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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?

Atticus Stonestrom
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User160
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1 Answers1

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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.

plop
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  • Why have you chosen the ideal $xA$? – User160 Nov 22 '21 at 17:15
  • @User160 The ideal $xA$ contains $x$ and the condition of $x$ being a unit is equivalent to $1\in xA$. So, $xA$ is related to what we want to prove. Moreover, since $xA$ is an ideal it gives us a way to relate it to maximal ideals of $A$, which is what the hypothesis is about. – plop Nov 22 '21 at 17:21
  • If you don’t want to use Zorn’s Lemma, observe that for $x$ not a unit, $A / (x)$ is not the zero ring, hence has a maximal ideal $V$, which by the correspondence theorem corresponds to a maximal ideal $U$ of $A$ which contains $x$. Though of course you still need Zorn’s Lemma for the fact that nonzero rings have a maximal ideal. – Mark Saving Nov 22 '21 at 17:29
  • @User160 That every nonunit $,n,$ has a max($\Rightarrow$ prime) ideal container ("divisor") $P$ is an ideal-theoretic generalization of: every nonunit integer $,n,$ has a prime divisor $,p,$ (recall contains = divides for principal ideals $,P=(p)\supseteq (a)\iff p\mid a)\ \ $ – Bill Dubuque Nov 22 '21 at 19:59