How to prove this is a field?

Question

Let $F=(\mathbb{Z}/5\mathbb{Z})[x]/(x^2+2x+3)$. How do I prove $F$ is a field?

I've shown its a commutative ring with an identity $\bar1$. Then we let $(\bar{a}x+\bar{b})^{-1}=(\bar{c}x+\bar{d}).$ Multiplying those together gives me and substituting $3x+2$ for $x^2$ gives me the following two equations...not sure what to do next.

$\bar{3ac}+\bar{ad}+\bar{ac}=\bar0$

$\bar{2ac}+\bar{ad}=\bar1$

I also need to prove that every element can be written as $\bar{a}x+\bar{b}$, although I feel like I already proved that somewhat.

Answer 1

$x^2+2x+3$ has no roots in $\mathbb{Z}_5$, so it is irreducible. Therefore $(x^2+2x+3)$ is maximal, so $\mathbb{Z}_5[x]/(x^2+2x+3)$ is a field.

Answer 2

Hint: Is $f(x) = x^2 + 2x +3$ irreducible in $\mathbb Z /5 \mathbb Z[x]$?

Answer 3

In $\mathbb Z/2 \mathbb Z$ we have $x^2 + 2x + 3$ = $x^2+1$, so $x + 1 \neq 0$ and (in the quotient) $(x+1)^2 = x^2+2x+1 = 2x = 0$. But in a field there are no zero divisors, a contradiction.

Answer 4

The quadratic $f$ has no roots in the field $\,\Bbb Z_5\,$ so it is irreducible. So, by Bezout

$\qquad\qquad\quad\ \ f\nmid g\ \Rightarrow\ a\, f + b\, g\, =\, 1\ \ {\rm for}\ \ a,b \in \Bbb Z_5[x]$

So $\ {\rm mod}\ f\!:\ g\not\equiv 0\,\Rightarrow\, a(0)+ bg\equiv 1,\,$ i.e. $\ g^{-1}\equiv 1$