Show that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})$ is a splitting field for $f(x)$.

abstract-algebrafield-theory

a) Show that $f(x)=x^6-2x^3-10$ is irreducible over $\mathbb{Q}$.

Proof: Using Eisenstein's Criterion, with $p=2$, we show $f(x)$ is irreducible over $\mathbb{Q}$.

b) Let $\alpha=\sqrt[3]{1+\sqrt{11}}$. Show that $|\mathbb{Q}(\alpha):\mathbb{Q}|=6$.

Proof: Note that $\alpha$ is a root of $f(x)$. By part (a), we can conclude that $f(x)$ is the minimal polynomial of $\mathbb{Q}(\alpha)$. Furthermore $|\mathbb{Q}(\alpha):\mathbb{Q}|=6$.

c) Show that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})$ is a splitting field for $f(x)$.

Proof: Note that $\beta=\sqrt[3]{1-\sqrt{11}}$ is also a root of $f(x)$. [Check by plugging it in $f(x)$.]
$$ \alpha \cdot \beta = \sqrt[3]{(1-\sqrt{11})(1-\sqrt{11})} = \sqrt[3]{-10} $$

d) Show that $\sqrt[3]{10}\not\in \mathbb{Q}(\sqrt{-3},\sqrt{11})$ and conclude that $|L:\mathbb{Q}|=12$, where $L=\mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})$. Moreover $K=L(\alpha)$ and thus $|K:\mathbb{Q}|=12$ or $36$.

  • How do I show $\sqrt[3]{10}\not\in \mathbb{Q}(\sqrt{-3},\sqrt{11})$?
  • Since the corresponding minimal polynomials of $\sqrt{-3},\sqrt{11},\sqrt[3]{10}$ have degree $2,2,3$, respectively, and share no roots, $|L:\mathbb{Q}|=12$.
  • $L(\alpha)= \mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})(\alpha)$. Does $\alpha$ get consumed by $\sqrt[3]{10}$ from what we shown in part (c) and the first part in (d)? Is this how we get $K=L(\alpha)$?
  • How can $|K:\mathbb{Q}|=36$?

Best Answer

Answers for (a) and (b) are in the post.

Proof of (c): Note that $\beta=\sqrt[3]{1-\sqrt{11}}$ is also a root of $f(x)$. [Check by plugging it in $f(x)$.]

$$\alpha\cdot\beta= \sqrt[3]{(1+\sqrt{11})(1-\sqrt{11})}=\sqrt[3]{-10}=-\sqrt[3]{10}$$ Since $\beta\in\mathbb{Q}(\alpha)$, $-\alpha\beta\in\mathbb{Q}(\alpha)$. Hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\sqrt[3]{10})$.

Recall the idea of roots of unity. The roots of $f(x)$ are $x=\alpha, \omega\alpha, \omega^2\alpha$ where $\omega=\dfrac{-1+i\sqrt{3}}{2}$. It follows that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})=\mathbb{Q}(\alpha,\sqrt[3]{10})$. Furthermore $K$ is a splitting field for $f(x)$.

Proof of (d): Since $|\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}|=3$, and $|\mathbb{Q}(\sqrt{-3},\sqrt{11}):\mathbb{Q}|=4$, $\sqrt[3]{10}\not\in\mathbb{Q}(\sqrt{-3},\sqrt{11})$. (why??) I know we said Tower Law, but I think I need a review of it.

Let $L=\mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})$. By Tower Law, $$ |L:\mathbb{Q}|=|L:\mathbb{Q}(\sqrt{-3},\sqrt{11}):\mathbb{Q}(\sqrt[3]{10})|\cdot |\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}| = 4\cdot 3 =12$$

Let $K=L(\alpha)$, if $\alpha$ is real, then just like the last part, we have $$|K:\mathbb{Q}|=12$$ Recall the $\alpha$ is a root of unity that, then we have 2 more roots to account more, so $$|K:\mathbb{Q}|=12\cdot 3 = 36.$$

Comments on the proofs will be greatly appreciated.

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