Group generator of ${F}_{{2}^{4}}$

Question

Is there any technique to find the generator of the group ${F}_{{2}^{4}}$? I know that this is a multiplicative finite group of order 15 whose elements are represented by the polynomial over ${F}_{2}$ of degree less than 4, where in this group the multiplication is performed modulo the irreducible polynomial $f(x)={x}^{4}+x+1$. The generator is given by the string (0010) which is the polynomial $f(x)=x$.

Answer 1

More accurately the generator is the coset $x+I$, where $I$ is the ideal of $\Bbb{F}_2[x]$ generated by $f(x)$. The fact that it is, indeed, a generator is easy to check. Such a calculation is done for example here, where I used the notation $\gamma=x+\langle x^4+x+1\rangle$. If you are given a finite field in the form $\Bbb{F}_[x]/\langle p(x)\rangle$, where $p(x)$ is an irreducible polynomial, there are methods for finding a generator (see Lidl & Niederreiter). In practice people often use a carefully chosen polynomial $p(x)$ such that the coset of $x$ can serve in that role. Such polynomials are called primitive. Finding them is a bit of work, so tables such as this one come in handy.

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As correctly pointed out by DonAntonio there will be several generators. If $g$ is one generator, then $g^j$ is another whenever $\gcd(j,p^n-1)=1$. Thus altogether there will be $\phi(p^n-1)$ generators.