Question

This exercise and the next generalize the Chinese remainder theorem to arbitrary rings. Suppose \(I\) and \(J\) are two ideals of a ring \(R\) such that \(I+J=R\). Show that the map \(\rho: R \rightarrow R / I \times R / J\) that sends \(a \in R\) to \(\left([a]_{I},[a]_{J}\right)\) is a surjective ring homomorphism with kernel \(I J\) (see Exercise 7.41 ). Conclude that \(R /(I J)\) is isomorphic to \(R / I \times R / J\)

Short answer

In this problem, we were given two ideals \(I\) and \(J\) of a ring \(R\) such that \(I+J=R\). We were asked to prove that the map \(\rho: R \rightarrow R / I \times R / J\) sending \(a \in R\) to \(\left([a]_{I},[a]_{J}\right)\) is a surjective ring homomorphism with kernel \(IJ\). We showed that \(\rho\) preserves addition and multiplication, making it a ring homomorphism. Then, we proved that the map is surjective by demonstrating that for every element \(\left([x]_I, [y]_J\right) \in R/I \times R/J\), there exists an element \(a\in R\) such that \(\rho(a) = \left([x]_I, [y]_J\right)\). Finally, we determined that the kernel of \(\rho\) is \(IJ\). In conclusion, using the First Isomorphism Theorem for rings, we found that \(R / (IJ) \cong R / I \times R / J\).

Step-by-step solution

Understanding the given information.

Given \(I\) and \(J\) are two ideals of a ring \(R\), such that \(I+J=R\). This means that for any element \(r\in R\), there exist elements \(i\in I\) and \(j\in J\) such that \(r=i+j\). Now, we need to show that the map \(\rho: R \rightarrow R / I \times R / J\) which sends \(a \in R\) to \(\left([a]_{I},[a]_{J}\right)\) is a surjective ring homomorphism with kernel \(I J.\)

Proving the map is a ring homomorphism.

To show that \(\rho\) is a ring homomorphism, we must verify that it preserves addition and multiplication. Let \(a,b\in R\). Then \(\rho(a+b)=\left([a+b]_{I},[a+b]_{J}\right)\), and \begin{align*} \rho(a)+\rho(b) &= \left([a]_{I},[a]_{J}\right)+\left([b]_{I},[b]_{J}\right) \\ &= \left([a]_{I}+[b]_{I},[a]_{J}+[b]_{J}\right) \\ &= \left([a+b]_{I},[a+b]_{J}\right). \end{align*} Since \(\rho(a+b)=\rho(a)+\rho(b)\), \(\rho\) preserves addition. Next, we verify that \(\rho\) preserves multiplication: \begin{align*} \rho(ab) &= \left([ab]_{I},[ab]_{J}\right) \\ \rho(a)\rho(b) &= \left([a]_{I},[a]_{J}\right)\left([b]_{I},[b]_{J}\right) \\ &= \left([a]_{I}[b]_{I},[a]_{J}[b]_{J}\right) \\ &= \left([ab]_{I},[ab]_{J}\right). \end{align*} Since \(\rho(ab)=\rho(a)\rho(b)\), \(\rho\) preserves multiplication. Therefore, \(\rho\) is a ring homomorphism.

Proving the map is surjective.

To show that \(\rho\) is surjective, we need to prove that for any element \(\left([x]_I, [y]_J\right) \in R/I \times R/J,\) there exists an element \(a\in R\) such that \(\rho(a) = \left([x]_I, [y]_J\right)\). Since \(I + J = R\), for any \(x,y\in R\), there exist \(i\in I\) and \(j\in J\) such that \(x+y=i+j\). Set \(a = i+j\). Then \(\rho(a) = \left([a]_I, [a]_J\right) = \left([i+j]_I, [i+j]_J\right) = \left([x]_I, [y]_J\right)\). Thus, \(\rho\) is surjective.

Determining the kernel of the map.

The kernel of a homomorphism is the set of elements that map to the identity element in the codomain. In this case, the identity element in the codomain is \(\left([0]_I, [0]_J\right)\). So, we want to find all \(a\in R\) such that \(\rho(a) = \left([0]_I, [0]_J\right)\). If \(\rho(a) = \left([0]_I, [0]_J\right)\), then \(a\in I\) and \(a\in J\), which implies that \(a\in I\cap J=IJ\) (as shown in Exercise 7.41). Conversely, if \(a\in IJ\), then \(a\in I\) and \(a\in J\), so \(\rho(a) = \left([0]_I, [0]_J\right)\). Therefore, the kernel of \(\rho\) is \(IJ\).

Concluding the isomorphism.

We have shown that \(\rho\) is a surjective ring homomorphism with kernel \(IJ\). By the First Isomorphism Theorem for rings, we can conclude that \(R / (IJ) \cong R / I \times R / J\).

Key concepts

Ring Homomorphism

A ring homomorphism is a function between two rings that respects the ring operations, addition and multiplication. Specifically, for two rings, say, R and S, a function f: R → S is considered a ring homomorphism if it satisfies the following conditions for all a, b in R:

• f(a + b) = f(a) + f(b), meaning f preserves addition.
• f(a * b) = f(a) * f(b), meaning f preserves multiplication.
• f(1_R) = 1_S (if R and S have multiplicative identities).

In the context of the Chinese remainder theorem generalization, the function ρ: R → R/I × R/J is shown to be a ring homomorphism. It maps every element a in R to a pair ([a]_I, [a]_J) in the product ring R/I × R/J, and preserving ring operations.

Ideals in Ring Theory

Ideals are fundamental components of ring theory. An ideal within a ring R is a subset that is closed under ring addition and absorbs products by elements from R. To clarify, a nonempty subset I of a ring R is an ideal when:

• If a, b are in I, then a + b is also in I.
• If a is in I and r is in R, then ra and ar are in I.

Ideals serve as the building blocks for constructing quotient rings and are central to the study of ring homomorphisms. In our exercise, the ideals I and J from the ring R are used to form the quotient rings R/I and R/J, which are the targets in the surjective ring homomorphism established by the map ρ.

First Isomorphism Theorem

The First Isomorphism Theorem is a fundamental result in algebra that connects homomorphisms with quotient structures and isomorphisms. For rings, it states that if φ: R → S is a surjective ring homomorphism with a kernel K, then the quotient ring R/K is isomorphic to S.

Isomorphism means there exists a one-to-one correspondence that preserves the ring operations, making two rings structurally identical. The theorem ensures a powerful way to deduce properties of one algebraic structure from another.

In our exercise, applying the First Isomorphism Theorem leads to the conclusion that the quotient ring R/(IJ) is isomorphic to the product ring R/I × R/J. This result generalizes the Chinese remainder theorem, showing a deep relationship between ideals and quotient rings.

Surjective Mapping

A surjective mapping, also known as a surjection, is a function that 'covers' every element of the function's codomain. In more technical language, a function f: A → B is surjective if for every b in B, there exists at least one a in A such that f(a) = b.

Surjections are particularly important in the study of homomorphisms since they ensure that the mapping reaches every element in the codomain. In the context of the exercise, the map ρ is demonstrated to be surjective, meaning the ring homomorphism covers the entire product ring R/I × R/J, a necessary condition for applying the First Isomorphism Theorem and thus, establishing the desired isomorphism.