# Element of given order in a finite field

Find an element of multiplicative order 4 and an element of order 5 in $$F_{121}$$ defined by $$x^{2} +x +7$$$$Z_{11}$$.

The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $$Z_{11}$$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?

I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $$5$$, we can note that $$5$$ divides $$11-1$$, so there is actually an element of order $$5$$ in $$\mathbb{F}_{11}$$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $$\mathbb{F}_{11}^*$$ is cyclic of order $$5$$ so any square besides $$0$$ and $$1$$ must have order $$5$$.
For an element of order $$4$$, we can save some work by noticing that an element of order $$4$$ is just a square root of $$-1$$. So, we just need to solve for $$a,b\in\mathbb{F}_{11}$$ such that $$(a+bx)^2=-1$$. Expanding out $$(a+bx)^2$$ using $$x^2+x+7=0$$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $$b=0$$ or $$2a=b$$. The first case does not work (since there is no square root of $$-1$$ in $$\mathbb{F}_7$$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $$\mathbb{F}_{11}$$ to get $$a=\pm 3$$. So, the elements of order $$4$$ are $$\pm(3+6x)$$.
When you form the splitting field of a quadratic polynomial (assuming characteristic $$\neq2$$) you essentially adjoin the square root of the discriminant.
By the quadratic formula the zeros of $$x^2+x+7$$ are $$x_{1,2}=\frac{-1\pm\sqrt{1^2-4\cdot7}}2=\frac{-1\pm\sqrt{6}}2$$ as $$1^2-4\cdot7=-27\equiv 6\pmod{11}$$. In other words, if $$\alpha=x+\langle x^2+x+7\rangle$$ is a zero of the quadratic, then $$2\alpha+1$$ will be a square root of six.
Here $$6\equiv-5$$ and $$-1$$ are both quadratic non-residues, so their ratio is a quadratic residue, and we can take advantage. It may be simplest to observe that $$4^2=16\equiv5$$, implying that $$\sqrt{-1}=\pm\frac14\sqrt{-5}=\pm\frac14(1+2\alpha)=\pm(3+6\alpha)$$ as $$1/4=3$$.