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I want to show that $x^4+x^3+2$ is primitive over $\mathbb{F}_3$. By definition, this means that $x^4+x^3+2$ is monic and has a root $\alpha$ that generates the multiplicative group of $\mathbb{F}_{3^4}$, namely $\mathbb{Z}_{80}$. Thus, one of its four roots in $\mathbb{F}_{3^4}$ must have an order coprime to $80$ in $\mathbb{F}_{3^4}$. This concept seems very abstract, and I'm struggling to grasp it properly.

Rodrigo de Azevedo
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doctor
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    It is possible to brute force this. I would not enjoy it, but I've seen (and done) worse. To get an idea of what that entails take a look at this old answer of mine where I show by brute force that a root $\gamma$ of $x^4+x+1$ in an extension field of $\Bbb{F}2$ is primitive (i.e. has order $2^4-1=15$). Undoubtedly you have already shown that $x^4+x^3+2$ is irreducible. This implies that $\alpha$ is an element of $\Bbb{F}{81}$ but not an element of a smaller field. – Jyrki Lahtonen May 20 '24 at 18:09
  • (cont'd) So we know that the order of $\alpha$ is a factor of $80$ (and is not a factor of $8$). As $80=2^4\cdot5$ your claim would follow, if you can show that neither $\alpha^{80/5}=\alpha^{16}$ nor $\alpha^{80/2}=\alpha^{40}$ is equal to $1$. That's not quite as taxing :-) – Jyrki Lahtonen May 20 '24 at 18:12
  • Hopefully you are aware of the good ole square-and-multiply. – Jyrki Lahtonen May 20 '24 at 18:17
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    Oh, and your next to last sentence is off. The task is to show that one of the roots has order exactly $80$ (rather than coprime to it). The (multiplicative) orders of all the (non-zero) elements of $\Bbb{F}_{81}$ are factors of $80$. By Lagrange's theorem from the first course on algebraic structures. – Jyrki Lahtonen May 21 '24 at 04:40

2 Answers2

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Test if $x^4 + x^3 + 2$ is primitive for $GF(3^4)$. Using $x^4 + x^3 + 2$ as field polynomial for $GF(3^4)$. Brute force check: for i = 1 to 79: $x^i \ne 1$, $x^{80} = 1$. Although not needed in this case, noting that prime factors of $3^4-1$ = $\{2,2,2,2,5\}$, then check that $x^{16} \ne 1$, $x^{40} \ne 1$, $x^{80} = 1$.

For the roots, I did a brute force search.

A list of the 8 primitive polynomials for $GF(3^4)$ and their roots shown as base 3 values.
$x^4 + x^3 + 2$ is the third line in this list:

poly    roots
10012 : 0010 1000 1110 1210
10022 : 0010 1000 1210 1110
11002 : 0010 1000 1220 1102
11122 : 0010 1000 2102 0220
11222 : 0010 1000 0021 2001
12002 : 0010 1000 1120 1201
12112 : 0010 1000 2201 0120
12212 : 0010 1000 0022 2002
rcgldr
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  • Alternatively, the sixteenth roots of unity are zeros of $$x^8+1=(x^8-2x^4+1)-x^4=(x^4-1)^2-(x^2)^2=(x^4-x^2-1)(x^4+x^2-1).$$ Those have obviously no common factors with the given quartic as the quartic polynomials in this factorization are irreducible. – Jyrki Lahtonen Feb 01 '25 at 19:54
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    @JyrkiLahtonen - thanks, By the way, this is the question I was looking for in my other question about efficient ways to find primitive polynomials. – rcgldr Feb 01 '25 at 19:54
  • Ok! I was contemplating writing something on the other question. But the current version is a bit more difficult, to say the least :-) – Jyrki Lahtonen Feb 01 '25 at 19:56
  • @JyrkiLahtonen - when looking for roots, I divide $x^4 + x^3 + 2$ by $x - a$, where $x$ and $a$ are field elements $c_3 z^3 + c_2 z^2 + c_1 z + c_0$, and check for remainder = 0. In my program, I create antilog and log tables, so I can use integers for field elements, using table for multiply and divide, mapping to poly and back for add and subtract, or using a 81 by 81 matrix. Is there a better way and better notation for how roots are found? – rcgldr Feb 01 '25 at 20:01
  • @JyrkiLahtonen - this is like a self-fulling prophecy, the field polynomial coefficients are $GF(3)$, but for the roots, everything is treated as $GF(3^4)$. Hamming codes are like this, 1 bit encoded values are treated as field elements to generate syndromes for decoding. – rcgldr Feb 01 '25 at 20:03
  • Undoubtedly you know this, but when you have found a root $\alpha$ of a polynomial (if I understand your notation $\alpha=0100$), then the other roots are always $\alpha^3$, $\alpha^9$ and $\alpha^{27}$ and you can form those using the log tables. Or the fact that in this cubing respects sums. – Jyrki Lahtonen Feb 01 '25 at 21:00
  • @JyrkiLahtonen - I used an old generic method, factor the polynomial by repeated division, poly[i+1] = poly[i] / $x - root[i]$ For one type isomorphic mapping like $GF(2^8)$ to $GF((2^4)^2$, the factoring is by irreducible quadratics. – rcgldr Feb 01 '25 at 21:17
  • Frobenius automorphis. If you knew about this, I apologize. – Jyrki Lahtonen Feb 01 '25 at 21:24
  • @JyrkiLahtonen - I'm still learning things as I read and occasionally answer questions here. Almost all of my finite field math dates back to the late 1980's for RS erasure and error correction on tape drives. – rcgldr Feb 01 '25 at 22:46
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Name $f(X) \in \mathbb{F}_3$; $f(X) = X^4+X^3+2$. And $E=\mathbb{F}_3(f)$ the descomposition field of $f$

You can see that $G =Gal(E/\mathbb{F}_3)=C_4$. Consider the following polynomial:

$\hat{f}(X) = f(X+2) = X^4 + 2X +2$. Then if ${x_1,x_2,x_3,x_4}$ are the roots for $f$, then $\alpha_i = x_i + 1$; $i=1,2,3,4$ are the roots for $\hat{f}$, and the descomposition field is the same.

Now you can see that $\Delta^2=Discr(\hat{f}) = 2$, and $\sqrt{2}$ it’s not in $\mathbb{F}_3$. And the cubic solver is $g(X) = X^3+X+2 = (X+1)(X^2+2X+2)$ that has an unic root in $\mathbb{F}_3$ and its descomposition field is $\mathbb{F}_3(\Delta)$. Then you can see $G$ is $C_4$ and $f$ primitive.

user_sion
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    Can you explain how the primitivity of $f$ is related to the fact that the splitting field of $g$ is $\mathbb F_3(\Delta)$? – Desperado May 23 '24 at 10:07
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    Perhaps you missed the OP's definition of a primitive element/polynomial? Your answer is, unfortunately, irrelevant for the purposes of this question. See my old CW answer explaining the two notions both called primitive. Also, every finite extension of every finite field automatically is a Galois extension with a cyclic Galois group. Hail the Frobenius automorphism that generates it! – Jyrki Lahtonen May 24 '24 at 07:56
  • I accidentally upvoted this answer, but it's not what I'm looking for! I'm not even familiar with some of the notations used. – doctor May 24 '24 at 14:39