1
$\begingroup$

The question reads: "Find the degree $[F: \mathbb{Q}]$ of the extension $F = \mathbb{Q} (i, \sqrt{3})$ over $\mathbb{Q}$. Find a basis of the vector space $F$ over $\mathbb{Q}$. Then find a primitive element of the extension."

So far, this is my answer:

Note this extension adds the roots of the polynomial $f(x) = x^2 + 3$, which is irreducible in $\mathbb{Q}$. By adding the roots in $F$, we can now reduce the polynomial $f(X)$ in $F$, and thus we can easily see that the polynomial $f(x)$ is the minimal polynomial for the extension $F/ \mathbb{Q}$. Thus the degree of the extension is $2$. A basis for this extension is $[1, i, \sqrt{3} , i \sqrt{3}]$.

I'm not sure on the correctness of my answer, and would like to double check it, as well as get help in finding the primitive element. Thanks for any help!

$\endgroup$
  • $\begingroup$ No, it's not correct. An immediate indication of an error is that your basis has $4$ elements, so the degree can't be $2$. $\endgroup$ – quasi Jul 15 '17 at 1:46
  • 1
    $\begingroup$ I would think about intermediate extensions as well. That is, think of the extension $\Bbb Q(\sqrt{3})/\Bbb Q$ and $\Bbb Q(i, \sqrt{3})/\Bbb Q(\sqrt{3})$. Then use that degrees of extensions are multiplicative. $\endgroup$ – ÍgjøgnumMeg Jul 15 '17 at 1:48
  • $\begingroup$ I should have seen that obvious mistake, thanks! I used that the degrees are multiplicative and got the degree of 4. Thanks! $\endgroup$ – obewanjacobi Jul 15 '17 at 18:28
2
$\begingroup$

The degree of a field extension is precisely the dimension over the base field. You are saying that the degree is 2 and the dimension is 4. So there is a mistake there. Namely, the degree is actually 4. As for a primitive element, the natural guess should be $i+\sqrt{3}$, so thats where you should start. One containment is obvious so you need to check only one containment.

$\endgroup$
  • $\begingroup$ Got it! You were correct in assuming that the primitive element would be $i + \sqrt{3}$, it just took some containment arguments. Thanks again! $\endgroup$ – obewanjacobi Jul 15 '17 at 19:21

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.