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Recently I started to read about finite fields. To see and better my understanding, I am trying to do some exercises:

Finite fields, if the amount of their elements is a prime, are easy to understand they are just equal to $\mathbb{Z}/p=:\mathbb{F}_p$.

Now if $|F|=q=p^n$ , such that $p$ is prime, every element $a \in F$ is a root of $x^q-x \in \mathbb{F}_p[x]$

Further,let $L$ be a field of characteristic = $p$, such that the polynomial $p(x)=x^q-x$ splits into linear factors. Then the set $S$ containing all roots of $p(x)$ is a subfield of $L$ such that $|S|=q$

Now I want to look at a problem: I want to construct $\mathbb{F}_9/\mathbb{F}_3$:

I am searching for a field $L$ with characteristics $3$, such that $x^9-x$ splits into linear factors. This means I need to find the splitting field of $x^9-x$ over $\mathbb{F}_3$.

As far as I understand, if $p$ is an irreducible polynomial $p(x) \in \mathbb{F}_3$ ,then the splitting field of $p(x) \in \mathbb{F}_3$ is isomorphic to $\mathbb{F}_3[x]/(p(x))$

But isn't $x^9-x$ reducible, I could just write $x (x^8-1)$ ? Is there something in what I wrote that is wrong or that I didn't understand correct? I don't really know how to continue from this point on, could someone show/explain it to me.

Edit:

Ok, since $|\mathbb{F}_9|=9$ and $|\mathbb{F}_3|=3$, then $\mathbb{F}_3[x]/(p(x))$ is a finite field with $3^2=9$ and thus $\mathbb{F}_3[x]/(p(x))=\mathbb{F}_9$ I need to find a irreducible polynomial with degree $2$, $p(x)=x^2+1$

$p(x)$ has no roots modulo $3$:

$p(0)=1$

$p(1)=2$

$p(2)=5=2$

Is the way I constructed $\mathbb{F}_9$ correct?

So, I have further questions: Is it possible to write $\mathbb{F}_9$ as:

$\mathbb{F}_9=\mathbb{F}_2(\alpha)$ for some $\alpha \ in \mathbb{F}_9$?

wanymose
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    Yes that is reducible, and actually you can factor out $x^3-x$ because the roots of that are the elements that are already in $\mathbb F_3$. The other factor might be what you are looking for. – Torsten Schoeneberg May 02 '23 at 20:22
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    You can keep factoring. $$x^8-1=(x^4+1)(x^4-1)=(x^4+1)(x^2+1)(x^2-1)=(x^4+1)(x^2+1)(x+1)(x-1).$$ What's more, over $\Bbb{F}_3$ we still have $$x^4+1=x^4+4=(x^4+4x^2+4)-4x^2=(x^2+2)^2-(2x)^2,$$ which factors in the well known way. The simplest irreducible factor to use is probably $p(x)=x^2+1$. In that case you get $\Bbb{F}_3[x]/(p(x))\simeq \Bbb{F}_3(i)$ with $i^2=-1$ :-) – Jyrki Lahtonen May 02 '23 at 20:51
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    @JyrkiLahtonen So is the approach to start with $x^9-x$ and to just keep factoring until I get an irreducible polynomial over $\mathbb{F}_3[x]$ ? Is this applicable in a general case? – wanymose May 02 '23 at 20:56
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    It is usually simpler to just try and find an irreducible $p(x)$ by whatever means. For example, a quadratic or a cubic $p(x)$ is irreducible if and only if it has no zeros in $\Bbb{F}_p$. Do keep in mind that the same does not apply to degree four polynomials. The polynomial $x^4+1$ above is a case in point. We saw that it is the product of two quadratics, but it has no zeros in $\Bbb{F}_3$. – Jyrki Lahtonen May 02 '23 at 21:26
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    For example $p(x)=x^2-2$ is irreducible over $\Bbb{F}_5$, because $p(0)=3$, $p(1)=4=p(4)$ and $p(2)=2=p(3)$. Hence you get a field with $25$ elements as $\Bbb{F}_5[x]/(x^2-2)$. You can also think of this field as $$\Bbb{F}_5(\sqrt2)={a+b\sqrt2\mid a,b\in\Bbb{F}_5}.$$ – Jyrki Lahtonen May 02 '23 at 21:29
  • @JyrkiLahtonen But does it work for more complicated examples? If I wanted to find a field with $8$ elements. And I know that $8=2^3$ ,then I can just find an irreducible polynomial of degree $3$, in this case $x^3+x+1$, and $\mathbb{F}_8=\mathbb{F}_2/(x^3+x+1)$. But to be able to write it as $\mathbb{F}_2(\alpha)$, I would have to solve $x^3+x+1=0$ which looks difficult. How would one do this? – wanymose May 02 '23 at 21:38
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    The trick is not to solve such an equation. Just postulate a zero, call it $\alpha$. Then use the equation $\alpha^3+\alpha+1=0$ to do the arithmetic. See this old answer of mine for examples how to use that particular $\alpha$. That extension is cubic, so the elements are of the form $b_0+b_1\alpha+b_2\alpha^2$ with $b_0,b_1,b_2\in{0,1}$. Higher degree terms ($\alpha^3,\alpha^4,\ldots$ you can reduce using the equation. – Jyrki Lahtonen May 02 '23 at 21:46
  • @JyrkiLahtonen Thank you, I will have a look. – wanymose May 02 '23 at 21:48
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    Much the same way you use $(\sqrt2)^2=2$ when doing arithmetic in $\Bbb{F}_5(\sqrt2)$. Or $\Bbb{Q}(\sqrt2)$ for that matter. – Jyrki Lahtonen May 02 '23 at 21:48
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    Good luck. Past my bedtime. Hope it becomes clear. Others can help you, too. – Jyrki Lahtonen May 02 '23 at 21:50

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