<s>[INST] 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...
 [/INST] 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...
</s>