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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?

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Why is $\alpha-i\alpha\in K$? –  Andrew Dec 3 '12 at 1:30
1  
Thanks, problem solved. Working with too many fields at once causes confusion with me apparently. –  Kyle Schlitt Dec 3 '12 at 1:32
    
Perhaps the best thing to do now is to post what you have found as an answer. Then, later, you can accept it. –  Gerry Myerson Dec 3 '12 at 6:29

1 Answer 1

$\alpha = \sqrt[4]{5}$ 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 $Q(\gamma, \bar{\gamma})=Q(\alpha,i)$.

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