Why is it true that $i \sqrt{3 + \sqrt{17}} \in \mathbb Q (\sqrt{3 + \sqrt{17}}, \sqrt{-3 + \sqrt{17}})$?

Question

I am going through some notes in Galois theory and i came accross this example:

consider the polynomial $$f(X) = X^4 + 6X^2 -8$$ in $F = \mathbb Q$. We want to show that, if $E$ is the splitting field of $f$, then $E/F$ is contained in a radical extention $E'/F$. The splitting field $E$ is $F(x_1,x_2,x_3,x_4)$ where $x_i$ is a root of $f$. Lets take $$E' = \mathbb Q(\sqrt{3 + \sqrt{17}}, \sqrt{-3 + \sqrt{17}})$$ The notes say that $E \subset E'$ but it seems to me that this is false: $x = i \sqrt{3 + \sqrt{17}}$ is a root of $f$ and for it to be in $E'$ we would need a rational funcion $g \in \mathbb Q(X_1,X_2)$ such that $$g(\sqrt{3 + \sqrt{17}}, \sqrt{-3 + \sqrt{17}}) = x$$ but this cannot happen since $g \in \mathbb R (X_1,X_2)$, so when evaluated in a point in $\mathbb R^2$ we get a real number.

I am sure i missed something somewhere but i dont seem to see where.

Answer 1

You are correct that the claim is false, although the first comment indicates you misplaced a negative sign under one of your radicals. In any event, let us factor the given polynomial:

$$X^4 + 6X^2 - 8 = (X^2+3)^2 - 17 = (X^2+3-\sqrt{17})(X^2+3+\sqrt{17})$$

To handle these last two terms, we will need $X = \pm\sqrt{-3 + \sqrt{17}}$ and $X = \pm\sqrt{-3 - \sqrt{17}}$.

Notice that the latter pair of factors are not real numbers; indeed, it is not clear to me where you found the $E'$ in your original post, and I suspect that this is where the error lies.