# Compute $T_{E|\mathbb{Q}}(\alpha)$ and $N_{E|\mathbb{Q}}(\alpha)$ for every $\alpha \in E$.

Consider the splitting field $E$ of $X^4 -2$.

Compute $T_{E|\mathbb{Q}}(\alpha)$ and $N_{E|\mathbb{Q}}(\alpha)$ for every $\alpha \in E$.

My attempt:

The splitting field $E = \mathbb{Q}(i,\sqrt{2},\sqrt{2})$ which is a degree 8 extension with basis $\mathcal{B} = \{1,i,\sqrt{2},\sqrt{2},i\sqrt{2},i\sqrt{2},\sqrt{8}\}$

We can use the definition of the trace in terms of the roots of the minimal polynomial over $\mathbb{Q}$ which is $X^4 -2$, since its an anihilating polynomial of order 4, which is monic. This polynomial is irreducible over $\mathbb{Q}$ by Einsestein criterion, then its the minimal polynomial.

Since the trace is the sum of all the zeros, and every zero is a conjugate of the other, the trace is Also the norm, can be obtained similarly.

I am right ?

## 2 Answers

First of all you say that "Hence, all the basis elements are the roots", which isn't true, as the basis has $8$ elements, as noted above, while the polynomial has $4$ roots. Nevertheless the trace can be easily calculated in terms of the representation in basis elements. So we have that

$$T(\alpha) = T(a + b\sqrt{2} + c\sqrt{2} + d\sqrt{8} + ei + fi\sqrt{2} + gi\sqrt{2} + fi\sqrt{8}) = aT(1) = 8a$$

This is true as the trace function is linear and the traces of the basis elements are $0$, except for the element $1$.

On the other side calculating the norm is substantially harder, as the norm is multiplicative, instead of linear. In particular you need to compute:

$$N(\alpha) = \prod_{\sigma \in \text{Gal}(E/\mathbb{Q})}\sigma(\alpha)$$

I guess doing this by hand is almost impossible, as there would be $8^8$ terms if you want to get an expression with respect to the basis representation of $\alpha$.

Note: This answer comes after a comment in a related post, 2819799, although it was already "green" as i started. Well, the dialog is important, this is my only interest...

Let us denote by $$t=\sqrt2$$ the generator of the subfield $$K=\Bbb Q(t)=\Bbb Q(\sqrt2)$$ of the given splitting field, $$E=K(i)=\Bbb Q(\sqrt2,i)\ ,$$ and since we will write this $t$ very often, so a short-hand was imperative above.

We will also identify it (agains my rules of always working algebraically) with the only complex root of the polynomial $X^4-2$, which is real, and positive. The conjugates of $t$ in $\Bbb C$ are $$t_1=t\ ,\ t_2=it\ ,\ t_3=i^2t\ ,\ t_4=i^3t\ ,$$ and corresponding to them we get embeedings $\sigma_j:\Bbb Q(t)\to\Bbb C$, $t\to t_j$.

The system $$\{\ 1,\ t,\ t^2,\ t^3\ \}$$ is a base of $K=\Bbb Q(t)$ over $\Bbb Q$, and computing the trace and the norm for the extension $K:\Bbb Q$ can be done in terms of it.

