## § Quotient by maximal ideal gives a field

#### § Quick proof

Use correspondence theorem. $R/m$ only has the images of $m, R$ as ideals which is the zero ideal and the full field.

#### § Element based proof

Let $x + m \neq 0$ be an element in $R/m$. Since $x + m \neq 0$, we have $x \not in m$. Consider $(x, m)$. By maximality of $m$, $(x, m) = R$. Hence there exist elements $a, b \in R$ such that $xa + mb = 1$. Modulo $m$, this read $xa \equiv 1 (\text{mod}~$m $)$. Thus $a$ is an inverse to $x$, hence every nonzero element is invertible.