We first show that $I$ is an ideal of $R$.
Let $f(x), g(x)\in R$ and $r\in R$.
Then the function $f(x)+rg(x)$ is a continuous function on $[0, 1]$ and we have
\begin{align*}
(f(x)+rg(x))(1/2)=f(1/2)+rg(1/2)=0+r\cdot 0=0
\end{align*}
since $f(1/2)=g(1/2)=0$.
Similarly, we have $(f(x)+rg(x))(1/3)=0$.
Hence $f(x)+rg(x)\in I$, and thus $I$ is an ideal of $R$.

Next, we show that the ideal $I$ is not a prime ideal.
Let us define
\[f(x)=x-\frac{1}{3} \text{ and } g(x)=x-\frac{1}{2}.\]
Then the functions $f(x), g(x)$ are continuous on $[0, 1]$, hence they are elements in $R$.

Since we have
\[f \left(\frac{1}{2}\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\neq 0 \text{ and } g\left(\frac{1}{3} \right)=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\neq 0,\]
the functions $f(x), g(x)$ are not in the ideal $I$.
However, the product $f(x)g(x)$ is in $I$ since we have
\begin{align*}
f\left(\frac{1}{2}\right)g \left(\frac{1}{2}\right)=\left(\frac{1}{2}-\frac{1}{3} \right)\left(\frac{1}{2}-\frac{1}{2}\right)=0\\
f\left(\frac{1}{3}\right)g \left(\frac{1}{3}\right)=\left(\frac{1}{3}-\frac{1}{3} \right)\left(\frac{1}{3}-\frac{1}{2}\right)=0.
\end{align*}

In summary, the functions $f(x)$ and $g(x)$ are not in $I$ but the product $f(x)(g)$ is in $I$. Thus the ideal $I$ is not a prime ideal.

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 […]

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 […]

A Prime Ideal in the Ring $\Z[\sqrt{10}]$
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 […]

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 […]

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 $t$ […]

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 $J \subset P$.
Proof. […]

Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]