Show that if $a\in R$ is not a zero divisor, then $a$ is unit.

Question

Let $R$ be a commutative ring with 1 which has exactly 3 ideals$\{0\},J$, and $R$. If $a\in R$ is not a zero divisor then show that $a$ is a unit.

My attempt: Suppose that $a$ is not a unit.

Then, $R.a=R.a^2$. This implies $R.a-R.a^2=0$ i.e. $Ra(1-Ra)=0$.

Since, $a$ is not a unit, then $1-Ra\neq 0$. This implies $Ra=0$. This is a contradiction.

Anyone please suggest me mistakes in this proof. This is first time for me do solve these types of problems.

Do you count $0$ as a zero divisor? (Some authors do not) – Arturo Magidin Jun 17 '20 at 22:48 — Arturo Magidin, Jun 17 '20 at 22:48

Nico2701 · Answer 1 · 2020-06-18T10:31:14.747

In your answer, what does it mean $R.a$? And why $R.a=R.a^2$? Anyway, here I post my answer:

Lemma 1. If $a\notin J$ then $a$ is unit.

Take $a\in R\backslash J,$ then take $\langle a \rangle$ the ideal generated by $a,$ that is, $\langle a \rangle =\{ax\mid x\in R\},$ (you can check it is an ideal). Therefore, $\langle a \rangle \in\{{0},J,R\}.$

As $a\notin J,$ then $a\notin \{0\},$ so $\langle a\rangle$ must be $R.$ Therefore, there exists $x\in R$ such that $ax=1.$

Lemma 2. The set $J\backslash \{0\}$ is the set of all zero divisors of $R.$

Take $a,b$ non zero elements of $J$ and suppose $a$ is not a zero divisor of $R.$ Clearly, $\langle a \rangle \in\{\{0\},J,R\}$ and $\langle a \rangle \subseteq J,$ so it is not $R,$ and since $a\ne 0,$ then $\langle a\rangle$ must be $J.$ Similarly, $\langle b \rangle=J.$ That is, $\langle a \rangle =\langle b \rangle.$ Hence, there exists $c,c'\in R$ such that $ac=b$ and $bc'=a.$

Then, $acc'=(ac)c'=bc'=a.$ Now, $a(cc'-1)=0.$ As $a$ is not a zero divisor, $cc'=1$ so $c$ is an unit. Look at what we have. If $ax\ne 0,$ then $ax\in\langle a\rangle\backslash\{0\}=J\backslash\{0\},$ so $x$ is a unit. Since $a$ is not an unit (otherwise $\langle a \rangle =R$), then $a^2$ must be $0.$

Final step. If $a\ne 0$ is not a zero divisor, by Lemma 2, $a\notin J,$ so by Lemma 1, $a$ is a unit.

$R.a$ is just $Ra$; same with $R.a^2$. Poor choice of typography. — Arturo Magidin, Jun 18 '20 at 02:17
This doesn’t prove the problem at hand: you did not show that if $a$ is not a zero divisor then it is a unit; you showed that if it is not in the unique maximal ideal then it is a unit. Different proposition. — Arturo Magidin, Jun 18 '20 at 02:18
Oh, that's right. I forgot proving that $J\backslash {0}$ is actually the set of all divisors of zero. — Nico2701, Jun 18 '20 at 09:46

Show that if $a\in R$ is not a zero divisor, then $a$ is unit.

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