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Is $1+X^2$ irreducible in $\mathbb{Z}_3$?
$N = 1 + X^2$ is irreducible in $\mathbb{Z}_3[X]$ , since
$1+0^2 = 1 $ and $1+1^2 = 1 + 2^2 = 2$.
Which means that $N$ can never be zero. Regarding the factor theorem, $N$ is irreducible.
Is this sufficient to prove the irreducibility?
In general, that is not enough. For example, consider the polynomial
$$ g(X) = 4X^2 + 4X + 1 \in \mathbb{Z}[X] $$
For every $a \in \mathbb{Z}$, you have $g(a) \neq 0$. However, you also have the factorization $g(X) = (2X+1)^2$.
The ring $\mathbb{Z} / 3 \mathbb{Z}$, however, is a field, and polynomials over a field are much better behaved. For example, every linear polynomial over a field has a root. Consequently, your example cannot have a linear factor.
*: I assume you are using the notation $\mathbb{Z}_3$ to mean the integers modulo $3$

In general, that is not enough. For example, consider the polynomial
$$ g(X) = 4X^2 + 4X + 1 \in \mathbb{Z}[X] $$
For every $a \in \mathbb{Z}$, you have $g(a) \neq 0$. However, you also have the factorization $g(X) = (2X+1)^2$.
The ring $\mathbb{Z} / 3 \mathbb{Z}$, however, is a field, and polynomials over a field are much better behaved. For example, every linear polynomial over a field has a root. Consequently, your example cannot have a linear factor.
*: I assume you are using the notation $\mathbb{Z}_3$ to mean the integers modulo $3$
20171203 05:50:59