criterion for maximal ideal

Theorem.  In a commutative ring $R$ with non-zero unity, an ideal $𝔪$is maximal if and only if

$\forall a\in R\setminus 𝔪\exists r\in R\mathit{\hspace{1em}}\text{such that}\mathrm{\hspace{0.33em}\hspace{0.33em}1}+ar\in 𝔪.$(1)

Proof.  $1^{\circ }$.  Let first $𝔪$ be a maximal ideal of $R$ and $a\in R\setminus 𝔪$.  Because  $𝔪+(a)=R$,  there exist some elements $m\in 𝔪$  and  $-r\in R$  such that  $m-ar=1$.  Consequently,  $1+ar=m\in 𝔪$.
$2^{\circ }$.  Assume secondly that the ideal $𝔪$ satisfies the condition (1).  Now there must be a maximal ideal ${𝔪}^{\prime }$ of $R$ such that

$$𝔪\subseteq {𝔪}^{\prime }\subset R.$$

Let us make the antithesis that ${𝔪}^{\prime }\setminus 𝔪$ is non-empty.  Choose an element

$$a\in {𝔪}^{\prime }\setminus 𝔪\subset R\setminus 𝔪.$$

By our assumption, we can choose another element $r$ of $R$ such that

$$s=\mathrm{\hspace{0.33em}1}+ar\in 𝔪\subset {𝔪}^{\prime }.$$

Then we have

$$1=s-ar\in {𝔪}^{\prime }+{𝔪}^{\prime }={𝔪}^{\prime }$$

which is impossible since with 1 the ideal ${𝔪}^{\prime }$ would contain the whole $R$.  Thus the antithesis is wrong and  $𝔪={𝔪}^{\prime }$  is maximal.