Arriving at a Contradiction with the Tower Law I keep arriving at a contradiction here, and I was hoping someone could point out my logic error, because it is pushing me to insanity....
Let $Q$ denote the rationals.
Consider $\alpha = \sqrt[4]{5}$.  Then $\gamma := \alpha + i\alpha$ is a root of the irreducible $f(x) = x^4 + 20$.  
Let $K = Q(\gamma)$, and we see that $[K:Q] = 4$.
Then we immediately have $K\subset Q(i,\alpha)$.
Since $\alpha + i\alpha\in K$, we obtain the following consequences:
$\begin{eqnarray*}
i\alpha &=& \frac{(\alpha + i\alpha) - (\alpha - i\alpha)}{2}&\in& K\text{(this is the problem, because }\alpha - i\alpha\notin K)\\
\alpha &=& (\alpha + i\alpha) - i\alpha &\in& K\\
i &=& \frac{i\alpha}{\alpha} &\in& K
\end{eqnarray*}$
Therefore $K$ contains $Q(i,\alpha)$ and thus we have equality.
That is, $Q(\alpha + i\alpha) = Q(i,\alpha)$.
But $\alpha$ is a root of the irreducible $x^4 - 5 = 0$, so $[Q(\alpha):Q] = 4$ and $i\notin Q(\alpha)$.  Therefore $[Q(\alpha,i):Q(\alpha)] = 2$.
Then by the Tower law, $[K:Q] = [Q(i,\alpha):Q] = [Q(i,\alpha):Q(\alpha)][Q(\alpha):Q] = 8$.
So $[K:Q]$ obviously cannot be two different values.  So which calculation is wrong?

$\alpha = \sqrt[4]{5}$ is a root of the irreducible $x^4 - 5 = 0$, so $[Q(\alpha):Q] = 4$.  Since  $Q(\alpha)$  is a real field,  $i\notin Q(\alpha)$  so  $[Q(\alpha,i):Q(\alpha)] = 2$ and  hence $[Q(i,\alpha):Q] = [Q(i,\alpha):Q(\alpha)][Q(\alpha):Q] = 8$.
Let $\gamma= \alpha + i\alpha=\alpha(1+i)$ is a root of the irreducible $f(x) = x^4 + 20$.  Let $K = Q(\gamma)$, thus $[K:Q] = 4$. The roots of $f$, $\pm \alpha (1\pm i)$, are elements of  $Q(i,\alpha)$. 
The field $L=Q(\gamma, \bar{\gamma})=K(i)$ has degree 2 over $K$ so it has dimension 8 over $Q$; $L$ is contained in $Q(\alpha,i)$ and has the same dimension so they are equal.
Hence the splitting field of $f$ is $Q(\gamma, \bar{\gamma})=Q(\alpha,i)$.
