# Showing that $[\mathbb{Q}(\sqrt[3]{2} \cdot \sqrt{3}) : \mathbb{Q}] = 6$

Well,

I want to show that $$f(x) = x^6 - 108$$ is irreducible over $$\mathbb{Q}$$, but I had no luck so far.

I tried to use Eisenstein criterion on $$f(x+1)$$ and $$f(x+2)$$ but it didn't work since the free coefficients doesn't suitable in both cases. I also tried to find a prime $$p$$ such that $$f(x)\bmod p$$ is irreducible, but it seems extremely inelegant.

Any other ideas?

• For the general case $f(X)=X^n-a$ see other posts, e.g., this one, or this one. Jun 5, 2021 at 8:07

The simplest thing to do is probably just to note that $$\newcommand\Q{\mathbb Q}\Q(\sqrt3)\subset\Q(\sqrt[6]{108})=:L$$ and $$\Q(\sqrt[3]2)\subset L$$. Thus, $$[L:\Q]$$ is divisible by $$2=[\Q(\sqrt3):\Q]$$ as well as by $$3=[\Q(\sqrt[3]2):\Q]$$, so $$6\mid[L:\Q]$$. At the same time, as you notices, $$[L:\Q]\le6$$ since it's generator satisfies a degree $$6$$ polynomial, so $$[L:\Q]=6$$ which means $$x^6-108$$ must be the minimal polynomial of $$\sqrt[6]{108}$$, and so must be irreducible.

Let's write $$\alpha = \sqrt[3]{2} \cdot \sqrt{3}$$.

Notice that $$\alpha^2 = 3 \cdot \sqrt[3]{4}$$ and $$\alpha^3 = 6 \cdot \sqrt{3}$$. Thus, $$\mathbb{Q}(\alpha)/\mathbb{Q}$$ has $$\mathbb{Q}(\sqrt[3]{4})$$ and $$\mathbb{Q}(\sqrt{3})$$ as intermediate extensions. Since these fields are of degree $$3$$ and $$2$$ over $$\mathbb{Q}$$, respectively, we conclude that $$6 | [\mathbb{Q}(\alpha) : \mathbb{Q}]$$.

On the other hand, since $$\alpha^6 - 108 = 0$$, we know $$[\mathbb{Q}(\alpha) : \mathbb{Q}]\leq 6$$, so we conclude $$[\mathbb{Q}(\alpha) : \mathbb{Q}] =6$$ and the minimum polynomial of $$\alpha$$ is degree 6. In particular, we must have that $$x^6-108$$ is the minimum polynomial.

This answer merely regards proving irreducibility of $$x^6 - 108$$ , not using Galois theory. That this particular fact proves the assertion is obvious (following a standard theorem that the degree of the minimal polynomial equals that of the field extension), since $$\alpha^6 = 108$$.

Eisenstein will not work to show the irreducbility of this polynomial. I'll expand on this in my answer : this part will require Galois theory.

There's a simple prime $$p$$ with respect to which this is irreducible , though.

However, partly motivated by applications to Galois theory, there exist plenty (and I mean plenty) of criterion for irreducibility of integer coefficient polynomials. I'll mention one of these as well.

### When doesn't Eisenstein work? (A general note)

Well, essentially speaking, we have this quantity called the discriminant of a polynomial : it's linked with the roots of the polynomial. If $$P(x) = a_nx^n + ... + a_0$$ has complex roots $$r_1,...,r_n$$ then : $$\Delta(P) = (-1)^{n(n-1)/2}a_n^{2n-2} \prod_{i

is basically some power of the leading coefficient, times all the squared differences between each different pair of roots. (Note : if $$P$$ has repeated roots like $$(x-1)^2$$ then the discriminant is zero).

A nice geometric interpretation allows for some comfort : one may realize the given quantity as a discriminant of the gram matrix of a certain set of vectors sitting in $$\mathbb C^m$$ for some $$m$$ (actually $$m$$ is the size of the Galois group, but we won't talk about that). Either formula is convenient.

