Let R be a ring and A be an ideal of R.
We say that an ideal B is prime in R if and only if B/A is prime in the quotient ring R/A.
I do not understand why the set of prime ideals of the quotient ring R/A is the set of ideals P/A where P is an ideal of R containing A?
May you help me, please?
Thank you in advance.
Best Answer
First, let us recall the following:
Proof. First, $f(P)$ is a subgroup of $R'$, since $f$ is a group homomorphism. Let $y\in f(P)$, there exists $x\in P$ such that $f(x)=y$ and let $r'\in R'$, since the map $f$ is surjective, there exists $r\in R$ such that $f(r)=r'$. Therefore, one gets: $$r'y=f(rx)\in f(P),$$ since $f$ is a ring homomorphism and $rx\in P$, using that $P$ is an ideal. Finally, one has to establish that $f(P)$ is prime in $R'$. Finally, it is left to show that $f(P)$ is prime, to do so let $(y_1,y_2)\in{R'}^2$ such that: $$y_1y_2\in f(P).$$ Since $f$ is surjective, there exists $r_i\in R$ such that $f(r_i)=y_i$, therefore using that $f$ is a ring homomorphism, one has: $$f(r_1r_2)\in f(P).$$ Hence, since $P$ is prime, $r_1\in P$ or $r_2\in P$, namely $y_1\in f(P)$ or $y_2\in f(P)$. Whence the result. $\Box$
Remark. The image of an ideal by a ring homomorphism is not an ideal, for example the image of $\mathbb{Z}$ by the inclusion $\mathbb{Z}\hookrightarrow\mathbb{Q}$.
Now, let $\pi\colon R\twoheadrightarrow R/A$ be the canonical surjection, then $P\mapsto \pi(P)$ is a bijection between the set prime ideals of $R$ containing $A$ and the set of prime ideals of $R/A$. The inverse is given by: $P\mapsto \pi^{-1}(P)$, since $\pi$ is surjective.