# Find the degree, the basis of the vector space, and a primitive element of the extension

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!

• No, it's not correct. An immediate indication of an error is that your basis has $4$ elements, so the degree can't be $2$. – quasi Jul 15 '17 at 1:46
• 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. – ÍgjøgnumMeg Jul 15 '17 at 1:48
• I should have seen that obvious mistake, thanks! I used that the degrees are multiplicative and got the degree of 4. Thanks! – obewanjacobi Jul 15 '17 at 18:28

## 1 Answer

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.

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