Showing $ \mathbb{Q}[\sqrt[4]{2}]$ is a field Here i want to prove
\begin{align}
  \mathbb{Q}[\sqrt[4]{2}] = \{ a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8} \mid a,b,c,d\in\mathbb{Q}\} 
\end{align}
this is a field. To do that i want to prove
\begin{align}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} \in \mathbb{Q}[\sqrt[4]{2}]
\end{align}
Is there any ideas for showing this?

[edit] Many of you prove $\mathbb{Q}[\sqrt[4]{2}]$ is a field in indirect way. Is there any ways for showing this in direct way? For $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt[3]{2}]$, using some formulas i can check that indeed every nonzero elementn is unit. But this case does not seem that easy...
 A: Idea: use that $\sqrt[4]{2} = \sqrt{ \sqrt{2}}$, so first rationalize to a denominator in $\mathbb{Q}(\sqrt{2})$, and then finish off.
$$\frac{1}{a+b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}}= \frac{1}{a + c \sqrt{2} + (b + d \sqrt{2}) \sqrt[4]{2}}= \\
=\frac{a + c \sqrt{2} - (b + d \sqrt{2}) \sqrt[4]{2}}{(a + c \sqrt{2})^2 -\sqrt{2}(b + d \sqrt{2})^2}$$ and rationalize by amplifying the fraction by $(a - c \sqrt{2})^2 +\sqrt{2}(b - d \sqrt{2})^2$
A: This is pretty standard and proved in most abstract algebra textbooks. Let $\alpha=\sqrt[4]{2}$ then $\alpha$ is root of $f(x) = x^{4}-2$ and this polynomial is irreducible over $\mathbb{Q} $. The expression $$\frac{1}{a+b\sqrt[4]{2}+c\sqrt[4]{4}+d\sqrt[4]{8}}$$ can be written as $1/g(\alpha)$ where $g(x) = a+bx+cx^{2}+dx^{3}$ is a polynomial of degree $3$ or less with rational coefficients. Since degree of $g$ is less than that of $f$ and $f$ is irreducible, it follows that the polynomials $f, g$ are relatively prime to each other.  In other words their GCD is $1$. Therefore there exist polynomials $a(x), b(x) $ with rational coefficients such that $a(x)f(x) +b(x) g(x) =1$. Putting $x=\alpha$ and noting that $f(\alpha) =0$, we see that $b(\alpha) g(\alpha) =1$ or $1/g(\alpha)=b(\alpha) \in\mathbb {Q} [\alpha ] $.

The argument above applies for any field $F$ and any element $\alpha\in K\supseteq F$ which is algebraic over $F$. Under these conditions $F[\alpha] =F(\alpha) $. In fact more is true: $\alpha$ is algebraic over $F$ if and only if $F(\alpha) =F[\alpha] $. The proof given above is based on the fact that $F[x] $ is an Euclidean domain with all the nice properties of GCD.

