<s>[INST] Find all monic irreducible quadratics in $\mathbb{Z}_3[x]$ The general form for any monic quadratic in $\mathbb{Z}_3[x]$ is: $x^2+Ax+C$
We get the following list of possibilities:
$x^2$
$x^2+1$
$x^2+2$
$x^2+x$
$x^2+x+1$
$x^2+x+2$
$x^2+2x$
$x^2+2x+1$
$x^2+2x+2$
A polynomial is said to be irreducible if it's degree is greater than or equal to 1 and if factored s.t. $f(x)=g(x)h(x)$ then one of the factors must be a constant polynomial. Also the factors must also be an element in the polynomialring.
Here's what I've done so far:
$x^2+2x$ is reducible since it can be factored of the form $x(x+2)$. Similar argument can be made for $x^2+x$. Then is $x^2$ also reducible? We can argue that $x^2=x*x$ and as such it is reducible, is this correct?
$x^2+x+1=(x+2)(x+2)=x^2+4x+4=x^2+x+1,\hspace{0.2cm} \text{ in } \hspace{0.2cm} \mathbb{Z}_3[x]$
$x^2+2x+1=(x+1)(x+1)=x^2+2x+1$
$x^2+2=(x+1)(x+2)=x^2+3x+2=x^2+2$
The ones I've found to be irreducible are:
$x^2+x+2$
$x^2+2x+2$
$x^2+1$
For the one with only a constant factor we can argue that one of the polynomials is simply the constant polynomial of $h(x)=1$.
For the other two I have simply not been able to find any such factorization, how can I argue for this in a better sense?
A statement my professor said was that if the polynomial has a degree less than or equal to three and there are no linear factors, it is irreducible. Then shouldn't $x^2+2$ and $x^2$ be irreducible?
I'm specifically looking for how I should argue for the irreducible ones, I hope I've atleast made an attempt on this exercise and found the correct answer. But I need help with the argumentative part!
 [/INST] $x^2+2$ is irreducible in for instance $\mathbb Z[x]$, but $\mathbb Z[x]/3\mathbb Z$ is a different domain. As you find in your answers $x^2+2\equiv(x+1)(x+2)\bmod 3$, and that settles reducibility. This factorization means while $x^2+2$ does not equal $0$ for any integer $x$, it can equal some other multiple of $3$ and then it will be $\equiv0\bmod3$.
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