I have to solve the following problem:
We say that a ring is "local" when it has only one maximal ideal. Prove that the elements that are out of the maximal ideal are units.
I know that if $A$ is a ring and $\mathfrak{a}$ its ideal, $\mathfrak{a}$ is a maximal ideal of $A$ when it is proper and when it satisfies the following: $\mathfrak{a}\subseteq\mathfrak{b}\subseteq A \Rightarrow \mathfrak{b}=\mathfrak{a}$ or $\mathfrak{b}=A $.
I have started like this:
Let we have $\mathfrak{a}$, the maixmal ideal of $A$ (the one and only). We know that $\mathfrak{a}\neq A $ and if $\mathfrak{a}\subseteq\mathfrak{b}\subseteq A$, $\mathfrak{b}=\mathfrak{a}$ or $\mathfrak{b}=A $.
Lets suppose first that $\mathfrak{b}=\mathfrak{a}$, and lets we take an element, $x$, which is in $A$ but which is not in $\mathfrak{a}$.
I don't know how to follow... I don't know if I'm going from the right way... Can someone help me?
Since $x\notin \mathfrak{a}$ and $\mathfrak{a}$ is the only maximal ideal, then there is no maximal ideal containing $x$.
Consider all ideals containing $x$ and not containing $1$.
If this set is empty, then $1$ is in the ideal $xA$ generated by $x$. This implies that there is $a\in A$ such that $1=ax$ and then $x$ is a unit.
If this set is non-empty you can apply Zorn's lemma to get a maximal element by inclusion. Observe that a maximal element $I$ of this set is also a maximal ideal of $A$. This is because if $I\subset J\subset A$ we would have $x\in J$ and if $J\neq A$, then $1\notin J$. Therefore, $J=I$.
Since $\mathfrak{a}$ is the only maximal ideal and $x\notin \mathfrak{a}$, such a maximal ideal $I\ni x$ cannot exist.