**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.9**

Solution: First we show that $I$ is an ideal. To That end, let $f,g \in I$. Then $$(f-g)(1/2) = f(1/2) - g(1/2) = 0-0 = 0$$ and similarly $(f-g)(1/3) = 0$. Thus $f-g \in I$. Since $0 \in I$, $I$ is an additive subgroup of $R$. Finally, let $h \in R$; then $$(fh)(1/2) = f(1/2)h(1/2) = 0 \cdot h(1/2) = 0$$ and similarly $$(hf)(1/2) = (fh)(1/3) = (hf)(1/3) = 0.$$ Thus $I$ is an ideal of $R$.

Now consider $f(x) = x-1/2$ and $g(x) = x-1/3$. Clearly $fg \in I$, but neither $f$ nor $g$ is in $I$. Thus $I$ is not a prime ideal of $R$.