall that $\left( a^2 - b^2 \right) = \left( a + b \right) \left( a - b \right)$ thus we have $\left( x^{9} - x \right) = x\left( x^{8} -1 \right) = x\left( x^{4} + 1 \right) \left( x^{4} - 1 \right)$ continuing, in a similar fashion, we have $x\left( x^{4} + 1 \right) \left( x^{2} + 1 \right) \left( x^{2} - 1 \right) = x\left( x^{4} + 1 \right) \left( x^{2} + 1 \right) \left( x + 1 \right) \left( x - 1 \right)$ we observe there are $\boxed{5}$ factors.

... but you need to show that $x^2+1$ and $x^4+1$ cannot be factorised as well...

Mark Hennings - 1 year, 10 months ago

Hmm yes I see now. I'm not sure if I have the tools to do so, can you give me a suggestion on how to continue?

Callum Cassidy-Nolan - 1 year, 10 months ago

Well, you could do it by hand...

If $x^2+1$ could be factorised over the integers, it would have to have a linear factor $x+a$ . But then $a$ would divide $1$ , and hence $a=\pm1$ . We can of course check that neither $x+1$ nor $x-1$ are factors. That handles the case of $x^2+1$ ,

$x^4+1$ could be factorised over the integers, it would either have a linear factor, or be the product of two quadratics. Just as in the previous paragraph, we can argue that it cannot have a linear factor. Try writing $x^4+1$ as the product of two quadratics. Since there are no $x^3$ and $x$ terms, and since the constant term is $1$ , these two quadratics would have to be $x^2 + ax + u$ and $x^2 - ax + u$ for some $a$ , where $u = \pm1$ . But then the coefficient of $x^2$ would have to be $0 = 2u - a^2$ , and this is not possible.

Eisenstein's Irreducibility Criterion says that if $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n -1} + a_nx^2$ is a polynomial with integer coefficients, and if $p$ is a prime number such that:

$p$ divides $a_0,a_1,a_2,...,a_{n-1}$ ,
$p$ does not divide $a_n$ ,
$p^2$ does not divide $a_0$ ,

then $f(x)$ is irreducible over the integers. That is what I used.

Mark Hennings - 1 year, 10 months ago

Basic Fundamentals...?

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