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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?

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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 $\pm(3+6x)$.

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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 $1/4=3$.

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