Splitting field of $x^3-10x+10$ over $\mathbb{Q}(\sqrt{13})$

Question

Is it possible to find the splitting field of $x^3 - 10x + 10$ over $\mathbb{Q}(\sqrt{13})$?

I think this polynomial can't be factorized.

Answer 1

As you guessed, $f$ is irreducible. Then its splitting field is Galois over $\mathbb{Q}$. To prove it's irreducible $\mathbb{Q}(\sqrt{13})$ is a slightly harder task, but I'll leave it to you.

Note that the this cubic is in the form $x^3+px+q$, the depressed cubic. The discriminant for this is $-4q^3-27p^2$.

Hence, the discriminant is $1300$.

Then $\sqrt{D}=\sqrt{1300}=10\sqrt{13}\in \mathbb{Q}(\sqrt{13})$.

Letting $K$ be the splitting field of $f$, any Galois group ${\rm Gal}(K/\mathbb{Q}(\sqrt{13}))$ must fix $\sqrt{D}$. But note that letting $x_1,x_2,x_3$ be the roots, transpositions of them do not fix the product

$$\sqrt{D}=(x_1-x_2)(x_1-x_3)(x_2-x_3)$$.

Hence, the Galois group must be $A_3$.

Answer 2

Discussing a couple ways of proving that $p(x)=x^3-10x+10$ is irreducible over $\Bbb{Q}(\sqrt{13})$.

Method 1. As William Chang observed the discriminant of $p(x)$ is a perfect square times $13$. Therefore the splitting field of $p(x)$ over $\Bbb{Q}$ actually contains $\sqrt{13}$. Therefore...

Method 2. (using bits and pieces of basic algebraic number theory) We see that the algebraic integers $(-1\pm\sqrt{13})/2$ have norm $-3$. Therefore they both generate prime ideals lying above the rational prime $p=3$. Let $\mathfrak{p}$ be one of those prime ideals. Clearly $\mathfrak{p}\cap\Bbb{Z}=(3)$. Also if $p(x)$ were reducible over $\Bbb{Q}(\sqrt{13})$ then those factors clearly would have coefficients in the ring of integers $\mathcal{O}=\Bbb{Z}[(-1+\sqrt{13})/2]$. Therefore the projection of $p(x)$ would be reducible also over $\mathcal{O}/\mathfrak{p}$. But this quotient ring is just $\Bbb{F}_3=\Bbb{Z}/3\Bbb{Z}$, so we can just as well studied $p(x)$ reduced modulo $3$.

But the polynomial $$ x^3-x+1\equiv p(x)\pmod 3 $$ is known to be irreducible over $\Bbb{F}_3$. It is cubic, so you can just verify that it has no zeros in $\Bbb{F}_3$. That is, unless you know the result that $x^p-x-a, p\nmid a$, is irreducible over $\Bbb{F}_p$ - an exercise that had been handled on our site umpteen times :-)

Answer 3

The discriminant of $x^3 - p x + q$ is $4 p^3 - 2 7 q^2$, so for $x^3 - 10x + 10$ we get $\Delta= 4000- 2700= 1300$. Recall that if $x_1$, $x_2$, $x_3$ are the roots of $x^3 - p x + q=0$ the discriminant is $\Delta= ((x_1-x_2)(x_1-x_3)(x_2-x_3))^2$. We conclude that $$(x_1-x_2)(x_2-x_3)(x_3-x_1) = \pm 10 \sqrt{13}$$

Now in general, the splitting field of a polynomial $P \in \mathbb{Q}[X]$ over a field $K\supset \mathbb {Q}$ is the field obtained by adjoining all the roots of $P$ to $K$. It is also the composite of the splitting field of $P$ over $\mathbb{Q}$ and $K$. In this case $K= \mathbb{Q}(\sqrt{13})$ is already contained in the splitting field over $\mathbb{Q}$ by the equality above, so it will equal the splitting field $L$ of $x^3 - 10 x + 10$ over $\mathbb{Q}$.

Note that the Galois group of $L$ over $\mathbb{Q}$ is not contained in $\mathbb{A}_3$ since $\Delta$ is not a root in $\mathbb{Q}$. Hence, $L \colon \mathbb{Q} = 6$. Moreover, since $\Delta > 0$, the copy of $L$ inside $\mathbb{C}$ lies in fact in $\mathbb{R}$ ( all the roots of the equation $P(x) = 0$ are real). Now $L$ is obtained from $\mathbb{Q}(\sqrt{13})$ by adjoining an element of degree $3$. In fact, it will be obtained by adjoining any root $x_k$ of the equation $P(x)=0$. Therefore,

$$L = \mathbb{Q}(x_1, \sqrt{13})$$

This is a normal extension of $\mathbb{Q}$ so also of $\mathbb{Q}(\sqrt{13})$. So it would be interesting to express inside $\mathbb{Q}(\sqrt{13})$ in terms of one of the roots the other roots of $P$, in a similar way we express the other root of an equation of second degree $x^2 - p x + q$ in terms of one of them ( $x' = p-x$). The Galois theory gives us these expressions that for the equation $x^3 - 10 x + 10 =0$ take form

$$x' = \frac{20}{\pm \sqrt{13}} - \frac{1}{2}\cdot ( 1 + \frac{9}{\pm \sqrt{13}})\ x - \frac{3}{\pm \sqrt{13}}\ x^2= \phi_{\pm}(x)$$

where the signs $\pm$ coordinate. We can test this numerically with the roots of $x^3 - 10 x + 10 =0$ which are approximately $x_1=-3.57709\ldots$, $x_2= 1.15347\ldots$, $x_3=2.42362\ldots$. We see that by $\phi_{+}$ produces the cycle $x_1 \to x_2\to x_3 \to x_1$, while $\phi_{-}$ the opposite cycle. Notice that with our choice we have $(x_1-x_2)(x_2-x_3)(x_3-x_1) = 10 \sqrt{13}$.