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I'm looking at how to find generators of finite fields, such as $$ \mathbb{F}_{25} := \frac{\mathbb{Z}_5[x]}{(x^2+2)} $$

or like $S := \dfrac{\mathbb{Z}_5[x]}{(x^2+x+2)}$ in Generators of a quotient ring.

For the second, DonAntonio shows, writing $w = x + (x^2+x+2)$ and equalities mod 3, that

$$\begin{align*}&w\\ &w^2=2w+1\\ &w^3=w(2w+1)=2w^2+w=2(2w+1)+w=w+2+w=2w+2\\ &w^4=w(w^3)=2w^2+2w=2(2w+1)+2w=w+2+2w=2=-1\end{align*}$$

Noting that we need only find an element $g \in S$ such that $g^4 \neq 1$ as by Lagrange the order is then 8.

That works out cleanly, but the image of $x$ by the quotient map isn't a general solution for a general group, which we can see from $y = x + (x^2+2) \in \mathbb{F}_{25}$ has (mod 5)

$$ \begin{align*} &y\\ &y^2 = -2\\ &y^4 = 4 = -1\\ &y^8 = 1\\ \end{align*}$$

so $y$ doesn't generate the units of $\mathbb{F}_{25}$. (Or directly from $y^2 \in \mathbb{Z}_5$ so $y^8 = (y^2)^{(5-1)} = 1$)

Is there a smarter way for either this specific example, or for a more general $\dfrac{\mathbb{Z}_p[x]}{(f(x))}$ or $\dfrac{\mathbb{Z}[x]}{(g(x))}$ for irreducible $f$ or $g$ than brute-forcing elements in the field until you find one that has the required order?

George
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  • I've linked a duplicate which explains why it's even hard to find primitive roots for $F_p$. – rschwieb Jun 08 '22 at 15:13
  • I wouldn't say you have to "brute-force all elements in the field" to discover one. You pick and element, and if it's not primitive you rule out all of its powers. The lower its order is, the less likely you are to select it at random. Ruling out an element is fairly rapid using intelligent exponentiation tricks. And then once you know one primitive root, you already know them all. – rschwieb Jun 08 '22 at 15:14
  • I see how that works well in the context of $F_p$, but if I'm looking for an explicit isomorphism into $F_{p^n}$ raising powers isn't necessarily an efficient task in the quotient - even in a small field like the order 25 one I gave, it seems tedious. I suppose I was looking more for efficient ways of testing elements in this specific case, or a way of not needing to test relatively arbitrary elements. – George Jun 09 '22 at 15:38
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    Can you elaborate? I don't see what finding an isomorphism has to do with finding primitive roots. – rschwieb Jun 09 '22 at 17:36
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    For $F_{25}$, i would do this to test an element: you'd just pick an element $a$, check $a^2$, square that to $a^4$, square that to $a^8$, multiply previously computed powers to get to get $a^{12}$. If $a^{12}=1$ you can accurately identify its order by checking $a^6$ and $a^3$, if necessary. At any point when you find the order to be less than 24, you can eliminate the powers of $a$ from consideration. There are 8 primitive roots, so you already have a probability of $\frac 13$ of picking one on the first try, and that goes up as you go. In these rings you might start simply with $x$. – rschwieb Jun 09 '22 at 17:42

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