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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following question:

Is there an easy way to prove that $x^6-72$ is irreducible over $\mathbb{Q}\ $?

I am trying to avoid reducing mod p and then having to calculate with some things like $(x^3+ax^2+bx+c)\cdot (x^3+dx^2+ex+f)$ and so on...

Thank you very much.

share|cite|improve this question
Eisenstein's Criterion can sometimes help with such questions. Unfortunately, it does not help in this particular example since $72 = 2^3 3^2$ (if a prime divides $72$, then so does its square). – Austin Mohr Jul 14 '13 at 20:17
This post might be helpful – Torsten Hĕrculĕ Cärlemän Jul 14 '13 at 20:17
Any factorization would have to yield a factorization of $x^6$ in $\mathbb Z_2[x]$ and $\mathbb Z_3[x]$. So $a,b,c,d,e,f$ in your "gross" approach must all be divisible by $6$. – Thomas Andrews Jul 14 '13 at 20:30
The multiplicative group of the residue class field $\mathbb{Z}_{73}$ is cyclic of order $72$. As $12\mid 72$ this means that $x^6+1$ splits into a product of linear factors modulo $73$. After all, its zeros are twelfth roots of unity (not all primitive). Modulo 73 we have $$x^6-72=(x+3)(x+24)(x+27)(x-27)(x-24)(x-3). $$ It is more helpful to try smaller primes, but Cocopuffs has already done that. – Jyrki Lahtonen Jul 27 '13 at 8:23
up vote 7 down vote accepted

I had posted a more general question on Brilliant before, namely asking when is $p_n(x) = x^6 + n $ reducible over the integers (which is equivalent to reducible over the rationals as the content of the polynomial is 1.)

Suppose $p_n(x)=g(x)\cdot h(x),$ where $g$ and $h$ are not constants. The sum of the degrees of $g$ and $h$ is $6$, and the product of the leading coefficients is $1.$ Because all coefficients are integers, this means that the leading coefficients of $g$ and $h$ are either both $1$ or both $-1.$ In the latter case, we multiply both $g$ and $h$ by $-1$ so that the leading coefficients are $1$. Also, we can assume, without loss of generality, that $\deg(g)\geq \deg(h).$

The polynomial $p_n$ has $6$ complex roots, all with absolute value $\sqrt[6]{|n|}$. Suppose the degree of $h$ is $k$, which can be $1,2,$ or $3$. Then the absolute value of the free term of $h$ is the product of absolute values of $k$ roots, thus it is $|n|^{k/6}.$ If this is an integer, then $|n|$ must be either a perfect square (if $k=3$) or a perfect cube (if $k=2$) or a perfect 6th power (if $k=1$, though this is also a perfect cube). Moreover, if $k=3$, then $n$ cannot be positive, because every cubic polynomial has a real root, and $x^6+n>0$ for positive $n$.

Hence $p_n(x)$ is reducible if and only if $n = -a^2 $ or $b^3$.

share|cite|improve this answer
Thank you very much for your answer and for the explanations therein! – Bernhard Boehmler Jul 14 '13 at 20:51

The problem becomes a lot simpler if you look at the field generated by a root of the polynomial. Let $\alpha^6 = 72$. Then $[{\mathbf Q}(\alpha):{\mathbf Q}] \leq 6$. Since $(\alpha^3/6)^2 = 2$ and $(\alpha^2/6)^3 = 1/3$, the field ${\mathbf Q}(\alpha)$ contains a square root of 2 and a cube root of 3. Thus $[{\mathbf Q}(\alpha):{\mathbf Q}]$ is divisible by 2 and by 3, hence by 6, so $[{\mathbf Q}(\alpha):{\mathbf Q}] = 6$, which is another way of saying the minimal polynomial of $\alpha$ over the rationals has degree 6. Therefore $x^6 - 72$ has to be the minimal polynomial of $\alpha$ over the rationals, so this polynomial is irreducible over $\mathbf Q$.

share|cite|improve this answer
Thanks! This is the most conceptual approach to this problem! – mathreader Jul 27 '13 at 19:53

It is helpful but you can look at smaller primes as well.

Reduction modulo $5$ gives $$X^6 - 72 = (X^2 + 2)(X^2 + X + 2)(X^2 - X + 2).$$ Reduction modulo $7$ gives $$X^6 - 72 = (X^3 + 3)(X^3 - 3).$$

The reduction modulo $5$ implies that there is no factor of degree $3$, and the reduction modulo $7$ implies that there is no factor of degree $2$.

share|cite|improve this answer
These factorizations look like magic to me! – mathreader Jul 27 '13 at 8:15
@mathreader Mod $7$: $X^6 - 72 = (X^3)^2 - 3^2 = (X^3 + 3)(X^3 - 3)$; mod $5$: $X^6 - 72 = (X^2)^3 + 2^3 = (X^2 + 2)(X^4 - 2X + 4)$. Maybe it is not obvious that $X^4 - 2X + 4$ factors further but that is not really important for the argument - we just need to know that there is no factor of degree $3$. In general there are good algorithms like Cantor-Zassenhaus – Cocopuffs Jul 27 '13 at 8:25
One way to get these factorisations in thos simple case is to look for prime factors of $72-N^2$ which gives $7$ for $N=3$, and likewise for cubes. – Mark Bennet Jul 27 '13 at 8:46

Following Thomas Andrews' remark, we observe that $x^6-72=f(x)g(x)$ with $f,g\in\mathbb Z[x]$ implies that all coefficients of $f,g$ except the leading $1$ are multiples of $6$ because reduction modulo $2$ or $3$ must give us a factorization of $x^6$. Clearly, $x^6-72$ has no linear factor. For a quadratic factor, we make the ansatz $$ x^6-72=(x^4+6ax^3+6bx^2+6cx+6d)(x^2+6ex+6f)$$ and find $d+6(ce+bf)=0$ from the coefficient of $x^2$, so $d$ is a multiple of $6$ and $-72=6^3df$, contradiction. Remains the cubic case $$ x^6-72=(x^3+6ax^2+6bx+6c)(x^3+6dx^2+6ex+6f).$$ From the constant term, we get $cf=-2$, hence $c+f=\pm1$ (one of them is $\pm2$,the other $\mp1$). Then from the coefficient of $x^3$ we get $c+f+6(bd+ae)=0$, i.e. $c+f\equiv0\pmod 6$, contradiction.

share|cite|improve this answer

Your Answer


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.