# Show that $\sqrt{7}\notin\mathbb{Q}(\alpha)$, with $\alpha$ a root of $X^7+6X^3+3X+15$

Let $f=X^7+6X^3+3X+15\in\mathbb{Q}[x]$, and $\alpha\in\mathbb{C}$ such that $f(\alpha)=0$. I want to show $\sqrt{7}\notin\mathbb{Q}(\alpha)$

First of all, by Eisenstein's criterion, $f$ is irreducible in $Q[x]$, so we can say that $\deg (\text{Irr}(\alpha,\mathbb{Q}))=7=\left[\mathbb{Q}(\alpha):\mathbb{Q}\right]$.

Then, $\left\{1,\alpha,\ldots,\alpha^6\right\}$ is a $\mathbb{Q}$-basis of $\mathbb{Q}(\alpha)$. Then, $\sqrt{7}\in\mathbb{Q}(\alpha)\Longleftrightarrow \sqrt{7}=\lambda_0 1+\lambda_1\alpha+\cdots+\lambda_6\alpha^6$ with $\lambda_i\in\mathbb{Q}$. But I get stuck here. How should I proceed?

• If $\sqrt 7\in \mathbb Q(\alpha)$, then $\mathbb Q\subseteq \mathbb Q(\sqrt 7)\subseteq \mathbb Q(\alpha)$. Think about the degrees of the extensions. – Git Gud Feb 7 '15 at 17:57

We take it as given, from user203327's work as expressed in the body of this question, that $[\Bbb Q(\alpha): \Bbb Q] = 7$.
Now suppose $\sqrt{7} \in \Bbb Q(\alpha)$; then we would have the tower of fields $\Bbb Q \subset \Bbb Q(\sqrt{7}) \subset \Bbb Q(\alpha)$. We note that the quadratic polynomial $X^2 - 7$ is irreducible over $\Bbb Q$, since $\sqrt{7} \notin \Bbb Q$; thus $[\Bbb Q(\sqrt{7}): \Bbb Q] = \deg(X^2 - 7) = 2$.
Since $\Bbb Q \subset \Bbb Q(\sqrt{7}) \subset \Bbb Q(\alpha)$, we have $[\Bbb Q(\alpha): \Bbb Q] = [\Bbb Q(\alpha): \Bbb Q(\sqrt{7})][\Bbb Q(\sqrt{7}): \Bbb Q]$; substituting in for $[\Bbb Q(\alpha): \Bbb Q]$ and $[\Bbb Q(\sqrt{7}): \Bbb Q]$ yields the equation $2[\Bbb Q(\alpha): \Bbb Q(\sqrt{7})] = 7$. But $2 \not \mid 7$; this contradiction shows that $\sqrt{7} \notin \Bbb Q(\alpha)$. QED.