Let $i,\sqrt{3}\in\mathbb{C}$. I know that both are algebraic over $\mathbb{Q}$.

Hence $[\mathbb{Q}(i\sqrt{3}):\mathbb{Q}]=\deg(i\sqrt{3},\mathbb{Q})$. This is equal to 2 since $\mathrm{irr}(i\sqrt{3},\mathbb{Q})=x^2+3$.

But now I am a little bit confused how to find the irreducible polynomial such that I can compute $[\mathbb{Q}(i,\sqrt{3}):\mathbb{Q}(i\sqrt{3})]$ which is equal the degree of that polynomial.

Any help is appreciated :)


First note that $\mathbb{Q}(i,\sqrt 3) = \mathbb{Q}(i\sqrt 3, i)$.

It remains hence to calculate the degree of the simple field extension $\mathbb{Q}(i\sqrt 3, i)/\mathbb{Q}(i\sqrt 3)$.

We find that $i$ is a zero of the polynomial $X^2+1$, which is still irreducible over $\mathbb{Q}(i\sqrt 3)$. Hence the degree is 2.

  • $\begingroup$ Is $irr((i,\sqrt{3}),\mathbb{Q}(i\sqrt{3}))=X^2+1$? If we apply $\sqrt{3}$ to $X$ it does not give $0$, only true for $i$. $\endgroup$ – Chen M Ling May 19 '16 at 15:26
  • $\begingroup$ But we are looking at the field $\mathbb{Q}(i\sqrt 3, i)$ over the field $\mathbb{Q}(i\sqrt 3)$, so there is no need to apply $\sqrt 3$ to $X$. More precisely, only the irreducible polynomial of an element is defined. It makes only sense to write $\mathrm{irr}(L, K)$ if $L/K$ is a simple field extension, i.e. there is $\alpha \in L$ such that $L=K(\alpha)$. My answer solves the problem by rewriting $\mathbb{Q}(i, \sqrt 3)$ in such a way that the required extension becomes a simple extension. $\endgroup$ – Orlando Marigliano May 19 '16 at 15:37
  • $\begingroup$ Oh I see... OK. Many thanks, that cleared my confusion :) $\endgroup$ – Chen M Ling May 19 '16 at 15:55

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.