Question

Let \(\rho: R \rightarrow R^{\prime}\) be a surjective ring homomorphism. Let \(S\) be the set of all ideals of \(R\) that contain \(\operatorname{Ker} \rho,\) and let \(S^{\prime}\) be the set of all ideals of \(R^{\prime} .\) Show that the sets \(S\) and \(S^{\prime}\) are in one-to-one correspondence, via the map that sends \(I \in S\) to \(\rho(I) \in S^{\prime} .\) Moreover, show that under this correspondence, prime ideals in \(S\) correspond to prime ideals in \(S^{\prime},\) and maximal ideals in \(S\) correspond to maximal ideals in \(S^{\prime}\). (See Exercise 7.38.)

Short answer

Question: Show that ideals of a ring \(R\) containing the kernel of a surjective ring homomorphism \(\rho: R \rightarrow R'\) are in one-to-one correspondence with ideals of \(R'\), and that this correspondence preserves prime and maximal ideals. Solution: We need to show that the map \(\Phi: S \rightarrow S'\) defined by \(\Phi(I) = \rho(I)\) is a bijection between ideals \(I\) in \(R\) containing the kernel of \(\rho\) and ideals in \(R'\), and that it preserves prime and maximal ideals. By following these steps: 1. Show the map \(\Phi\) is well-defined. 2. Show the map \(\Phi\) is injective. 3. Show the map \(\Phi\) is surjective. 4. Show that \(\Phi\) preserves prime ideals. 5. Show that \(\Phi\) preserves maximal ideals. We can conclude that the map \(\Phi\) is a bijection between ideals of \(R\) containing the kernel of \(\rho\) and ideals of \(R'\), and that it preserves prime and maximal ideals.

Step-by-step solution

Show That the Map \(\Phi\) Is Well-Defined

To show that \(\Phi(I)\) is indeed an ideal of \(R^{\prime}\) whenever \(I\) is in \(S\), we will look at how \(\rho\) behaves on an ideal. Let \(I \in S\), which means that \(\operatorname{Ker} \rho \subseteq I\). Then, it can be seen that: 1. \(\rho(I)\) is nonempty, since \(\rho(0) = 0 \in \rho(I)\). 2. \(\rho(I)\) is closed under addition: if \(a^{\prime} = \rho(a) \in \rho(I)\) and \(b^{\prime} = \rho(b) \in \rho(I)\), then \(a^{\prime} + b^{\prime} = \rho(a + b) \in \rho(I)\), since \(a + b \in I\). 3. \(\rho(I)\) is closed under multiplication by elements of \(R^{\prime}\): if \(a^{\prime} = \rho(a) \in \rho(I)\) and \(r^{\prime} \in R^{\prime}\), then there exists some \(r \in R\) such that \(\rho(r) = r^{\prime}\). Then, \(r^{\prime} a^{\prime} = \rho(r) \rho(a) = \rho(ra) \in \rho(I)\), since \(ra \in I\). Thus, \(\rho(I)\) is indeed an ideal of \(R^{\prime}\) for each ideal \(I \in S\), and \(\Phi\) is well-defined.

Show That \(\Phi\) Is Injective

To show that \(\Phi\) is injective, we need to prove that if \(\Phi(I) = \Phi(J)\) for some \(I, J \in S\), then \(I = J\). We have \(\rho(I) = \rho(J)\), which means that for each \(a \in I\), there exists a \(b \in J\) such that \(\rho(a) = \rho(b)\), and also for each \(b^{\prime} \in J\), there exists an \(a^{\prime} \in I\) such that \(\rho(a^{\prime}) = \rho(b^{\prime})\). Thus, we get that for all \(a \in I\), \(\rho(a)\) can be written as \(\rho(b)\), where \(b \in J\), implying \(a - b \in \operatorname{Ker} \rho \subseteq I\). This means \(b = a - (a - b) \in I\), so \(J \subseteq I\). Similarly, we can show that \(I \subseteq J\), and therefore, we conclude that \(I = J\). This implies that \(\Phi\) is injective.

Show That \(\Phi\) Is Surjective

To show that \(\Phi\) is surjective, we need to prove that for any \(I^{\prime} \in S^{\prime}\), there exists some ideal \(I \in S\) such that \(\rho(I) = I^{\prime}\). Let \(I^{\prime}\) be an ideal in \(S^{\prime}\). Define \(I = \rho^{-1}(I^{\prime})\). Because \(\rho\) is a surjective ring homomorphism, it is clear that \(\rho(I) = I^{\prime}\). Now we have to show that \(I\) is an ideal of \(R\) and contains \(\operatorname{Ker} \rho\). Define \(I = \rho^{-1}(I^{\prime})\). Since \(\rho\) is a surjective ring homomorphism, the inverse map $\rho^{-}' 1}'$ preserves both operations: 1. Let \(a, b \in I\). Then \(\rho(a), \rho(b) \in I^{\prime}\). Since \(I^{\prime}\) is an ideal, then \(\rho(a) - \rho(b) = \rho(a - b) \in I^{\prime}\). Thus, \(a - b \in I\) because \(I = \rho^{-1}(I^{\prime})\). 2. Let \(a \in I\) and \(r \in R\). Since \(\rho(a) \in I^{\prime}\) and \(I^{\prime}\) is an ideal, then \(\rho(r) \rho(a) = \rho(ra) \in I^{\prime}\). Therefore, \(ra \in I\) because \(I = \rho^{-1}(I^{\prime})\). Also, if \(x \in \operatorname{Ker} \rho\), \(\rho(x) = 0 \in I^{\prime}\). Thus, \(x \in I\), so \(\operatorname{Ker} \rho \subseteq I\). Therefore, \(I \in S\), and \(\Phi\) is surjective.

