Help with Atiyah-McDonald proof. $1 + \hat{m} = $ units $\implies A$ is a local ring.

Question

Prop 1.6 i) Let $A$ be a ring and $\hat{m} \neq (1)$ an ideal of $A$ such that every $x \in A - \hat{m}$ (set difference) is a unit in $A$. Then $A$ is a local ring and $\hat{m}$ its maximal ideal.

ii) Let $A$ be a ring and $\hat{m}$ a maximal ideal of $A$, such that every element of $1 + \hat{m}$ (ie. every $1 + x,$ where $x \in \hat{m}$ is a unit in $A$. Then $A$ is a local ring.

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The proof of ii) goes: Let $x \in A - \hat{m}.$ Since $\hat{m}$ is maximal, the ideal generated by $x$ and $\hat{m}$ is $(1)$, hence there exist $y \in A, t \in \hat{m}$ such that $xy + t = 1$ hence $xy = 1 - t$ belongs to $1 + \hat{m}$ and therefore is a unit. Now use use i). QED

I'm only having trouble with part ii). I'm not sure how to prove that all of $A - \hat{m}$ are units, since the above says that for all $x \in A - \hat{m}$ there is $y \in A$ such that $xy$ is a unit. It doesn't say that $x$ is neccessarily a unit.

Answer 1

$$xy\;\;\text{is a unit}\implies \exists\,u\in A\;\;s.t.\;\;1=(xy)u=x(yu)\implies x\;\;\text{is a unit}$$

Answer 2

Units $\,u\,$ (i.e. $\,u\mid 1)$ are closed under divisors: $\ x\mid u\mid 1\,\Rightarrow\,x\mid 1\,$ by transitivity of "divides".

i.e. in words, a divisor of a divisor of foo, remains a divisor of foo. As here, it is often helpful to view units as the divisors of $\,1,$ since it allows one to employ well-known divisibility properties.

Answer 3

If $a$ is not an unit, then $1 \notin (a)$, where $(a)$ is the ideal generated by $a$. Therefore $(a) \neq A$ which implies that $(a)$ can be included in a maximal ideal. But there is only one maximal ideal. Therefore $a \in (a) \subset m$.