## § 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.