Question

Let \(\rho: R \rightarrow R^{\prime}\) be a ring homomorphism with kernel \(K\). Let \(I\) be an ideal of \(R\). Show that we have a ring isomorphism \(R /(I+K) \cong \rho(R) / \rho(I)\).

Short answer

In this exercise, we are given a ring homomorphism ρ:R→R' and asked to show that there is a ring isomorphism between the quotient rings R/(I+K) and ρ(R)/ρ(I), where I is an ideal in ring R and K is the kernel of the homomorphism. To do this, we follow these steps: 1. Define a new homomorphism 𝜌~ : R / (I+K) → ρ(R) / ρ(I) by 𝜌~(r + (I+K)) = ρ(r) + ρ(I) for every r in R. 2. Verify that 𝜌~ is well-defined by showing it is independent of the choice of the representative r of the coset r + (I+K). 3. Show that 𝜌~ is a homomorphism by verifying it preserves addition and multiplication. 4. Find the kernel and image of 𝜌~: Ker(𝜌~) = I+K and Im(𝜌~) = ρ(R) / ρ(I). 5. Apply the First Isomorphism Theorem to 𝜌~: R / (I+K) ≅ ρ(R) / ρ(I), proving that there is a ring isomorphism between the two quotient rings.

Step-by-step solution

Define the new homomorphism

Denote the given ring homomorphism by \(\rho : R \rightarrow R'\). Define the new homomorphism \(\overline{\rho} : R / (I+K) \rightarrow \rho(R) / \rho(I)\) by: \(\overline{\rho}(r + (I+K)) = \rho(r) + \rho(I) \quad \text{ for every } r \in R.\)

Well-defined

To prove that \(\overline{\rho}\) is well-defined, we have to show that it is independent of the choice of the representative \(r\) of the coset \(r + (I+K)\). Let \(r_1, r_2 \in R\) such that \(r_1 + (I+K) = r_2 + (I+K)\). Then we have \(r_1 - r_2 \in (I+K).\) Since \((I+K)\) is an ideal, \(r_1 - r_2 \in I + K\). This implies that \(\rho(r_1 - r_2) \in \rho(I) + K = \rho(I)\), where we have used the fact that \(K\) is the kernel of \(\rho\). Hence, \(\rho(r_1) + \rho(I) = \rho(r_2) + \rho(I).\) So, \(\overline{\rho}\) is well-defined.

Homomorphism

Now we will show that \(\overline{\rho}\) is a homomorphism. For all \(r_1, r_2 \in R.\) we have \(\overline{\rho}((r_1 + (I+K)) + (r_2 + (I+K))) =\overline{\rho}((r_1 + r_2) + (I+K)) = \rho(r_1 + r_2) + \rho(I) = (\rho(r_1) + \rho(I)) + (\rho(r_2) + \rho(I)) = \overline{\rho}(r_1 + (I+K)) + \overline{\rho}(r_2 + (I+K)).\) Similarly, \(\overline{\rho}((r_1 + (I+K)) (r_2 + (I+K))) = \overline{\rho}((r_1 r_2) + (I+K)) = \rho(r_1 r_2) + \rho(I) = (\rho(r_1) + \rho(I)) (\rho(r_2) + \rho(I)) = \overline{\rho}(r_1 + (I+K)) \overline{\rho}(r_2 + (I+K)).\) Hence, \(\overline{\rho}\) is a homomorphism.

Kernel and Image

Now we will find the kernel and image of \(\overline{\rho}\). For the kernel, let \(r \in R\) be such that \(\overline{\rho}(r + (I+K)) = \rho(r) + \rho(I) = 0 + \rho(I)\). This means that \(\rho(r) \in \rho(I)\). Since \(I\) is an ideal, it is clear that \(K \subseteq I\). Therefore, \(r \in I+K\), which implies that \(\operatorname{Ker}(\overline{\rho}) = I+K\). For the image, we have \(\overline{\rho}(R/(I+K)) = \rho(R) / \rho(I)\). Thus, the image of \(\overline{\rho}\) is \(\rho(R) / \rho(I)\).

Apply the First Isomorphism Theorem

We can now apply the First Isomorphism Theorem (FIT) to \(\overline{\rho}\): \(R / \operatorname{Ker}(\overline{\rho}) \cong \operatorname{Im}(\overline{\rho}).\) So we have \(R/(I+K) \cong \rho(R) / \rho(I).\) This means that there is a ring isomorphism between the two quotient rings \(R / (I+K) \cong \rho(R) / \rho(I)\).

Key concepts

Ring Homomorphism

A ring homomorphism is a function that maps elements from one ring to another, preserving the underlying ring operations — addition and multiplication. When working with ring homomorphisms, it's essential that

• the homomorphism preserves addition: for any elements \(a, b\) in the ring, \( \rho(a + b) = \rho(a) + \rho(b) \),
• and multiplication: for any elements \(a, b\), \( \rho(a \cdot b) = \rho(a) \cdot \rho(b) \).

These properties ensure that the structure of the original ring is respected in the image ring. A crucial aspect of ring homomorphisms is their impact on ideals, subsets of rings that are closed under addition and under multiplication by ring elements, which can lead to intriguing properties in quotient rings, as shown in the exercise.

First Isomorphism Theorem

The First Isomorphism Theorem is a cornerstone in the study of ring theory. It states that if there's a ring homomorphism \( \rho: R \rightarrow R' \), then the image of this homomorphism \( \text{Im}(\rho) \) is isomorphic to the quotient of \( R \) by the kernel of \( \rho \). Formally, it is written as: \[ R / \text{Ker}(\rho) \cong \text{Im}(\rho) \]This means that we can take any ring \( R \) and partition it by its kernel to obtain something that looks much like the image of \( \rho \) in the other ring. This theorem is very powerful because it enables mathematicians to study complicated structures by looking at simpler, equivalent structures. In the given exercise, we use this theorem to establish an isomorphism between the rings \( R/(I+K) \) and \( \rho(R)/\rho(I) \), by considering the kernel and image of a constructed homomorphism.

Kernel of a Homomorphism

The kernel of a ring homomorphism \( \rho: R \rightarrow R' \) is a set consisting of all the elements in \( R \) that are mapped to the zero element in \( R' \). It is defined as:\[ \text{Ker}(\rho) = \{ r \in R \mid \rho(r) = 0_{R'} \} \]The kernel plays a vital role because it tells us how much information is "lost" in the mapping process. It is a type of subring, known as an ideal, and it provides a key to understanding the structure of \( R \). If the kernel is small, meaning it contains few elements, the homomorphism carries a lot of information from \( R \). Conversely, a large kernel suggests more information compression.

• The kernel is crucial when the First Isomorphism Theorem is applied, as it provides a natural candidate for the quotient ring.
• In the exercise, we utilized the kernel to identify an isomorphism between different quotient rings, demonstrating its utility in simplifying complex ring structures.