Is $\mathbb{Q}[x]/(x^2-5x+6)$ a field?

Question

Is $\mathbb{Q}[x]/(x^2-5x+6)$ a field? Why? I know when $\mathbb{Q}[x]/(x^2-2)$ or irreducible polynomial $p(x)$ in $\mathbb{Q}[x]$ will make the extention field such that root of $p(x)$ in that field. But when fraction by reducible polynomial Will concept be the same?

Answer 1

Given a field $k$ and $p(x)\in k[x]$, the following are equivalent:

(1) $p(x)$ is irreducible in $k[x]$,

(2) $I=(p(x))$ is a maximal ideal in $k[x]$,

(3) $k[x]/(p(x))$ is a field.

So, $\mathbb{Q}[x]/(x^2-5x+6)$ cannot be a field as $x^2 - 5x + 6$ is reducible over $\mathbb{Q}$.

Answer 2

No; $(x-3)(x-2)\equiv0\pmod{x^2-5x+6},$ and fields don't have non-zero zero divisors.

The roots of $x^2-5x+6$ are in $\mathbb Q.$