Question

Show that if \(R=R_{1} \times R_{2}\) for rings \(R_{1}\) and \(R_{2},\) and \(I_{1}\) is an ideal of \(R_{1}\) and \(I_{2}\) is an ideal of \(R_{2},\) then we have a ring isomorphism \(R /\left(I_{1} \times I_{2}\right) \cong\) \(R_{1} / I_{1} \times R_{2} / I_{2}\)

Short answer

**Short Answer:** In order to show that the quotient rings \(R/\left(I_{1}\times I_{2}\right)\) and \(R_{1}/I_{1} \times R_{2}/I_{2}\) are isomorphic, we defined a function \(\phi: R_{1}/I_{1}\times R_{2}/I_{2} \rightarrow R/\left(I_{1}\times I_{2}\right)\), and we showed that \(\phi\) is both a bijection and preserves the additive and multiplicative structure of the rings. Therefore, \(\phi\) is a ring isomorphism, and we can conclude that \(R/\left(I_{1}\times I_{2}\right) \cong R_{1}/I_{1}\times R_{2}/I_{2}\).

Step-by-step solution

Defining the isomorphism

Let's define a function \(\phi: R_{1}/I_{1}\times R_{2}/I_{2} \rightarrow R/\left(I_{1}\times I_{2}\right)\) such that \(\phi\) respects the additive and multiplicative structure of the rings. That is: $$\phi\left((a+I_{1}, b+I_{2}\right) = (a, b) + \left(I_{1}\times I_{2}\right)$$

Show that \(\phi\) maps into the quotient ring

We need to show that the image of \(\phi\) is in the quotient ring, that is \(\phi(a+I_{1}, b+I_{2}) \in R/\left(I_{1}\times I_{2}\right)\). Note that \((a, b) \in R_{1}\times R_{2}\) and \(\left(I_{1}\times I_{2}\right)\) is an ideal of \(R\). Thus, \((a,b) + \left(I_{1}\times I_{2}\right)\) is an element of the quotient ring \(R/\left(I_{1}\times I_{2}\right)\). So, \(\phi\) maps into the quotient ring.

Prove that \(\phi\) is a bijection

We need to show that \(\phi\) is both injective and surjective. Injective: If \(\phi(a+I_{1}, b+I_{2}) = \phi(c+I_{1}, d+I_{2})\), then $$(a, b) + \left(I_{1}\times I_{2}\right) = (c, d) + \left(I_{1}\times I_{2}\right)$$ By definition of the ideal product, we have that \((a-c, b-d) \in \left(I_{1}\times I_{2}\right)\). It follows that \(a-c \in I_{1}\) and \(b-d \in I_{2}\). Thus, \(a+I_{1} = c+I_{1}\) and \(b+I_{2} = d+I_{2}\). Surjective: For any \((x,y) + \left(I_{1}\times I_{2}\right) \in R/\left(I_{1}\times I_{2}\right)\), we have that \(\phi(x+I_{1}, y+I_{2}) = (x, y) + \left(I_{1}\times I_{2}\right)\). Therefore, \(\phi\) is surjective. Now that we have proved \(\phi\) is a bijection, our next step is to show that it preserves addition and multiplication.

Prove that \(\phi\) preserves addition

Let \((a+I_{1}, b+I_{2})\) and \((c+I_{1}, d+I_{2})\) be any elements in \(R_{1}/I_{1}\times R_{2}/I_{2}\). Then: $$\phi\left((a+I_{1} + c+I_{1}, b+I_{2} + d+I_{2}\right)) = \phi(a+I_{1} + c+I_{1}, b+I_{2} + d+I_{2})$$ $$= (a+c, b+d) + \left(I_{1}\times I_{2}\right)$$ $$= (a, b) + (c, d) + \left(I_{1}\times I_{2}\right)$$ $$= \phi(a+I_{1}, b+I_{2}) + \phi(c+I_{1}, d+I_{2})$$ So, \(\phi\) preserves addition.

Prove that \(\phi\) preserves multiplication

Using the same elements as in Step 4, we have: $$\phi\left((a+I_{1})(c+I_{1}), (b+I_{2})(d+I_{2})\right) = \phi(ac+I_{1}, bd+I_{2})$$ $$= (ac, bd) + \left(I_{1}\times I_{2}\right)$$ $$= (a, b)(c, d) + \left(I_{1}\times I_{2}\right)$$ $$= \phi(a+I_{1}, b+I_{2})\phi(c+I_{1}, d+I_{2})$$ So, \(\phi\) preserves multiplication. Since \(\phi\) is a bijection and preserves both addition and multiplication, it is a ring isomorphism. Thus, we have showed that \(R/\left(I_{1}\times I_{2}\right) \cong R_{1}/I_{1}\times R_{2}/I_{2}\).

Key concepts

Ideal

In ring theory, an ideal is a special subset of a ring. An ideal is closed under addition and multiplication by elements from the ring. Also, for any element in the ring and any element in the ideal, multiplying them results in another element in the ideal. This makes ideals quite similar to normal subgroups in group theory, providing structure to the ring that allows us to construct other mathematical objects like quotient rings.

To understand this, consider the ring \( R_1 \) and its ideal \( I_1 \). If you take any element \( i \) from \( I_1 \) and any element \( r \) from \( R_1 \), \( r \cdot i \) will also be in \( I_1 \).

• Closed under addition: If \( a, b \in I_1 \), then \( a + b \in I_1 \).
• Closed under ring multiplication: If \( r \in R_1 \) and \( a \in I_1 \), then \( r \, a \in I_1 \).

By understanding ideals, we lay the groundwork to explore more complex equations and structures, such as quotient rings and ring homomorphisms.

Quotient Ring

A quotient ring is formed by "dividing" a ring by one of its ideals, similar to how you might form a quotient group in group theory. In order to form a quotient ring \( R/I \), we take the set of all possible cosets of an ideal \( I \) in \( R \).

A coset is formed by adding each element of the ideal \( I \) to an element \( a \) from the ring \( R \), written as \( a + I \). All of these cosets together form the set that is the quotient ring.

• Addition: Defined as \( (a + I) + (b + I) = (a + b) + I \)
• Multiplication: Defined as \( (a + I) \cdot (b + I) = (a \cdot b) + I \)

Quotient rings provide important insights into the structure of the ring itself and help with working on problems related to abstract algebra and other mathematical fields.

Bijection

A bijection is a function between two sets that is both injective (one-to-one) and surjective (onto). This means every element in the first set corresponds to exactly one element in the second set, and every element in the second set is the image of an element in the first set.

In the context of ring isomorphisms, showing that a function is a bijection is a crucial step. It ensures that each element of one ring has a unique, well-defined corresponding element in another ring.

• Injective: Different elements in the domain map to different elements in the codomain, i.e., \( f(a) = f(b) \) implies \( a = b \).
• Surjective: Every element in the codomain is the image of some element from the domain.

By establishing a bijection, we demonstrate a perfect "pairing" between two sets, which in turn is essential for proving the existence of an isomorphism between rings, showing they are structurally identical even if they look different individually.