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Hey I have this exercise where I have problems understanding the solutions.

Is the field $L = \mathbb{Q}( \sqrt{2 + \sqrt{2}})$ normal over $K=\mathbb{Q}$?

What we have done is: $a=\sqrt{2 + \sqrt{2}}$ is a root of the polynomial $f(x)=(x^2-2)^2-2=x^4-4x^2+2 \in \mathbb{Q}[x]$ With Eisenstein we see that is irreducible and therefore $[\mathbb{Q}(a):\mathbb{Q}]=4$

The polynomial $f(x)$ has following roots: $a,-a,\sqrt{2-\sqrt{2}},-\sqrt{2-\sqrt{2}}$ so we can write $f(x)$ as linear factors: $f(x)=(x-a)(x+a)(x-\sqrt{2-\sqrt{2}})(x+\sqrt{2-\sqrt{2}})$.

Each of this roots are in $L$ and therefore is the field normal over $\mathbb{Q}$

Now, what I am not nunderstanding is how we can say that $\sqrt{2-\sqrt{2}}$ is in $L$.

Marco Di Giacomo
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1 Answers1

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Consider the product $$ \sqrt{2+\sqrt2}\sqrt{2-\sqrt2} = \sqrt{2^2-\sqrt2^2} = \sqrt2 $$ Clearly this is an element of $L$ (it's equal to $a^2 - 2$). And since $\sqrt{2+\sqrt2}\in L$ as well, then $$ \sqrt{2-\sqrt2} = \frac{\sqrt2}{\sqrt{2+\sqrt2}} $$ must also be in $L$

Arthur
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