Since $a$ is nilpotent, we have $a^n=0$ for some positive integer $n$.
Then for any $b \in R$, we have $(ab)^n=a^nb^n=0$ since $R$ is commutative.

Then we have the following equality:
\[(1-ab)(1+(ab)+(ab)^2+\cdots+(ab)^{n-1})=1.\]
Therefore $1-ab$ is a unit in $R$.

Proof 2.

There exists $n \in \N$ such that $a^n=0$ since $a$ is nilpotent.
Assume that $1-ab$ is not a unit for some $b \in R$.
Then there exists a prime ideal $\frakp$ of $R$ such that $\frakp \ni 1-ab$.

Since $a^n=0\in \frakp$, we have $a\in \frakp$ since $\frakp$ is an prime ideal.
Then $ab \in \frakp$ and we have
\[1=(1-ab)+ab \in \frakp.\]

However, this implies that $\frakp=R$ and this is a contradiction. Thus $1-ab$ is a unit of the ring $R$ for all $b \in R$.

Is the Set of Nilpotent Element an Ideal?
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If so, prove it. Otherwise give a counterexample.
Proof.
We give a counterexample.
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Boolean Rings Do Not Have Nonzero Nilpotent Elements
Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a Boolean ring.)
(a) Prove that $x^n=x$ for any positive integer $n$.
(b) Prove that $R$ does not have a nonzero nilpotent […]

Equivalent Conditions For a Prime Ideal in a Commutative Ring
Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:
(a) The ideal $P$ is a prime ideal.
(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.
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Prime Ideal is Irreducible in a Commutative Ring
Let $R$ be a commutative ring. An ideal $I$ of $R$ is said to be irreducible if it cannot be written as an intersection of two ideals of $R$ which are strictly larger than $I$.
Prove that if $\frakp$ is a prime ideal of the commutative ring $R$, then $\frakp$ is […]

If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
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Consider the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}\]
and its ideal
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}.\]
Show that $p$ is a prime ideal of the ring $\Z[\sqrt{10}]$.
Definition of a prime ideal.
An ideal $P$ of a ring $R$ is […]

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Let $R$ be a principal ideal domain (PID) and let $P$ be a nonzero prime ideal in $R$.
Show that $P$ is a maximal ideal in $R$.
Definition
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