The characteristic of a commutative ring $R$ with $1$ is defined as follows.
Let us define the map $\phi: \Z \to R$ by sending $n \in \Z$ to
\[\phi(n)= \begin{cases}
\underbrace{1+\cdots+1}_{n\text{ times}} \text{ if } n>0\\
0 \text{ if } n=0\\
– (\underbrace{1+\cdots+1}_{-n\text{ times}}) \text{ if } n<0.
\end{cases}
\]
Then this map $\phi$ is a ring homomorphism and we define the characteristic $c$ of $R$ to be the integer $c$ such that
\[\ker(\phi)=(c).\]
(Note that the kernel of $\phi$ is an ideal in $\Z$, and $Z$ is a principal ideal domain (PID), thus such an integer $c$ exists.)

Proof.

Let us now prove the problem.
Let $c$ be the characteristic of an integral domain $R$.

Then by the first isomorphism theorem with the ring homomorphism $\phi: \Z\to R$ as above, we have an injective homomorphism
\begin{align*}
\Zmod{c}=\Z/\ker(\phi) \to R.
\end{align*}
Since $R$ is an integral domain, $\Zmod{c}$ is also an integral domain.

This yields that $c\Z$ is a prime ideal of $\Z$.
Therefore $c=0$ or $c$ is a prime number.

$(x^3-y^2)$ 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 […]

In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal
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
A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element […]

A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine […]

Every Prime Ideal in a PID is Maximal / A Quotient of a PID by a Prime Ideal is a PID
(a) Prove that every prime ideal of a Principal Ideal Domain (PID) is a maximal ideal.
(b) Prove that a quotient ring of a PID by a prime ideal is a PID.
Proof.
(a) Prove that every PID is a maximal ideal.
Let $R$ be a Principal Ideal Domain (PID) and let $P$ […]

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 […]

Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain
Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.
Prove that the polynomial
\[f(x)=x^n-t\]
in the ring $S[x]$ is irreducible in $S[x]$.
Proof.
Consider the principal ideal $(t)$ generated by $t$ […]

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.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral […]