Algebraic Number Theory,Marcus, Chapter 2, Question 16 In question 16 of chapter 2 in Marcus Book, I have to show that $\sqrt{3}\not\in\mathbb{Q}(\alpha)$,where $\alpha=\sqrt[4]{2}$ using the trace idea.
the proof starts by assuming that $\sqrt{3}=a+b\alpha+c\alpha^2+d\alpha^3$ (I know the basis is $\{1,\alpha,\alpha^2,\alpha^3\}$).
I found the trace of $x=A+B\alpha+C\alpha^2+D\alpha^3$ and it equals $4A$. so $Tr(\sqrt{3})=4a,Tr(\alpha)=Tr(\alpha^2)=Tr(\alpha^3)=0$
The question asks about what is the trace of $\frac{\sqrt{3}}{\alpha}$. 
I don't know how to calculate it with $\alpha$ in denominator.
if I divide $\sqrt{3}$ by $\alpha$, then how to find $Tr(\frac{a}{\alpha})$?
I am wondering what is the trace of $\sqrt{3}/\alpha,\sqrt{3}/\alpha^2,\sqrt{3}/\alpha^3$
According to Marcus, this is will leads to a contradiction. 
 A: Recall that for $K/\mathbb Q$, the trace is defined as $$\text{Tr}(x) = \sum_{\sigma} \sigma(x),$$
where $\sigma$ runs over all embeddings of $K$ into $\mathbb C$.
You have already shown that $\text{Tr}\left(\frac{\sqrt 3}{\alpha}\right) = 4b$. Let me calculate it another way. There are four embeddings of $\mathbb Q(\alpha)$ into $\mathbb C$ given by whether $\alpha$ goes to $\alpha, i\alpha, -\alpha$ or $-i\alpha$. More succintly the group of embeddings is generated by $\sigma$ which takes $\alpha\mapsto i\alpha$. Each of these embeddings must take $\sqrt 3$ to either itself or $-\sqrt 3$. Either way $\sigma^2(\sqrt 3)=\sqrt 3$ and $\sigma^3(\sqrt 3)=\sigma(\sqrt 3)$. Now,
$$\text{Tr}\left(\frac{\sqrt 3}{\alpha}\right) = \frac{\sqrt 3}{\alpha}+\frac{\sigma(\sqrt 3)}{i\alpha}+\frac{\sqrt 3}{-\alpha}+\frac{\sigma(\sqrt 3)}{-i\alpha}=0.$$
So, $b=0$. 
A similar song and dance with $\frac{\sqrt 3}{\alpha ^3}$ shows that $\text{Tr}\left(\frac{\sqrt 3}{\alpha ^2}\right) = 4d$ and $$\text{Tr}\left(\frac{\sqrt 3}{\alpha^3}\right) = \frac{\sqrt 3}{\alpha^3} + \frac{\sigma(\sqrt 3)}{-i\alpha^3} + \frac{\sqrt 3}{-\alpha^3} + \frac{\sigma(\sqrt 3)}{i\alpha^3} = 0.$$
So $d = 0$.
Thus $\sqrt 3 = a + c \alpha^2 = a + c \sqrt 2$. This is a laughable equation as is seen by squaring it to get $2ac \sqrt 2 = 0$. So either $a$ is zero in which case we are claiming that $\sqrt 6$ is rational or $c = 0$ in which case $\sqrt 3$ is rational. Both are ridiculous so $\sqrt 3 \notin \mathbb Q(\alpha)$.