Show That \(\Phi\) Preserves Prime Ideals

To show that \(\Phi\) preserves prime ideals, let \(I \in S\) be a prime ideal, and let \(a^{\prime}, b^{\prime} \in R^{\prime}\) be such that \(a^{\prime} b^{\prime} \in \rho(I)\). Since \(\rho\) is surjective, there exist \(a, b \in R\) for which \(\rho(a) = a^{\prime}\) and \(\rho(b) = b^{\prime}\). Then, \(\rho(ab) = \rho(a) \rho(b) = a^{\prime} b^{\prime} \in \rho(I)\), which implies that \(ab \in I\). As \(I\) is a prime ideal, either \(a \in I\) or \(b \in I\), so either \(a^{\prime} = \rho(a) \in \rho(I)\) or \(b^{\prime} = \rho(b) \in \rho(I)\). Thus, \(\rho(I)\) is a prime ideal in \(R^{\prime}\).

Show That \(\Phi\) Preserves Maximal Ideals

To show that \(\Phi\) preserves maximal ideals, let \(I \in S\) be a maximal ideal. Let \(I^{\prime}\) be an ideal of \(R^{\prime}\) such that \(\rho(I) \subset I^{\prime}\). By the surjectivity of \(\Phi\), there exists an ideal \(J \in S\) such that \(\Phi(J) = I^{\prime}\), i.e., \(\rho(J) = I^{\prime}\). Since \(\rho(I) \subseteq \rho(J)\), we have \(\rho^{-1}(\rho(I)) \subseteq \rho^{-1}(\rho(J))\), which implies that \(I \subseteq J\). As \(I\) is a maximal ideal, we have \(I = J\) or \(J = R\). If \(J = R\), then since \(\operatorname{Ker} \rho \subseteq J\), \(\Phi(J) = \rho(J) = \rho(R) = R^{\prime}\), which implies that \(I^{\prime} = R^{\prime}\). Therefore, \(\rho(I)\) is a maximal ideal in \(R^{\prime}\). Thus, we have shown that the map \(\Phi: S \rightarrow S^{\prime}\) defined by \(\Phi(I) = \rho(I)\) is a bijection between ideals of \(R\) containing the kernel of \(\rho\) and ideals of \(R^{\prime}\), and that it preserves prime and maximal ideals.

Key concepts

Ideals in Rings

An ideal in a ring is a special subset that satisfies specific properties making it an integral part of ring theory. Picture an ideal as a subring that interacts nicely with its parent ring. Ideals are crucial because they allow us to construct new rings through a process called quotienting.
To understand ideals better, consider:

• An ideal must be closed under both addition and multiplication by any element of its parent ring. This means if you have two elements within an ideal, adding them or multiplying them by any element in the ring must also yield a result that stays within the ideal.
• There is a special subset of ideals called "principal ideals," which are generated by a single element. These ideals offer a straightforward way to build and understand more complex ideals.

In the exercise above, we explore ideals within the context of a surjective ring homomorphism. Specifically, we identify ideals in the ring \( R \) that contain the kernel of \( \rho \). This setup leads to important insights about how these ideals behave when mapped to another ring, \( R' \).

Surjective Homomorphisms

Surjective homomorphisms are a specific type of ring homomorphism that perfectly map one ring onto another. This means every element in the target ring \( R' \) has a preimage in the source ring \( R \). Think about covering every possible location in \( R' \) with points from \( R \).
A surjective homomorphism has several implications:

• Any property defined in the domain can be translated into the codomain, given that every element in the codomain comes "from somewhere" in the domain.
• These homomorphisms allow for the rearrangement or transformation of structures while maintaining a one-to-one correspondence between certain substructures, like ideals.

In our exercise, \( \rho \) is a surjective homomorphism that maps ring \( R \) onto \( R' \). Such surjectivity ensures that corresponding ideals in \( R \), especially those containing the kernel, can adequately reflect this structure in \( R' \).

Correspondence of Ideals

The correspondence of ideals refers to the clear mapping of one set of ideals in one ring to another set of ideals in a different ring. This is particularly interesting when dealing with homomorphisms, which allow transformations between rings.
For example, when using a surjective homomorphism like \( \rho \) in the exercise, there's a direct and well-maintained linkage between the ideals of two rings, \( R \) and \( R' \):

• This correspondence ensures ideals in \( R \) containing \( \operatorname{Ker} \rho \) match with ideals in \( R' \). Essentially, each ideal in \( R \) maps to one and only one ideal in \( R' \), and vice versa.
• Notably, this mapping respects prime and maximal ideals, meaning a prime ideal in one ring remains a prime ideal in the corresponding position of the second ring, and similarly for maximal ideals.

Such a relationship showcases the elegance and utility of homomorphisms in algebra, providing a bridge to understand complex interactions in structured mathematical systems.