But the key result is the following (Theorem 3.3 from Conrad's notes present here) :

Suppose that $$f(x)$$ is a $$p$$-Eisenstein polynomial of degree $$n$$. Then if $$p$$ divides $$n$$, we must have $$p^n$$ divides $$\Delta(P)$$. On the other hand, if $$p$$ doesn't divide $$n$$, we must have $$p^{n-1}$$ divides $$\Delta(P)$$.

This helps us instantly eliminate a LOT of primes $$p$$ with respect to which this polynomial can be Eisenstein. In case you are worried about shifts : the discriminant doesn't change under shifts, nothing to worry about.

So all we have to do is find the discriminant. Easier said than done?

Yes : the discriminant is $$2^{16}3^{21}$$. If you're wondering, there is an easier way of finding the discriminant in this case : that's because you can write the roots of $$P$$ as $$\omega^i \beta,i=0,1,...,5$$ where $$\omega$$ is the sixth root of unity and $$\beta$$ is the unique positive sixth root of $$108$$. Using the root-based formula, the $$\beta$$s come out and the product simplifies into something involving the roots of unity, and you can use roots-of-unity type tricks to work things out.

But what this proves is : if the polynomial is Eisenstein, then it must be Eisenstein either modulo $$2$$ or $$3$$. Which may have looked trivial at first glance , but was worth expounding on anyway.

But clearly the polynomial can't be Eisenstein modulo either : recall that Eisenstein polynomials modulo $$p$$ must be irreducible modulo $$p$$ : however, whether it's modulo $$2$$ or modulo $$3$$, the polynomial $$x^6-108$$ reduces to $$x^6$$, which of course is reducible.

It follows that $$x^6-108$$ isn't $$p$$-Eisenstein for any prime $$p$$!

### A remarkably simple criteria for irreducibility

The polynomial $$x^6-108$$ is irreducible. For this, we use another criteria using Vieta's formula.

Suppose that a monic polynomial $$P$$ has complex roots $$r_1,r_2,...,r_n$$, and suppose that for each non-empty subset $$S \subsetneq \{1,2,...,n\}$$, we have that $$\prod_{i \in S} r_i \notin \mathbb Q$$. Then $$P$$ is irreducible.

Why? Well, suppose that $$P(x) = p(x)q(x)$$ for polynomials of smaller degree $$p$$ and $$q$$. Then, using Vieta's formulas , we know that $$p(x) = \prod_{i \in S'} (x-r_i)$$ for some non-empty subset $$S' \subsetneq \{1,2,...,n\}$$, implying that $$|p(0)| = \prod_{i\in S'} |r_i|$$. The right hand side is irrational but the left hand side is rational, a contradiction.

In fact, it's enough to prove this for any $$S \subset \{1,2,...,n\}$$ with $$|S| \leq \frac n2$$. That's because we can WLOG consider the polynomial $$p$$ or $$q$$ of smaller degree, which can have at most $$\frac n2$$ roots.

To use this, we have to go for $$|S| = 1,2,3$$. For $$|S| =1$$ it's obvious that every root is irrational (since $$108$$ is not a perfect sixth power).

For $$|S| = 2$$, you have a product of the form $$\omega^{i+j}\beta^{\frac 26}$$. Each such number is a root of $$x^6= 108^3$$ : using the rational root theorem, such numbers are irrational.

For $$|S| = 3$$, you have a product of the form $$\omega^{i+j+k}\beta^{\frac 36}$$. Each such number is a root of $$x^6 = 108^2$$ : which is again always irrational by the rational root theorem!

The result follows. Note that it would have been difficult if , for example, some combination of roots would have given us trouble e.g. for $$x^4+4$$, the roots are given by $$i^{j}\sqrt[4]{4}, j=0,1,2,3$$, but if you multiply the roots corresponding to $$j=1,3$$ you'd get $$2$$. So you know that the criterion fails, and as it turns out the polynomial is reducible.

