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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 :)

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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.

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  • $\begingroup$ 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

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