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I am solving an ex.: Find basis of $\mathbb{Q}(\sqrt{1+\sqrt3})$ and degree of $\mathbb{Q}(\sqrt{1+\sqrt3}):\mathbb{Q}$. SO first of all I have showed that $1+\sqrt{3}$ is not a square in $\mathbb{Q}(\sqrt3)$. Also I think that degree in this exercise we find with Short Tower Law. So, degree of $\mathbb{Q}(\sqrt3):\mathbb{Q}$ is $2$. How to show that degree of $\mathbb{Q}(\sqrt{1+\sqrt3}):\mathbb{Q}(\sqrt3)$ is also $2$? And what is the basis of $\mathbb{Q}(\sqrt{1+\sqrt3})$ over $\mathbb{Q}$?

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    $\begingroup$ Let $K$ be a field. $\alpha$ is a root of some irreducible polynomial of degree $n$ with coefficients in $K$ if and only if $\{1, \alpha, \dots, \alpha^{n-1}\}$ is a $K$-basis for $K(\alpha)$. If you understand this result, then you will be able to answer both of your questions. $\endgroup$ – Gunnar Sveinsson Mar 25 at 8:30
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Since $1+\sqrt{3}$ is not a square in $\mathbb{Q}(\sqrt{3})$, it follows that $$ [\mathbb{Q}(\sqrt{1+\sqrt{3}}):\mathbb{Q}(\sqrt{3})]=2 $$ because clearly $\sqrt{1+\sqrt{3}}$ is a root of $x^2-(1+\sqrt{3})\in\mathbb{Q}(\sqrt{3})[x]$.

Now, by general theory, you know that a basis of $\mathbb{Q}(\sqrt{1+\sqrt{3}})$ over $\mathbb{Q}$ is given by $$ 1,\quad\sqrt{3},\quad\sqrt{1+\sqrt{3}},\quad\sqrt{3}\sqrt{1+\sqrt{3}} $$

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$$\alpha =\sqrt{1+\sqrt 3}\implies \alpha ^2=1+\sqrt 3\implies \alpha ^4-2\alpha ^2-2=0,$$ this polynomial is irreducible by Eisenstein criterion. Therefore, $$[\mathbb Q(\alpha ):\mathbb Q]=4.$$

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