I need to prove the two statements below:
Given a commutative ring $R$ with unity, if $m$ is a proper ideal in $R$ then: $$m \,\,\text{is maximal } \Leftrightarrow a\in R-m \,\,\text{implies} \,\,Ra+m=R$$ $$ m \,\,\text{is maximal } \Leftrightarrow \forall a\in R-m \,\, \exists\,\, r\in R,\,\, k\in m \,\,\text{such that}\,\, ra+k=1.$$
First it is unclear if by $R-m$ it is meant the complement of $m$ in $R.$ If we suppose that is the case, let m be a maximal ideal (i.e. for all proper ideals $a$ in $R,\,\,$ $m \subset a$ implies $a=m$ ) and $a\in R-m.$ Let $Ra$ be the ideal generated by $a.$ Since $m$ is supposed maximal, it is not contained in $Ra.$ Thus, $\exists \,s \in m$ and $s \notin Ra,\,\, $such that $a+s \in R, $ thus $Ra+m \subset R.$ Is this correct ? So the inclusion $"\subset "$ is more or less trivial. To prove the other inclusion, $"\supset", \,\,$ let $r\in R, \,a\in R-m$. Since $r\in m \,\,$ will contradict the assumption $a\in R-m,\,$ lets suppose $r\notin m.$ It implies that $r\in R-m.$ Now I am not quite sure what should follow. I am tempting to state that $r$ belongs to the same ideal as $a$ does (since $a\in R-m$). Is this correct ? If it were correct, then I guess eather $r\in Ra$ or $Rr\,\, \cap \,\,Ra\neq \varnothing.$ I dont understand how I could decompose $r$ into two componenets. Also so far I didnt use the maximality of $m.$
Can somebody help me out to correct and finish the proof, and also propose a solution to the second statement. Many thanks.
You are doing a lot more work than you have to. Remember that the sum of ideals is an ideal, that ideals are additive subgroups of $R$, and that they are closed under multiplication with elements of $R$, and the arguments for the first equivalence can be made a lot shorter:
$\implies$: $aR+m$ is an ideal of $R$. It contains $m$ and is strictly larger than $m$ because it also contains $a$. By maximality of $m$, we must therefore have $aR+m=R$.
$\Longleftarrow$ (contrapositive): let $m$ be non-maximal. Then there is a proper ideal $n\subset R$ with $m\subsetneq n$. Let $a\in n-m$ (the complement of $m$ in $n$). Then $aR+m\subseteq n\subsetneq R$.