The prime ideals of a quotient ring

Question

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.

Answer 1

First, let us recall the following:

Proposition. Let $f\colon R\twoheadrightarrow R'$ be a surjective ring homomorphism and let $P$ be a prime ideal of the ring $R$ containing $\text{ker}(f)$, then $f(P)$ is a prime ideal of $R'$.

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 $R'$. Since $f(1_A) = 1_B$, if it were $1_B\in f(P)$ then $1_A\in P$ from the assumption $\text{ker}(f)\subseteq P$, which is impossible. So $1_B\notin f(P)$ so $f(P)$ is a proper ideal. 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 $\text{ker}(f)\subseteq P$ and $P$ 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.

Answer 2

Let $u: R \to R/A$ be the canonical projection map.

If $C$ is a prime ideal in $R/A$, then its contraction $u^{-1}(C)$ is a prime ideal in $R$. This is true for all ring homomorphism.

If $B$ is a prime ideal in $R$, then its image $u(B)$ is a prime ideal in $R/A$. This is not true in general. It is not even true that the image of an ideal is an ideal.

Notice that this is part of the correspondence theorem, that is, there exists an one-to-one order-preserving correspondence between the set of ideals in $R$ containing $A$ and the set of ideals in $R/A$.

Answer 3

Let $f:R\to R/A$ be the projection map, which is surjective. For any ideal $\mathfrak{q}\subseteq R/A$, with $\mathfrak{p}=f^{-1}(\mathfrak{q})$, it is easy to show that $R/\mathfrak{p}\simeq (R/A)/\mathfrak{q}$. Thus, $\mathfrak{p}$ is prime (resp. maximal) iff $\mathfrak{q}$ is. By the correspondence theorem, the prime ideals of $R/A$ correspond to precisely the prime ideals in $R$ containing $A$.