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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.

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    $\begingroup$ Have you tried showing it using the definition of prime ideal? What problem did you face while proving it? $\endgroup$ – Sahiba Arora Jun 2 '17 at 15:11
  • $\begingroup$ Please show your work and the point where you get stuck. $\endgroup$ – Sahiba Arora Jun 2 '17 at 15:17
  • $\begingroup$ Let J' be a prime ideal of R/A. Let (a+A), (b+A) be two elements of R/A such that (a+A)(b+A) = ab+A is in J'. Then, a+A is in J' or b+A is in J', since J' is prime. J' is of the form J/A for an ideal J of A which contains J. So, we have that if ab is in J, then either a is J or b is in J. Hence J is a prime ideal of R containing A, so J' = J/A, for a prime ideal of R containing A. I do not know how to prove that all the ideals of R/A are of the form B/A for an ideal B that contains A. $\endgroup$ – User1999 Jun 2 '17 at 15:30
  • $\begingroup$ @User1999 Use latex (click on edit to see what the people typed). If $J$ is an ideal of $R/A$ then $I = \{ x \in R, x+A \in J\}$ is an ideal of $R$ containing $A$. $\endgroup$ – reuns Jun 2 '17 at 15:41
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    $\begingroup$ I meant with $\pi : x\mapsto x+A$, take any subset $E \subset R$ such that $\pi(E) = J$ and let $ I=(A,E)= (A,e_1,e_2,\ldots)$. It is clearly an ideal of $R$ containing $A$ and $\pi(I)=J$ $\endgroup$ – reuns Jun 2 '17 at 17:14
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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 $R$, 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 $R'$. 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.

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  • $\begingroup$ at the end, you meant a bijection between the (prime) ideals of $R$ containing $A$ and the (prime) ideals of $R/A$ $\endgroup$ – reuns Jun 2 '17 at 15:35
  • $\begingroup$ @user1952009 Sure, thank you! I fixed it. $\endgroup$ – C. Falcon Jun 2 '17 at 15:36
  • $\begingroup$ @ C. Falcon Thank very much! Now, I can see it, since there exists a bijection between the set of prime ideals of R/A and the set of prime ideals of R containing A. By your remark, you mean that if f is a surjective ring homomorphism, then the image of an ideal is always an ideal. In the case of the inclusion, f is injective. $\endgroup$ – User1999 Jun 2 '17 at 16:46
  • $\begingroup$ You are welcome! Regarding my remark, yes, this is exactly what I meant! $\endgroup$ – C. Falcon Jun 2 '17 at 16:59
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    $\begingroup$ @C.Falcon In the proposition, you do not assume that the prime ideal needs to contain $A$? Consider $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$. $3\mathbb{Z}$ is mapped to the whole $\mathbb{Z}/2\mathbb{Z}$ which is not a prime. Therefore, the condition that $\mathfrak{p}$ contains the kernel is important. $\endgroup$ – CO2 Aug 12 '19 at 13:07
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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 $R/A$.

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