It is precisely this particular proof which left a deep impression of abstract algebra on me. Before this I used to think that abstract algeba consisted of boring axioms and playing around with them and that the classical theory of equations was the real algebra. For me this was the first non-trivial result where the classical manipulation techniques failed and GCD came to the rescue.
A: Through the magic of upper triangulation and backsubstitution...
\begin{align*}
      \frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} &= e + f \sqrt[4]{2} + g \sqrt[4]{4} + h \sqrt[4]{8}  \\
\iff  1 &= (a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8})(e + f \sqrt[4]{2} + g \sqrt[4]{4} + h \sqrt[4]{8})  \\
        &= (ae + 2bh + 2cg + 2df) \\ &\qquad + (af + be + 2ch + 2dg)\sqrt[4]{2} \\ &\qquad + ( ag + bf + ce + 2dh )\sqrt[4]{4} \\ &\qquad + ( ah + bg + cf + de )\sqrt[4]{8}  \\
\iff \{ 1 &= ae + 2bh + 2cg + 2df,  \\
        0 &= af + be + 2ch + 2dg,  \\
        0 &= ag + bf + ce + 2dh,  \\
        0 &= ah + bg + cf + de \}  \\
\implies \{ h &= \frac{-bg - \dots}{a}, \\
            g &= \frac{-bf- \dots}{a}, \\
            f &= \frac{-be- \dots}{a}, \\
            e &= \frac{a^3-2 a \left(2 b d+c^2\right)+2 c \left(b^2+2
   d^2\right)}{a^4-8 b d \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2 \left(2 a
   c+b^2\right)+8 a b^2 c-2 b^4+4 c^4-8 d^4} \}  \\
\implies f &= \frac{-b \left(a^2+2 c^2\right)+2 d \left(2 a
   c+b^2\right)-4 d^3}{a^4-8 b d \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2
   \left(2 a c+b^2\right)+8 a b^2 c-2 b^4+4 c^4-8 d^4}  \\
\implies g &= \frac{-a^2 c+a b^2+2 a d^2-4 b c d+2 c^3}{a^4-8 b d
   \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2 \left(2 a c+b^2\right)+8 a b^2 c-2
   b^4+4 c^4-8 d^4}  \\
\implies h &= \frac{-d \left(a^2+2 c^2\right)+2 b \left(a
   c+d^2\right)-b^3}{a^4-8 b d \left(a^2+2 c^2\right)-4 a^2 c^2+8 d^2
   \left(2 a c+b^2\right)+8 a b^2 c-2 b^4+4 c^4-8 d^4}
\end{align*}
Maybe this isn't the best way to go...
A: So take any element of $Q[\sqrt[4]{2}]$. It will be algebraic over $Q$. You write down the algebraic equation. Say it looks like $x^n+a_{n-1}x^{n-1}+\dots+a_0=0$ with $a_0\in Q$. Define $g(x)=x^{n-1}+a_{n-1}x^{n-2}+\dots+a_1$. Then $x(g(x))=-a_0$. Note that $a_0$ is invertible. So you have found the inverse of $x$ by $a_1^{-1}g(x)$ by simply plugging in $x$ value of that element. 
A: Let $R = \mathbb{Q}[\sqrt[4]{2}]$. 

To show that $R$ is a field, rather than trying to directly show that
$$\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}} \in R$$
for all $a,b,c,d \in \mathbb{Q}$, other than $(a,b,c,d) = (0,0,0,0)$, instead show that $R$ is isomorphic to some ring which is clearly a field.

Thus, consider the principal ideal $(x^4-2)$ of $\mathbb{Q}[x]$.

It's not hard to prove that $x^4-2$ is irreducible in $\mathbb{Q}[x]$. Hence, since $\mathbb{Q}[x]$ is a PID, it follows that $(x^4-2)$ is a maximal ideal, so $\mathbb{Q}[x]/(x^4-2)$ is a field.

Next, consider map $f:\mathbb{Q}[x] \to R$ defined by $f(p(x)) = f(p(\sqrt[4]{2}))$.

Argue that $f$ is a surjective homomorphism from $\mathbb{Q}[x]$ onto $R$, with kernel $(x^4-2)$.

It follows that $R$ is isomorphic to $\mathbb{Q}[x]/(x^4-2)$, hence $R$ is a field.

Of course then it follows, without actually computing the inverse that
$$a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}$$
is a unit of $R$, for all $a,b,c,d \in \mathbb{Q}$, other than $(a,b,c,d) = (0,0,0,0)$.
A: If $$\alpha \in \mathbb{Q}[2^{1/4}], \qquad \alpha = \sum_{m=0}^3 b_m 2^{m/4}, \qquad b_m \in \mathbb{Q}$$
Then $$N_{K/\mathbb{Q}}(\alpha) = \prod_{k=1}^4 (\sum_{m=0}^3 b_m (i^k 2^{1/4})^m) \in \mathbb{Q}$$
(proof : it is invariant by $2^{m/4} \mapsto (i^k 2^{1/4})^m$, where $i^{k} 2^{1/4}$ are the other roots of $x^4-2$) 
And hence for $\alpha \ne 0 \implies N_{K/\mathbb{Q}}(\alpha) \ne 0$
$$\alpha^{-1} = \frac{\prod_{k=1}^3 (\sum_{m=0}^3 b_m (i^k 2^{1/4})^m)}{N_{K/\mathbb{Q}}(\alpha)} =  \sum_{l=0}^3 c_l 2^{l/4} \in \mathbb{Q}[2^{1/4}]$$
where the $c_l $ are rational functions of the $b_m$.