The trace is linear, so for computing it, it is enough to compute \begin{aligned} \operatorname{Trace}(1) &= \sum_{1\le 4\le j}\sigma_j(1) &&=1+1+1+1 &&=4\ ,\\ \operatorname{Trace}(t) &= \sum_{1\le 4\le j}\sigma_j(t) &&=t+it+-t-it&&=0\ ,\\ \operatorname{Trace}(t^2) &= \sum_{1\le 4\le j}\sigma_j(t^2) &&=t^2-t^2+t^2-t^2 &&=0\ ,\\ \operatorname{Trace}(t^3) &= \sum_{1\le 4\le j}\sigma_j(t^3) &&=t^3-it^3-t^3+it^3 &&=0\ . \end{aligned} So for a general $$\xi = a+bt+ct^2+dt^3\in K=\Bbb Q(t)\ ,\qquad a,b,c,d\in \Bbb Q\ ,$$ we have $$\operatorname{Trace}_{K:\Bbb Q}\xi = 4a\ .$$ The norm is relatively simple, too. First, we need an identity: \begin{aligned} &(X+Y+Z+T) (X+iY-Z-iT) (X-Y+Z-T) (X-iY-Z+iT) \\[3mm] &\qquad= \Big(\ (X+Z)^2-(Y+T)^2\ \Big) \Big(\ (X-Z)^2+(Y-T)^2\ \Big) \\[3mm] &\qquad= \Big(\ (X^2+Z^2-2YT)-(Y^2+T^2-2XZ)\ \Big) \\ &\qquad\ \phantom{=} \Big(\ (X^2+Z^2-2YT)+(Y^2+T^2-2XZ)\ \Big) \\[3mm] &\qquad= (X^2+Z^2-2YT)^2-(Y^2+T^2-2XZ)^2 \\[3mm] &\qquad= X^4-Y^4+Z^4-T^4 \\ &\qquad\qquad -2X^2Z^2 + 2Y^2T^2 \\ &\qquad\qquad\qquad - 4 X^2 YT + 4Y^2XZ - 4 Z^2 YT + 4T^2XZ\ . \end{aligned} Now we can simply compute the norm for the "smaller extension" of an element of the form $\xi=a+bt+ct^2+dt^3$ \begin{aligned} \operatorname{Norm}_{K:\Bbb Q}\xi &= \operatorname{Norm}_{K:\Bbb Q}(a+bt+ct^2+dt^3) \\[3mm] &= (a+bt+ct^2+dt^3) (a+ibt-ct^2-idt^3) (a-bt+ct^2-dt^3) (a-ibt-ct^2+idt^3) \\[3mm] &=a^4-b^4t^4+c^4t^8-d^4t^{12}\\ &\qquad -2a^2c^2t^4 + 2b^2d^2t^8\\ &\qquad\qquad-4a^2bdt^4 + 4b^2act^4-4c^2bdt^8+4d^2act^8 \\[3mm] &=a^4-2b^4+4c^4-8d^4\\ &\qquad -4a^2c^2 + 8b^2d^2\\ &\qquad\qquad-8a^2bd + 8b^2ac - 16c^2bd + 16d^2ac\ . \end{aligned} We can ask the computer for a confirmation of the above:

var('a,b,c,d')
t = 2^(1/4)
prod( [ a + b*i^k*t + c*i^(2*k)*t^2 + d*i^(3*k)*t^3 for k in [0,1,2,3] ] ).expand()
latex(_)


and we get the result:

a^4 - 2*b^4 + 8*a*b^2*c - 4*a^2*c^2 + 4*c^4 - 8*a^2*b*d
- 16*b*c^2*d + 8*b^2*d^2 + 16*a*c*d^2 - 8*d^4


(Result was manually broken in two lines.) In LaTeX: $$a^{4} - 2 \, b^{4} + 8 \, a b^{2} c - 4 \, a^{2} c^{2} + 4 \, c^{4} - 8 \, a^{2} b d - 16 \, b c^{2} d + 8 \, b^{2} d^{2} + 16 \, a c d^{2} - 8 \, d^{4}$$ Now each element in $E=K(i)$ can be written uniquely in the form $$\xi+i\xi'\ .$$ We can use the above computation and the "transitivity of the related norms". The norm for the extension $E:K$ for the above element is $$(\xi+i\xi')(\xi-i\xi')\ .$$ This brings in scene $\xi^2$ and ${\xi'}^2$. We have to isolate the many components of them w.r.t. the basis $1,t,t^2,t^3$, then apply the above formula.

(For the trace, the game is simpler, the trace over $E:K$ of $\xi+i\xi'$ is the sum of the elements $\xi\pm i\xi'$, so it is $2\xi$. From here we apply the already computed trace.

It is hard to say more in detail for "generic elements".

• I love how you transform the problem to avoid the roots. Thanks. – Richard Clare Jun 16 '18 at 20:02