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.)

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