Number field question Show that $\sqrt3$ $\notin$ $\Bbb Q[\sqrt[4]2]$
The problem wishes us to use the trace function of a number field, and hints to write $\sqrt 3$ as $a+bx+bx^2+cx^3$.
Why can we write it like this, and where do we go from there?
 A: Note that the matrix of the multiplication $m_\alpha:\mathbb{Q}(\sqrt[4]{2})\to \mathbb{Q}(\sqrt[4]{2})$ map with respect to a generic element $\alpha=a+b\sqrt[4]{2}+c\sqrt[4]{2^2}+d\sqrt[4]{2^3}$ is (in the canonical basis)
$$\begin{pmatrix}a & 2d & 2b & 2b \\ b & a & 2d & 2c \\ c & b & a & 2d \\ d & c & b & a \end{pmatrix}$$
so, the trace of the matrix is $4a$. Now, note that if $\sqrt{3}\in\mathbb{Q}(\sqrt[4]{2})$ then 
$$\text{Tr}_{\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}}(\sqrt{3})=\text{Tr}_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}\left(\text{Tr}_{\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{3})}(\sqrt{3})\right)=\text{Tr}_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}(2\sqrt{3})=0$$
(this is just using the transitivity of the field trace--a similar result can be arrived at calculating the minimal polynomial of $\sqrt{3}$). So, $a=0$. Thus, by dividing both sides by $\sqrt[4]{2}$ we get 
$$\sqrt[4]{\frac{9}{2}}=b+c\sqrt[4]{2^2}+d\sqrt[4]{2^3}$$
Taking the minimal polynomial of the left hand side shows that $\text{Tr}_{\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}})\left(\sqrt[4]{\frac{9}{2}}\right)=0$. But, the trace of the right hand side is $b$. Continuing this process you find that $a=b=c=d=0$, which is clearly impossible.
EDIT: Just for funsies, here is another number theoretic way to approach this problem. This is actually what initially occurred to me--not the trace trick. I hope you find it enlightening.
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Derivation of the ring of integers of $\mathbb{Q}(\sqrt[4]{2})$--skip if you're willing to believe it's $\mathbb{Z}[\sqrt[4]{2}]$.

Let $K=\mathbb{Q}(\sqrt[4]{2})$. It's obvious that $\mathbb{Z}[\sqrt[4]{2}]\subseteq\mathcal{O}_K$, and so to show that $\mathcal{O}_K=\mathbb{Z}[\sqrt[4]{2}]$ it suffices to show that $\mathbb{Z}[\sqrt[4]{2}]$ is integrally closed, or equivalently, that its locally integrally closed. 
But, since $\mathbb{Z}[\sqrt[4]{2}]$ is a simple integral extension of $\mathbb{Z}$, with generating having minimal polynomial $f(x)=x^4-2$, it suffices to check that its integrally closed only at maximal ideals containing $f'(\sqrt[4]{2})=4 \sqrt[4]{2^3}$. In particular, we see that if $f'(\sqrt[4]{2})\in M$, then $8=\sqrt[4]{2}f'(\sqrt[4]{2})\in M$, so that $M$ lies over $2$. But, since $x^4-2\equiv x^4 \mod 2$, the only max ideal of $\mathbb{Z}[\sqrt[4]{2}]$ lying over $2$ is $M=(2,\sqrt[4]{2})$. But, since this max ideal is already principal $M=(\sqrt[4]{2})$ it's trivial that the only maximal ideal, and thus the only prime ideal (since $\mathbb{Z}[\sqrt[4]{2}]$ is dimension $1$, and thus so is $\mathbb{Z}[\sqrt[4]{2}]_M$) of $\mathbb{Z}[\sqrt[4]{2}]_M$ is principal. Thus, $\mathbb{Z}[\sqrt[4]{2}]_M$ is a PID, and so, in particular, integrally closed. It follows from previous discussion that $\mathbb{Z}[\sqrt[4]{2}]$ is integrally closed, and thus $\mathcal{O}_K=\mathbb{Z}[\sqrt[4]{2}]$. 

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Now, note that we can use the Dedekind-Kummer theorem to factor $3\mathcal{O}_K$. Namely, since $(x^2+x+2)(x^2+2x+2)$ is the factorization into irreducibles of $x^4-2$ in $\mathbb{F}_3[x]$ we see that $(3,\sqrt[4]{2^2}+\sqrt[4]{2}+2)(3,\sqrt[4]{2^2}+2\sqrt[4]{2}+2)$ is the factorization of $3\mathcal{O}_K$ into primes. In particular, $3$ doesn't ramify in $K$ (this could have also been determined by finding $d_K$, but I thought it would be nice to have the actual factorization).
But, since $3$ ramifies in $\mathbb{Q}(\sqrt{3})$ we must have that $\mathbb{Q}(\sqrt{3})\not\subseteq\mathbb{Q}(\sqrt[4]{2})$ as desired.
The above is all kind of misleading. It makes it seem like the technique involving traces is much more natural than the above, but this is only because some more sophisticated machinery went into it--and that I fully fleshed it out. The idea is simpler, and is the one I would imagine would immediately occur to most people. Namely, it's simple--$3$ ramifies in $\mathbb{Q}(\sqrt{3})$ but not in $\mathbb{Q}(\sqrt[4]{2})$, so we can't possibly have $\sqrt{3}\in\mathbb{Q}(\sqrt[4]{2})$. 
A: But more simply, if $\sqrt3$ is an element of $Q[\sqrt[4]2]$, then $Q[\sqrt2,\sqrt3]$ is a subfield of $Q[\sqrt[4]2]$. However, since both are order 4, they must be isometric. But the first is a splitting field and the second is not.
