I propose the following: suppose $\,U\,$ is not prime, thus there exist $$\,x,y\in R \text{ such that }x,y\notin U\,,\,xy\in U\,.$$ Define now $B:=U+\langle y\rangle$.
By maximality of $\,U\,$ we have that $\,B\,$ is f.g., say $$\,B=\Bigl\langle u_i+r_iy\,,\,1\leq i\leq k\,,\,k\in\mathbb{N}\,\,;\,\,u_i\in U\,,\,r_i\in R\Bigr\rangle$$ and let now $$U_y:=\{s\in R\,\,;\,\,sy\in U\}.$$ (1) Check that $\,U_y\,$ is a proper ideal in $\,R\,$.
(2) Show that $\,U_y\,$ is f.g.
Put $\,U_y=\langle s_1,\ldots,s_m\rangle\,$, and take $\,u\in U\Longrightarrow\,\exists v_1,\ldots,v_k, t_1,\ldots,t_k\in R\,\,s.t.$$$u=\sum_{n=1}^kv_nu_n+\sum_{n=1}^kt_nr_ny.$$
(3) Show that $\displaystyle{\sum_{n=1}^kt_nr_n}\in U_y.$
(4) Putting $\,\Omega:=\{u_i,\ldots,u_k,ys_1,\ldots,ys_m\}\,$, derive the contradiction $\,U=\langle\Omega\rangle$.
For a commutative ring with unity $R$ and an ideal $I$, we have the following:
- $I$ is prime if and only if $R/I$ is an integral domain.
- $I$ is maximal if and only if $R/I$ is a field.
This shows that every maximal ideal is a prime ideal since all fields are integral domains (every unit is not a zero-divisor).
The isomorphism $R[x]/(x) \cong R$ also provides a way to construct a prime ideal which is not maximal. For instance, pick any polynomial ring over a domain which is not a field and look at the ideal $(x)$. In particular, $\mathbb{Z}[x]$ for instance, $(x)$ is prime but not maximal since $\mathbb{Z} \cong \mathbb{Z}[x]/(x)$ is an integral domain, but not a field.
If the ring does not have an identity, it is actually possible. For instance $2 \mathbb{Z}$, the even integers. The ideal $(4)$ is maximal but not prime since $2\cdot 2 \in (4)$ but $2\notin (4)$.
Best Answer
It is equivalent to show that every non-minimal ideal is maximal. If $P\subsetneq Q\subset R$ are prime ideals, we can view $Q$ as an ideal of the integral domain $R/P$. The additive group of $R/P$ is also finitely generated, so it suffices to show that if $R$ is an integral domain with finitely generated additive group, then every non-zero prime ideal of $R$ is maximal.
Suppose $R$ is an integral domain whose additive group is finitely generated, and let $n$ be the rank of the additive group. Suppose $P\subset R$ is a non-zero prime ideal, and take $x\in P\backslash\{0\}$. Since multiplication by $x$ is a bijection of $R$ onto $xR$, the rank of $xR$ is $n$, so the rank of $P$ is also $n$. The quotient of a finitely generated abelian group by a subgroup with the same rank is finite, so $R/P$ is a finite integral domain, hence a field. It follows that $P$ is maximal.