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I know that there are 9 distinct monic polynomials of degree 2 in $F_3$. To find which are irreducible, should I just list them all out and check each one, or is there a better way of checking this?

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A polynomial of degree $2$ is irreducible if and only if it has not roots in $\mathbf{F}_3$, which is equivalent to say that its discriminant is not a square in $\mathbf{F}_3$. Note $X^2 - s X + p$ your monic polynomial, so that its discriminant, $\Delta = s^2 - p$, must be equal to the only element of $\mathbf{F}_3$ which is not a square, that is $2$. Can you finish by solving $s^2 - p = 2$ in $\mathbf{F}_3^2$ ?

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  • $\begingroup$ Hi Robert, is p just an element of F? $\endgroup$ – jstnchng Feb 15 '15 at 4:50
  • $\begingroup$ Also, isn't the discriminant $s^2 - 4p$? $\endgroup$ – jstnchng Feb 15 '15 at 4:58
  • $\begingroup$ $s$ and $p$ are the coefficients of your monic polynomial with coefficients in $\mathbf{F}_3$, so that the are in $\mathbf{F}_3$ indeed. The discriminant is $s^2 - 4p$ indeed, but remember that in $\mathbf{F}_3$, as $3 = 0$, you have that $-4 = -1$. $\endgroup$ – Olórin Feb 15 '15 at 10:46
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Another way to do this is to think of what the quadratic irrationalities look like over $\Bbb F_3$. You can write them down: $\pm i$, $1\pm i$, and $-1\pm i$, where by $i$ I mean a square root of $-1=2$ in the extended field $\Bbb F_9$. Each conjugate pair that I’ve written gives you a monic quadratic irreducible. This is very constructive, as you see.

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  • $\begingroup$ Aren't there three of these though? I did them by hand and found $x^2+1$, $x^2+2$, $x^2+x+2$. How exactly do these pair up? $\endgroup$ – jstnchng Feb 15 '15 at 4:56
  • $\begingroup$ Right: for instance the pair $1\pm i$ gives you the polynomial $(x-(1+i))(x-(1-i))=x^2+x+2$. $\endgroup$ – Lubin Feb 15 '15 at 20:43

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