A ring $R$ is called Artinian if it satisfies the defending chain condition on ideals.
That is, whenever we have ideals $I_n$ of $R$ satisfying
\[I_1\supset I_2 \supset \cdots \supset I_n \supset \cdots,\]
there is an integer $N$ such that
\[I_N=I_{N+1}=I_{N+2}=\cdots.\]

Proof.

Let $x\in R$ be a nonzero element. To prove $R$ is a field, we show that the inverse of $x$ exists in $R$.
Consider the ideal $(x)=xR$ generated by the element $x$. Then we have a descending chain of ideals of $R$:
\[(x) \supset (x^2) \supset \cdots \supset (x^i) \supset (x^{i+1})\supset \cdots.\]

In fact, if $r\in (x^{i+1})$, then we write it as $r=x^{i+1}s$ for some $s\in R$.
Then we have
\[r=x^i\cdot xs\in (x^i)\]
since $(x^i)$ is an ideal and $xs\in R$.
Hence $(x^{i+1})\subset (x^i)$ for any positive integer $i$.

Since $R$ is an Artinian ring by assumption, the descending chain of ideals terminates.
That is, there is an integer $N$ such that we have
\[(x^N)=(x^{N+1})=\cdots.\]

It follows from the equality $(x^N)=(x^{N+1})$ that there is $y\in R$ such that
\[x^N=x^{N+1}y.\]
It yields that
\[x^N(1-xy)=0.\]

Since $R$ is an integral domain, we have either $x^N=0$ or $1-xy=0$.
Since $x$ is a nonzero element and $R$ is an integral domain, we know that $x^N\neq 0$.

Thus, we must have $1-xy=0$, or equivalently $xy=1$.
This means that $y$ is the inverse of $x$, and hence $R$ is a field.

Torsion Submodule, Integral Domain, and Zero Divisors
Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.
The set of torsion elements is denoted
\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\]
(a) Prove that if $R$ is an […]

Every Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring
Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal […]

Every Maximal Ideal of a Commutative Ring is a Prime Ideal
Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.
We give two proofs.
Proof 1.
The first proof uses the following facts.
Fact 1. An ideal $I$ of $R$ is a prime ideal if and only if $R/I$ is an integral […]

Finite Integral Domain is a Field
Show that any finite integral domain $R$ is a field.
Definition.
A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors.
That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$.
Proof.
We give two proofs.
Proof […]

$(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain.
Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.
Proof.
Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending […]

If a Prime Ideal Contains No Nonzero Zero Divisors, then the Ring is an Integral Domain
Let $R$ be a commutative ring. Suppose that $P$ is a prime ideal of $R$ containing no nonzero zero divisor. Then show that the ring $R$ is an integral domain.
Definitions: zero divisor, integral domain
An element $a$ of a commutative ring $R$ is called a zero divisor […]

No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field
(a) Let $F$ be a field. Show that $F$ does not have a nonzero zero divisor.
(b) Let $R$ and $S$ be nonzero rings with identities.
Prove that the direct product $R\times S$ cannot be a field.
Proof.
(a) Show that $F$ does not have a nonzero zero divisor.
[…]