Show if $\Bbb Q[x] / x^2+3x+2$ is a field or not

Question

How do I prove that $\Bbb Q[x] / x^2+3x+2$ is a field or not?

The polynomial is reducible and has the solutions $-2 $ and $-1$ which shows me that it is not maximal in $\Bbb Q[x]$. How do I continue from here?

Answer 1

If you want something more concrete, your observation is enough to say that $(x+1)(x+2)=0$ in the quotient ring. Since neither $x+1$ nor $x+2$ is zero (clearly neither is a multiple of $x^2+3x+2$), this means they are zero divisors, which can't exist in a field.

Answer 2

Let $p(x)\in\Bbb Q[x]$. Then $\Bbb Q[x]/\langle p(x)\rangle$ is a field if and only if $\langle p(x)\rangle$ is a maximal ideal of $\Bbb Q[x]$. Another way to see this: $\Bbb Q[x]/\langle p(x)\rangle$ is a field if and only if $p(x)$ is irreducible in $\Bbb Q[x]$. You have shown that $x^2+3x+2$ is reducible in $\Bbb Q[x]$. Hence, $\Bbb Q[x]/\langle x^2+3x+2\rangle$ is not a field.

In general, for a commutative ring $R$ with identity, an ideal $M$ of $R$ is maximal if and only if $R/M$ is a field.

Answer 3

Alternatively: a quadratic has at most two roots in a field, but yours has three $\,-2,-1,\,\bar x$.