Are the ideals of any integral domain $R$ trivial only? Are the ideals of any integral domain $R$ trivial only?
 A: No. The fact that $R$ is an integral domain is equivalent to say that the ideal $(0)$ is prime, but it doesn't tell you anything on the other ideals.
As examples you can consider:


*

*The ring $\mathbb{Z}$ of integers, where $n\mathbb{Z}$ is a proper ideal for every $n\neq0,1$.

*Given an integral domain $R$, then the ring $R[x]$ is again an integral domain which has infinitely many ideals (consider e.g. $(x^n)$ for $n\ge1$) independently of the ideals of $R$ (i.e. you can also take $R$ to be a field).

A: No, if $R$ is not a trivial ring, i.e. $R \ne \{0 \}$, since for any $0 \ne a \in R$ the principal ideal $\langle a \rangle = aR \ne \{ 0\}$, by virtue of the fact that $R$ has no zero divisors.
It is perhaps worth noting in this context that if $R$ is an integral domain the only ideals of which are $\{ 0 \}$ and $R$ itself, then $R$ must be a field; for if take any $a \in R$, the principal ideal $\langle a \rangle = aR \ne \{ 0 \}$, so we must have $aR = R$.  Then there must exist $b \in R$ with $ab = 1$.  Thus $b = a^{-1}$ and $R$ is a field.  Conversely, any integral domain which is not a field has non-trivial, proper ideals, viz. the $aR$ for non-unit $a$.   
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
