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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.

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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

4 Answers 4

up vote 5 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 $b^3$.

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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 $\mathbf Q$.

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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 $2$.

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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.

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