Non-Prime Ideal of Continuous Functions

Problems and solutions of ring theory in abstract algebra

Problem 199

Let $R$ be the ring of all continuous functions on the interval $[0,1]$.
Let $I$ be the set of functions $f(x)$ in $R$ such that $f(1/2)=f(1/3)=0$.

Show that the set $I$ is an ideal of $R$ but is not a prime ideal.

 
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Proof.

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.


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