Polynomial irreducibility over $\mathbb{F} _{7^n}$

Question

I recently came across this question from a while ago.

In it, an affirmation is made: The smallest extension of $\mathbb{F}_7$ in which an irreducible cubic polynomial has a root is $\mathbb{F}_{7^3}$. Therefore, since the polynomial $X^3+2$ is irreducible in $\mathbb{F}_7$ (easy to check), it is irreducible in $\mathbb{F}_{7^2}$ and reducible in $\mathbb{F}_{7^3}$.

On the other hand, I have made a program that checks all the possible roots for that polynomial in $\mathbb{F}_{7^3}=\mathbb{F}_{343}$, and it has not found any roots.

Since I do not understand the argument, can anyone give me a hand? Is the reasoning in the question linked wrong?

Answer 1

$X^3 + 2$ does have a root in $\mathbb F_{7^3}$. There are a few ways to see this. For example:

• $\mathbb F_{7^3}$ is the splitting field for the polynomial $X^{7^3} - X$ over $\mathbb F_7$. (Indeed, many books construct $\mathbb F_{7^3}$ this way.) Now suppose that $L$ is an extension of $\mathbb F_{7^3}$ containing an element $\theta$ that is a root of $X^3 + 2$ (e.g. $L$ could be the splitting field of $X^3 + 2$ over $\mathbb F_{7^3}$). Since $\theta^3 + 2 = 0$, it follows that $\theta^{7^3} = \theta \times (\theta^3)^{114} = \theta \times (-2)^{114} = \theta \times ((-2)^6)^{19} = \theta \times 1^{19} = \theta$. Thus $\theta$ is a root of $X^{7^3} - X$. Since $X^{7^3} - X$ splits completely in $\mathbb F_{7^3}$, it must be the case that $\theta \in \mathbb F_{7^3}$.
• Alternatively, we could consider the field $E = \mathbb F_7 [x] / (x^3 + 2)$. (We know $E$ is a field because $X^3 + 2$ is irreducible in $F_7 [x]$.) Now $E$ contains a root of $X^3 + 2$; indeed, the element $\theta := x \in E$ is a root of $X^3 + 2$. Since $E = \mathbb F_7(\theta)$ and the minimal polynomial of $\theta$ over $\mathbb F_7$ is the cubic polynomial $X^3 + 2$, it follows that $[E : \mathbb F_7] = 3$, i.e. $E$ contains $7^3$ elements. But $\mathbb F_{7^3}$ is the unique field with $7^3$ elements! So $E$ is isomorphic to $\mathbb F_{7^3}$. If $f : E \to \mathbb F_{7^3}$ is this isomorphism, then the fact that $\theta^3 + 2 = 0$ in $E$ implies that $f(\theta)^3 + 2 = 0$ in $\mathbb F_{7^3}$. Thus $f(\theta)$ is a root of $X^3 + 2$ in $\mathbb F_{7^3}$. (Furthermore, $\mathbb F_{7^3}$ is a Galois extension of $\mathbb F_{7^3}$, so we can also infer that $X^3 + 2$ splits completely in $\mathbb F_{7^3}$.)

Anyway, both of these arguments show that $X^3 + 2$ has a root in $\mathbb F_{7^3}$.

How did you write computer program to do arithmetic in $\mathbb F_{7^3}$? It's hard to see how you could have figured out how arithmetic works in $\mathbb F_{7^3}$ without arguing along the lines of what I've written out. [Also, remember that $\mathbb F_{7^3}$ is not the same thing as $\mathbb Z_{7^3}$ - this is a common source of confusion!]

Answer 2

Here is a simple argument. We need to show that $\mathbb{F}_{343}$ contains a cube root of $a=-2$. The element $a$ has order $6$ in the multiplicative group $\mathbb{F}_{343}^{\times}$, which is cyclic of order $343-1=342$. Now since $\mathbb{F}_{343}^{\times}$ is cyclic, and the quotient $342/6 = 57$ is divisible by three, we can divide any element of $\mathbb{F}_{343}^{\times}$ of order $6$ by $3$.

Indeed, the elements of $\mathbb{F}_7^{\times}$ within $\mathbb{F}_{343}^{\times}$ are exactly the elements of the form $\alpha^{57}$ for $\alpha \in \mathbb{F}_{343}^{\times}$. Hence there exists an $\alpha$ with $\alpha^{57}=a$, and then $\beta=\alpha^{19}$ satisfies $\beta^3=a$. Since $2^3=1$ in $\mathbb{F}_7$, the other two cube roots of $a$ are given by $2\beta$ and $4\beta$.