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Suppose I want to find all irreducible polynomials in $P_3(\mathbb{F_3})$.

From what I understand, a polynomial is not irreducible if it can be written as a product of two simpler polynomials where those degrees are smaller than the one I started with.

$P_3=a_3x^3+a_2x^2+a_1x+a_0$
$\mathbb{F_3}=\{1,2,3\}$

Now I'm a bit confused as to how I'm suppose to do this. There are quite a bit of polynomials, and even if I do find all possible polynomials, I'm not sure how to test if they are irreducible or not.

Any help would be appreciated. I believe there may be a trick to this to speed things up.

Thanks.

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    If such a polynomial is reducible then it will factor into at least one linear term of the form $x-a$. This means that the condition is that $P_3(x)\neq 0$ for $x\in \mathbb{F}_3$ – 2017-01-14
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    Are you actually being asked to carry out this calculation or to design an algorithm to do it? In either case, an approach like the sieve of Eratosthenes is likely to be the most efficient way to do it, but it would be useful to know what you are actually trying to do. – 2017-01-14
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    I would like to carry out the calculation. But I think it could be faster if I only look at monic polynomials – 2017-01-14
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    Emma's answer gives a workable method for verifying that a given polynomial is irreducible. The search can be speeded up if you know a few basic facts. 1) The formula telling you how many monic irreducible polynomials of a given degree over $\Bbb{F}_p$ there are (or any finite field for that matter). Search for the formula e.g. on our site! Anyway, there are 3 irreducible monic quadratics and 8 irreducible monic cubics, so you know when you can stop looking for more. 2) If $P(x)$ is irreducible so are $P(x+1)$, $P(x-1)$ and $x^3P(1/x)$, so if you find one irreducible you usually get others. – 2017-01-14

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From the Factor Theorem (for $f(x) \in F[x]$, $F$ a field, $a \in F$ is a root of $f(x)$ iff $x-a$ is a factor of $f(x)$ in $F[x]$), we can derive a corollary that states that if $f(x)$ is a polynomial of degree 2 or 3, then $f(x)$ is irreducible in $F[x]$ iff $f(x)$ has no roots in $F$.

So, in short, you just need to check for each polynomial if it has any roots in $\mathbb{F}_3$.

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    "just"? That's quite a lot of work: $3^4 = 81$ polynomials and $3$ candidate roots for each polynomial. – 2017-01-14
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    @RobArthan Fair enough - though (and maybe I'm misunderstanding here) is there a reason we're not using $\mathbb{F}_3 = \{0,1,2\}$? That would bring us down to 54 polynomials, and testing 1 would be simple, and testing 0 would be extremely simple. Better yet, use -1, 0, and 1. – 2017-01-14
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    Whichever way you represent the coefficients, you get $81$ polynomials of degree at most $3$ and $54$ of degree exactly $3$ (I agree that the OP's choice of $\{1, 2, 3\}$ for the coefficients is a bit strange). My point was that you will want to cut down the space a bit for practical calculation. e.g., (for a start) by only looking at monic polynomials. – 2017-01-15