§ Quotient by maximal ideal gives a field

§ Quick proof

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

§ Element based proof

Let x+m0x + m \neq 0 be an element in R/mR/m. Since x+m0x + m \neq 0, we have xnmx \not in m. Consider (x,m)(x, m). By maximality of mm, (x,m)=R(x, m) = R. Hence there exist elements a,bRa, b \in R such that xa+mb=1xa + mb = 1. Modulo mm, this read xa1(mod xa \equiv 1 (\text{mod}~m)). Thus aa is an inverse to xx, hence every nonzero element is invertible.