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# Is $1+X^2$ irreducible in $\mathbb{Z}_3$?

2017-12-03 03:08:31

$N = 1 + X^2$ 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 $3$

2017-12-03 05:50:59