Given a number field $\mathbb{Q}[\beta]$, where the minimal polynomial of $\beta$ in $\mathbb[Z][x]$ has degree $n$, I would like to calculate the norm of the general element $a_0+a_1\beta+\cdots+a_{n-1}\beta^{n-1}.$
In particular, here is my attempt when $\alpha=2^\frac{1}{3}$:
Let $K=Q[2^\frac{1}{3}]$. Using the definition of the norm, it is the determinant of the linear transformation. Consider $\alpha =a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ acting by multiplication on the element $d+2^{\frac{1}{3}}e+2^{\frac{2}{3}}f$. Since $\left(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c\right)\left(d+2^{\frac{1}{3}}e+2^{\frac{2}{3}}f\right)=ad+2bf+2ce+2^{\frac{1}{3}}\left(ae+bd+2cf\right)+2^{\frac{2}{3}}\left(af+dc+be\right) $ in the basis $[1,2^{\frac{1}{3}},2^{\frac{2}{3}}$ we may view multiplication by $\alpha$ as a linear transform $\alpha\left[\begin{array}{c} d\\ e\\ f \end{array}\right]=\left[\begin{array}{c} ad+2bf+2ce\\ ae+bd+2cf\\ af+dc+be \end{array}\right].$ Using the above, we see that $\alpha=\left[\begin{array}{ccc} a & 2c & 2b\\ b & a & 2c\\ c & b & a \end{array}\right] $ in this basis. Taking the determinant we find $\det \left[\begin{array}{ccc} a & 2c & 2b\\ b & a & 2c\\ c & b & a \end{array}\right] =a\left(a^{2}-2bc\right)-2c\left(ba-2c^{2}\right)+2b(b^{2}-ac) $ $=a^{3}+2b^{3}+4c^{3}-6abc. $
This means we have shown that $N_K(\alpha)=a^{3}+2b^{3}+4c^{3}-6abc$ for $\alpha=a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$
Questions:
(1) Was the above calculation correct? Can we conclude that the norm of a general element in that space is $a^{3}+2b^{3}+4c^{3}-6abc$?
(2) Is there a better way to do this computation? What about if the extension is Galois?