6
$\begingroup$

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?

$\endgroup$
  • 9
    $\begingroup$ 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. $\endgroup$ – Git Gud Feb 7 '15 at 17:57
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.