## Compute a discriminant

Let $\theta = \sqrt[4]{5}$ and let $K = \mathbb{Q}(\theta)$.

1. Compute $\Delta[1,\theta,\theta^2,\theta^3]$.
2. Let $\alpha_i = 1 + \theta^i$. Compute $\Delta[\alpha_1,\alpha_2,\alpha_3,\alpha_4]$.
3. Let $\beta_1 = 1$, $\beta_2 = \theta$, $\beta_3 = (1+\theta^2)/2$, and $\beta_4 = (\theta+\theta^3)/2$. Compute $\Delta[\beta_1,\beta_2,\beta_3,\beta_4]$.

Note that the conjugates of $\theta$ are $\pm \theta$ and $\pm i \theta$.

We may use the Vandermonde’s identity to compute $\Delta[1,\theta,\theta^2,\theta^3] = \prod_{1 \leq i < j \leq 4}(\theta_{(i)} - \theta_{(j)})^2$, where $\theta_{(1)} = \theta$, $\theta_{(2)} = - \theta$, $\theta_{(3)} = i\theta$, and $\theta_{(4)} = -i\theta$. Alternately, we may compute

$\Delta[1,\theta,\theta^2,\theta^3] = \mathsf{det} \left( \left[ \begin{array}{cccc} 1 & \theta & \theta^2 & \theta^3 \\ 1 & -\theta & \theta^2 & -\theta^3 \\ 1 & i\theta & -\theta^2 & -i\theta^3 \\ 1 & -i\theta & -\theta^2 & i\theta^3 \end{array} \right] \right)^2$.

Evidently, $\Delta[1,\theta,\theta^2,\theta^3] = -32000$. (WolframAlpha agrees.)

Now $\alpha_1 = p_1(\theta)$, where $p_1(x) = 1 + x$. So the conjugates of $\alpha_1$ are $1\pm\theta$ and $1 \pm i\theta$. Similarly, we see that the conjugates of $\alpha_2$ are $1 \pm \theta^2$, of $\alpha_3$ are $1 \pm \theta^3$ and $1 \pm i\theta^3$, and of $\alpha_4$ are all 6. So we have

$\Delta[\alpha_1,\alpha_2,\alpha_3,\alpha_4] = \left( \left[ \begin{array}{cccc} 1+\theta & 1+\theta^2 & 1 +\theta^3 & 6 \\ 1 - \theta & 1 + \theta^2 & 1 - \theta^3 & 6 \\ 1 + i\theta & 1 - \theta^2 & 1 - i\theta^3 & 6 \\ 1 - i\theta & 1 - \theta^2 & 1 + i\theta^3 & 6 \end{array} \right] \right)^2$.

Evidently, $\Delta[\alpha_1,\alpha_2,\alpha_3,\alpha_4] = -1152000$. (WolframAlpha agrees.)

Likewise, we can see that the conjugates of $\beta_1$ are all 1, of $\beta_2$ are $\pm \theta$ and $\pm i\theta$, of $\beta_3$ are $1/2 \pm 1/2 \theta^2$, and of $\beta_4$ are $\frac{1}{2}\theta + \frac{1}{2}\theta^3$, $-\frac{1}{2}\theta - \frac{1}{2} \theta^3$, $\frac{i}{2}\theta - \frac{i}{2} \theta^3$, and $-\frac{i}{2}\theta + \frac{i}{2}\theta^3$. So

$\Delta[\beta_1,\beta_2,\beta_3,\beta_4] = \left( \left[ \begin{array}{cccc} 1 & \theta & \frac{1}{2} + \frac{1}{2} \theta^2 & \frac{1}{2} + \frac{1}{2} \theta^3 \\ 1 & -\theta & \frac{1}{2} + \frac{1}{2} \theta^2 & -\frac{1}{2} - \frac{1}{2} \theta^3 \\ 1 & i\theta & \frac{1}{2} - \frac{1}{2} \theta^2 & \frac{i}{2} - \frac{i}{2} \theta^3 \\ 1 & -i\theta & \frac{1}{2} - \frac{1}{2} \theta^2 & -\frac{i}{2} + \frac{1}{2} \theta^3 \end{array} \right] \right)^2$.

Evidently, $\Delta[\beta_1,\beta_2,\beta_3,\beta_4] = -2000$. (WolframAlpha agrees.)