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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?

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Just some minor suggestions: The $1/3$ exponents are quite hard to read, I thought they were $1/s.$ Perhaps just use $1/3$ instead of using \frac{} ? Or \sqrt[3]{2} ? Also, in the first line of the post, should that [Z][x] be $\mathbb{Z}[x]$? The last line before the grey area, I think it should be $\beta$ and not $\alpha$ there. And the first line of the grey area I think the Q should be $\mathbb{Q}.$ –  Ragib Zaman Jul 2 '12 at 15:01
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1 Answer

up vote 2 down vote accepted

1) Yes, the calculation was correct. A decent way to check you haven't made any arithmetic errors is to try some small integers for $a,b,c,d,e,f$ and check the norm is multiplicative.

2) This is probably the best way to do the computation. If we have a Galois extension, then the norm of an element is the product of all of its Galois conjugates. There are two ways I can think of that we can use this fact:

  • Produce the elements of $G(K/\mathbb{Q}).$ Sometimes one has to do this anyway, so in those cases that could be a short cut.
  • Assume for now your general element has the relevant coefficient non-zero so that its degree is the degree of the extension. Then since the list of distinct Galois conjugates being $\alpha_1, \alpha_2, \cdots \alpha_n$ implies the minimal polynomial is $m(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n),$ we can read the norm of $\alpha$ off the constant term of the minimal polynomial. Then the norm for lower degree elements follows by a limit argument.
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