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Find all the monic irreducible polynomials of degree less than equal to $3$ in $F_2[X]$, and the same in $F_3[x]$.

My Attempt:

We need to have $1$ in the expression of our polynomial since otherwise our polynomial will have $0$ as a root and will therefore not be irreducible. So starting with $1$ we can take $x,x^2$ and $x^3$ as the following expression. We can't stop here since the polynomials $1+x,1+x^2$ and $1+x^3$ will have $1$ as root or in other words polynomials of "length" $4$ is reducible. So we carry on: we can now add $x^2$ or $x^3$ to $1+x$ and similarly, we can add $x^3$ to the polynomial $1+x^2.$ Thus we get all the irreducible polynomials $$1+x,1+x+x^2,1+x+x^3,1+x^2+x^3.$$ Note we do not extend the polynomial $1+x^3$ since it would repeat the above cases.

For $F_3[x]$ we have in the first step $$1+x,1+x^2,1+x^3.$$ Note that $1+x^3$ is reducible. In the second step $$1+x+x^2,1+x+x^3,1+x^2+x^3.$$ If we add one more term then each polynomial will be divisible by $x+1.$ So 4 terms are out. Three terms are also out since $x+2$ divides all of them. So the final list of polynomials should be $$x+1,x^2+1.$$

Is this correct?

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Think about it this way: Every element of $GF(p)$ (the finite field of order 3), $GF(p^2)$, and $GF(p^3)$ has a minimal polynomial of degree 1, 2, or 3, and this gives all the monic irreducibles. So, for example, for $GF(3)$, you have 3 degree 1 irreducibles corresponding to the elements of $GF(3)$, $3=\frac{9-3}{2}$ irreducibles of degree 2 corresponding to the elements of $GF(9)$ not in $GF(3)$ (dividing by 2 because conjugates share the same minimal polynomial) and $8=\frac{27-3}{3}$ cubic monic irreducibles corresponding to the 24 elements of $GF(27)$ not in $GF(3)$ dividng by 3 since each minimal is shared by 3 conugates. Thus, for $GF(3)$ you should have found 14 (3+3+8) irreducible monics of degree 3 of less. (By similar reasoning, for $GF(2)$ you should have found 5 (2+1+2) irreducible monic polynomials of degree 3 or less.)

Note that this makes use of the fact that finite extensions of finite fields are separable.

Since these aren't that hard to enumerate in the case of $GF(2)$, we have: $x, x+1, x^2+x+1, x^3+x+1, x^3+x^2+1$.

In the case of $GF(3)$ you have $$x, x-1, x+1, x^2+x-1, x^2+1, x^2-x-1, x^3-x-1, x^3-x+1,$$$$ x^3+x^2-1, x^3-x^2+1, x^3+x^2+x-1,$$$$ x^3-x^2+x+1, x^3+x^2-x+1, x^3-x^2-x-1$$ as you can tell by inspection.

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  • $\begingroup$ Added the actual polynomials, which is pretty straightforward since for quadratic and cubic polynomial, irreducible is equivalent to not having a root in the ground field. (It also always helps to know how any you are looking for, of course...) $\endgroup$ – C Monsour Jun 24 '18 at 21:24

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