Brian Bi
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Section 15.4. Finding the Irreducible Polynomial

Exercise 15.4.1 We know that $$\{1, \alpha, \alpha^2\}$$ forms a $$\Q$$-basis of $$\Q(\alpha)$$, and we know that $$\alpha^3 = \alpha + 1$$. Therefore \begin{align*} \gamma^2 &= 1 + 2\alpha^2 + \alpha^4 \\ &= 1 + 2\alpha^2 + \alpha(\alpha + 1) \\ &= 1 + \alpha + 3\alpha^2 \end{align*} and if $$p\gamma^2 + q\gamma + r = 0$$ with $$p, q, r \in \Q$$ then we have $$0 = p(1 + \alpha + 3\alpha^2) + q(1 + \alpha^2) + r = (p + q + r) + p\alpha + (3p + q)\alpha^2$$ implying $$p + q + r = p = 3p + q = 0$$ so $$p = q = r = 0$$. So the minimal polynomial has degree at least 3. Now \begin{align*} \gamma^3 &= (1 + \alpha^2)(1 + \alpha + 3\alpha^2) \\ &= 1 + \alpha + 3\alpha^2 + \alpha^2 + \alpha^3 + 3\alpha^4 \\ &= 1 + \alpha + 4\alpha^2 + \alpha + 1 + 3\alpha^2 + 3\alpha \\ &= 2 + 5\alpha + 7\alpha^2 \end{align*} and by inspection \begin{equation*} \gamma^3 - 5\gamma^2 = -8\alpha^2 - 3 = -8\gamma + 5 \end{equation*} so $$\gamma^3 - 5\gamma^2 + 8\gamma - 5 = 0$$ and the minimal polynomial is $$x^3 - 5x^2 + 8x - 5$$.

Exercise 15.4.2

1. By a similar argument to that used in Exercise 11.3.3(d) we find that if $$f(\sqrt{3} + \sqrt{5}) = 0$$ and $$f \in \Q[x]$$ then we must have $$f(\pm \sqrt{3} \pm \sqrt{5}) = 0$$ so that $$f$$ has degree at least 4. In fact $$\prod (x \pm \sqrt{3} \pm \sqrt{5}) = x^4 - 16x^2 + 4 \in \Q[x]$$ so this is the minimal polynomial.

2. Since $$\sqrt{5}$$ generates a quadratic extension of $$\Q$$ while $$\alpha$$ generates an extension of degree 4, it is not possible that $$\alpha \in \Q(\sqrt{5})$$, so the minimal polynomial has degree at least 2. Now $$\alpha^2 = 8 + 2\sqrt{15}$$ and $$\sqrt{5}\alpha = \sqrt{15} + 5$$ so we have $$\alpha^2 - 2\sqrt{5}\alpha = -2$$ so the minimal polynomial is $$x^2 - 2\sqrt{5}x + 2$$.

3. We first observe that $$\Q(\sqrt{3}, \sqrt{5})$$ is a field extension of $$\Q$$ of degree 4 that contains $$\Q(\sqrt{3} + \sqrt{5})$$, so the two fields are equal.

Now we'll show that $$\sqrt{10} \notin \Q(\sqrt{3}, \sqrt{5})$$. Suppose for the sake of contradiction that $$\sqrt{10} = a + b\sqrt{5}$$ where $$a, b \in \Q(\sqrt{3})$$. Then squaring both sides gives $$10 = a^2 + 5b^2 + 2ab\sqrt{5}$$, and since $$\{1, \sqrt{5}\}$$ is a $$\Q(\sqrt{3})$$-basis of $$\Q(\sqrt{3}, \sqrt{5})$$, this implies $$2ab = 0$$. So either $$a = 0$$ or $$b = 0$$. If $$a = 0$$, then $$b\sqrt{5} = \sqrt{10}$$ so $$b = \sqrt{2}$$, but $$\sqrt{2} \notin \Q(\sqrt{3})$$. If on the other hand $$b = 0$$, then $$\sqrt{10} = a$$, but $$\sqrt{10} \notin \Q(\sqrt{3})$$ so this is not possible either. So in fact our assumption that $$\sqrt{10} \in \Q(\sqrt{3}, \sqrt{5})$$ must have been wrong.

This implies that $$\Q(\alpha, \sqrt{10})$$ is a quadratic extension of $$\Q(\alpha)$$ so it has degree 8 over $$\Q$$. Then $$[\Q(\alpha, \sqrt{10}) : \Q(\sqrt{10})] = [\Q(\alpha, \sqrt{10}) : \Q] / [\Q(\sqrt{10}) : \Q] = 8/2 = 4$$. So the minimal polynomial of $$\alpha$$ over $$\Q(\sqrt{10})$$ is the same as over $$\Q$$.

4. By the same argument as in part (b), the minimal polynomial has degree at least 2 and it is easy to see that it is simply $$x^2 - (8 + 2\sqrt{15})$$.

Exercise 15.4.3 The basis given in Example 15.4.4(b) can also be written in the form $$\{1, \sqrt[3]{2}, \sqrt[3]{4}, \omega, \omega\sqrt[3]{2}, \omega\sqrt[3]{4}\}$$ where $$\omega = e^{2\pi i/3}$$. Using this it is easy to compute \begin{align*} \alpha_1 + \alpha_2 &= \sqrt[3]{2} + \omega\sqrt[3]{2} \\ (\alpha_1 + \alpha_2)^2 &= \omega\sqrt[3]{4} \\ (\alpha_1 + \alpha_2)^3 &= -2 \end{align*} and to see that $$\{1, \alpha_1 + \alpha_2, (\alpha_1 + \alpha_2)^2\}$$ is linearly independent over $$\Q$$ but $$(\alpha_1 + \alpha_2)^3 + 2 = 0$$, so the minimal polynomial is $$x^3 + 2$$.