### When does reducibility modulo $$p$$ not work? (A general note)

Reducibility modulo $$p$$, has also got a Galois flavour to it as well. This is a result that will look surprising. Of course, note that if $$p$$ divides the discriminant of $$P$$, then modulo $$p$$, the discriminant is zero i.e. there are repeated roots. The discriminant being zero will imply that $$P$$ will have a common root with $$P'$$, and hence a linear factor, a contradiction. So a polynomial $$P$$ can only be irreducible modulo $$p$$ for $$p$$ not dividing the discriminant of $$P$$. For those, we have this result.

Suppose that $$P$$ is an irreducible integer coefficient polynomial, and $$K$$ is the field generated by the roots of $$P$$ (We know that $$K/\mathbb Q$$ is a Galois extension). Suppose that $$p$$ is a prime and that $$P$$ is irreducible modulo $$p$$. Then the Galois group of $$P$$ must contain an element of order $$\deg f$$.

So if the Galois group did not contain an element of order $$\deg f$$, then the polynomial is in fact reducible modulo $$p$$ for all $$p$$.

Can we write down the Galois group for $$x^6-108$$? After some effort, you'll notice that it's a group of size $$12$$ which , as it happens, contains a group element of order $$6$$ : Fail!

As a matter of fact, the polynomial $$x^6-108$$ is irreducible modulo $$7$$.

Here is an example where the criteria actually worked. I couldn't believe that it did. The technique used to prove irreducibility there, doesn't work here because you can't rule out simple factors. You may read the reference present there as well.

### A lovely little side note

Using complex root positions and Vieta's formulas, you can try obtaining the following interesting result , Theorem 4.1 from the paper here:

Suppose that $$P$$ is a polynomial and $$H = \max_{0 \leq i \leq n-1} \left|\frac{a_i}{a_n}\right|$$ ($$H$$ is called the height). Then if there is $$m \geq H+1$$ such that $$P(m)$$ is a prime, $$P$$ is an irreducible polynomial.

The height of $$x^6-108$$ is $$108$$. So we can try to see if values of $$P(m)$$ are prime from $$m=109$$ onwards. As luck would have it, $$P(125) = 125^6 -108 = 3814697265517$$ is prime. Done!

### Irreducibility proving

Regarding irreducibility, essentially speaking there are some standard techniques that are used to prove this for polynomials. Very briefly :

• Reducibility modulo $$p$$ for a prime $$p$$ (Includes Eisenstein).

• Positions of the roots of the polynomial induced by abnormally large coefficients (like Perron's criterion).

• Constraining coefficients of factors using prime values at large inputs (Cohn's criterion, the height criterion, and so on).

• Newton polynomials , which use results that link root placement with the p-adic values of the coefficients to constrain coefficients of linked polynomials such as compositions, products etc. (Explained here).

All these techniques contribute to Galois theory at our level, since they will be able to give us computational as well as ad hoc criteria for testing irreducibility.

• it seems you've shown that $x^6-108$ does not have any roots modulo $7$, but this doesn't show irreducibility. Jun 5, 2021 at 2:34
• @BrianShin Good point : I'll edit the post. I can't believe I missed it. It's true, but I should have been careful. Jun 5, 2021 at 2:45
• @BrianShin I completed my proof using another simple criteria. I chose to focus on irreducibility. I wished to expound on other things regarding irreducibility, since the attempts made were in this direction. Jun 5, 2021 at 3:44
• Very cool, thank you! Jun 5, 2021 at 12:06
• @r0nnie Oh yes, thanks, my intention was to make this a post that was different from the others. Others use purely Galois-theoretic methods of proving irreducibility : I resolve to instead using simpler and more computable criteria. I hope my answer will do more justice than this mere question : you can try to apply the criteria I mention freely to a very large class of polynomials. Jun 5, 2021 at 12:12