Integral basis for $\mathbb{Q}(\theta)$ where $\theta^3 - 4 \theta + 2 =0$

Question

I am trying to find an integral basis for the field $K =\mathbb{Q}(\theta)$ where $\theta^3 - 4 \theta + 2 =0$. I suspect that $\{1, \theta, \theta^2 \}$ is a potential candidate. For $a,b,c \in \mathbb{Z}$ it is easy to see that $a + b\theta + c\theta^2 \in O_K$.

Let $\alpha = y_0 + y_1\theta + y_2\theta^2 \in O_K $ where $y_i (i =1,2,3)$ are rationals. Let $\theta_1 = \theta, \theta_2, \theta_3$ denote conjugates of $\theta$ and $\alpha, \alpha', \alpha''$ denote conjugates of $\alpha$ i.e. \begin{align} \alpha = y_0 + y_1\theta_1 + y_2\theta_1^2 \\ \alpha' = y_0 +y_1\theta_2 + y_2\theta_2^2 \\ \alpha'' = y_0 + y_1\theta_3 + y_2\theta_3^2 \end{align}

We see that $\alpha + \alpha' + \alpha'', \alpha\alpha' + \alpha\alpha'' + \alpha'\alpha''$ and $\alpha\alpha'\alpha''$ are symmetric polynomial in terms of $\theta_1, \theta_2, \theta_3$ and so they are expressible in terms of elementary symmetric polynomials which in this case are coeffcients of $(x-\theta_1)(x-\theta_2)(x-\theta_3)$. It can be easily seen that these combination of $\alpha$s must be rationals using fundamental theorem of symmetric polynomials and hence rational integers since they are in $O_K$. Indeed we have \begin{align} \alpha + \alpha' + \alpha'' = 3y_0 + 8y_2 \\ \alpha\alpha' + \alpha\alpha'' + \alpha'\alpha'' = 3y_0^2 - 4y_1^2 + 16y_2^2 + 16y_0y_2 + 6y_1y_2 \\ \alpha\alpha'\alpha'' = y_0^3 - 2y_1^3 + 4y_2^3 - 4y_0y_1^2 + 16y_0y_2^3 + 8y_0^2y_2 + 8y_1y_2^2 + 6y_0y_1y_2 \end{align}

Knowing that the above rational numbers are integers, I am not able to prove that $y_i$ are integers. How should I proceed? Are there are any other approaches?

Answer 1

Note that the discriminant of the polynomial $x^3-4x+2$ is $148=2^2 \cdot 37$. As $2$ is the only prime factor whose square divides this discriminat we get that $\mathcal{O}_K \subset \frac{1}{2} \Bbb Z[\theta]$, so assume that $\alpha=\frac{z_0}{2}+\frac{z_1}{2}\theta+\frac{z_2}{2}\theta^2 \in \mathcal O_K$, where $z_0, z_1, z_3 \in \Bbb Z$. We have to show that $z_0,z_1,z_2$ are even. For this, we can use the equations you derived with $y_i=\frac{z_i}{2}$

The first equation tells us that $3\frac{z_0}{2}+4z_2$ is an integer, which clearly implies that $z_0$ is even.

The second equation gives $\frac{3}{4}z_0^2-z_1^2+4z_2+4z_0z_2+\frac{3}{2}z_1z_2 \in \Bbb Z$
Since we already know that $z_0$ is even, this implies that either $z_1$ or $z_2$ is even.

The third equation yields $$\frac{1}{8}z_0^3-\frac{1}{4}z_1^3+\frac{1}{2}z_2^3-\frac{1}{2}z_0z_1^2+2z_0z_2^2+z_0^2z_2+z_1z_2^2+\frac{3}{4}z_0z_1z_2\in \Bbb Z$$ knowing that both $z_0$ and at least one of $z_1$ and $z_2$ is even, the interesting part of this statement is that $$-\frac{1}{4}z_1^3+\frac{1}{2}z_2^3 \in \Bbb Z$$ Now we just make a case distinction (knowing that at least one of $z_1$ and $z_2$ is even): if $z_1$ is even, then the above statement clearly implies that $z_2$ is even, too. On the other hand, suppose that $z_2$ is even, then from the statement above we get that $z_1$ must be even, too.

Thus all $z_0,z_1,z_2$ are even and thus $\alpha \in \Bbb Z[\theta]$.

Answer 2

Here’s one way of doing it:

The discriminant of this polynomial is $-148=-4\cdot37$. You know that the extension has to be ramified at (above) $2$, because the polynomial is Eisenstein for the prime $2$. You also use the fact that if your basis $\{1,\theta,\theta^2\}$ is not the integral basis, the discriminant of the correct basis will be of the form $-148/n^2$, that is, the discriminant $-148$ will be off by a square. But if you divide by $37^2$, you’ll get a noninteger. You can’t divide by $4$ either, ’cause then you’d show no ramification at two. So $-138$ it is, for the discriminant, and your guess is correct.

(Let me know in case you aren’t familiar with the fact that if $\theta$ is the root of an Eisenstein at a prime $p$, then the powers of $\theta$ are, locally at $p$, an integral basis. I can add an edit in that case.)