Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the four roots but I am unsure how to proceed next, and how to show the degree of the extension over Q is 8.

share|improve this question
2  
Do you know what "splitting field" means? –  Chris Eagle Oct 23 '12 at 20:12
    
it's the smallest field extension in which the polynomial splits, right? I have the 4 roots (via factoring), but not sure what to do next –  Mark S Oct 23 '12 at 20:32
    
Hint: $(x^4+13)(x^4-13) = x^8 -169$. So a root of your polynomial is also... –  Derek Allums Oct 23 '12 at 20:40

1 Answer 1

By looking at the roots $$\sqrt[4]{-13},i\sqrt[4]{-13},-\sqrt[4]{-13},-i\sqrt[4]{-13}$$ it is clear that the splitting field is $$\mathbb{Q}(i,\sqrt[4]{-13})$$ that is since both $i\sqrt[4]{-13},\sqrt[4]{-13}$ is in the splitting field hence $\frac{i\sqrt[4]{-13}}{\sqrt[4]{-13}}=i$ is.

Now we need to find $[\mathbb{Q}(i,\sqrt[4]{-13}):\mathbb{Q}]$.

Argue that $i\not\in\mathbb{Q}(\sqrt[4]{-13})$ by using that fact that $\sqrt{13}\not\in\mathbb{Q}$, deduce that the degree is $8$ by noting $[\mathbb{Q}(\sqrt[4]{-13}):\mathbb{Q}]=4$ since it is a simple extension and $x^{4}+13\in\mathbb{Q}[x]$ is irreducible

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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