*Algebra*

## 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

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.

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 \(x^2 - 2\sqrt{5}x + 2\).

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 \(\Q\).

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 \(x^3 + 2\).