normal basis of GF(16)/GF(2) and field trace [duplicate]

This is an easy special case where a so called optimal normal basis is available.

Let $\alpha$ be a root of unity of order $61$. Because $2$ generates the multiplicative group $\Bbb{Z}_{61}^*$, we know that

• $GF(2^{60})=GF(2)[\alpha]$,
• The powers $\alpha,\alpha^2,\alpha^3,\cdots,\alpha^{60}$ are all conjugates of each other. In other words, they are the same set as $\alpha,\alpha^2,\alpha^4,\alpha^8,\cdots,\alpha^{2^{59}}$.
• Because the minimal polynomial of $\alpha$, the cyclotomic polynomial $\Phi_{61}(x)$, is of degree $60$, we see that the list of the preceding bullet is a basis of this field extension. Because they are also conjugates of each other they form a normal basis.
• Unless $i+j=61$ the product $\alpha^i\cdot\alpha^j=\alpha^{i+j}$ is another basis element. When $i+j=61$ we get $\alpha^i\cdot\alpha^j=1=\sum_{k=1}^{60}\alpha^k$ is a sum of $60$ basis elements. Together these imply that the multiplication table of this normal basis has the minimum possible number of non-zero coefficients.