Let $\frac{a}{b}\in R_p$, with $\gcd(b,p)=1$. Prove that if $\gcd(a,p)=1$, then $\frac{a}{b}$ is a unit in $R_p$, and so cannot be contained in any proper ideal of $R_p$.
Conclude that if $I$ is a proper ideal of $R_p$, then every element of $I$ can be written with a denominator that is prime to $p$, and the numerator divisible by $p$. Deduce that if $I$ is a proper ideal, then $I\subseteq I_p$.
(For extra points, use the idea above to describe all ideals of $R_p$).
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 $\,U=\langle\Omega\rangle$.
Best Answer
Here is a possibly different way to prove the result for prime ideals, and it will help us determine a sufficient condition for the inverse image of a maximal ideal to be maximal.
Let $f:A\to B$ be any commutative ring homomorphism, and let $\mathfrak b$ be any ideal of $B$. Then $\mathfrak a=f^{-1}\mathfrak b$ is an ideal of $A$ such that $f$ determines an injection $$\varphi:A/\mathfrak a\to B/\mathfrak b$$ This is basically the lattice theorem from group theory. Now, an ideal $\mathfrak q\subset B$ is prime if and only if $B/\mathfrak q$ is an integral domain. Let $\mathfrak p=f^{-1}\mathfrak q$ and note that $A/\mathfrak p$ embeds $B/\mathfrak q$. Since a subring of an integral domain is in turn an integral domain, $\mathfrak p$ is necessarily prime.
Now, an ideal $\mathfrak n\subset B$ is maximal if and only if $B/\mathfrak n$ is a field. Moreover, if we set $\mathfrak m$ to be the preimage of $\mathfrak n$, $A/\mathfrak m$ embeds the field $B/\mathfrak n$. So far so good. But the problem is that subrings of fields are not necessarily fields; this is the only reason why we cannot draw the same conclusion with maximal ideals as we did with prime ideals. However, it can be seen that if $f$ is surjective, then the embedding $\varphi$ is surjective--hence, $\varphi$ is an isomorphism--and $A/\mathfrak m$ is a field.