Problem 30

Question

Let $I_{1}$ be an ideal of $R_{1},$ and $I_{2}$ an ideal of $R_{2} .$ Show that $I_{1} \times I_{2}$ is an ideal of $R_{1} \times R_{2}$

Step-by-Step Solution

Verified

Answer

Question: Prove that if $I_1$ is an ideal of a ring $R_1$ and $I_2$ is an ideal of a ring $R_2$, then $I_1 \times I_2$ is an ideal of the product ring $R_1 \times R_2$. Answer: We have shown that $I_1 \times I_2$ is closed under addition and multiplication, which are the two properties required for it to be an ideal. Therefore, $I_1 \times I_2$ is an ideal of the product ring $R_1 \times R_2$.

1Step 1: Show that $I_1 \times I_2$ is closed under addition

Let $(a_1,b_1)$ and $(a_2,b_2) \in I_1 \times I_2$. This means that $a_1, a_2 \in I_1$ and $b_1, b_2 \in I_2$. Now consider the sum of the two elements: $$(a_1,b_1) + (a_2,b_2) = (a_1 + a_2, b_1 + b_2).$$ Since $I_1$ and $I_2$ are ideals, they are closed under addition, which means $a_1 + a_2 \in I_1$ and $b_1 + b_2 \in I_2$. So, we have: $$(a_1 + a_2, b_1 + b_2) \in I_1 \times I_2.$$ Thus, $I_1 \times I_2$ is closed under addition.

2Step 2: Show that $I_1 \times I_2$ is closed under multiplication

Now let's show that $I_1 \times I_2$ is closed under multiplication. Let $(a,b) \in I_1 \times I_2$ and $(r,s) \in R_1 \times R_2$. This means that $a \in I_1$, $b \in I_2$, $r \in R_1$, and $s \in R_2$. Consider their multiplication: $$(a,b) \cdot (r,s) = (ar,bs).$$ Since $I_1$ and $I_2$ are ideals, they are closed under multiplication by elements from their respective rings: $ar \in I_1$ and $bs \in I_2$. So, we have: $$(ar,bs) \in I_1 \times I_2.$$ Thus, $I_1 \times I_2$ is closed under multiplication.

3Step 3: Conclusion

We have shown that $I_1 \times I_2$ is closed under addition and multiplication, which are the two properties required for it to be an ideal. Therefore, $I_1 \times I_2$ is an ideal of the product ring $R_1 \times R_2$.

Key Concepts

Closure Under AdditionClosure Under MultiplicationProduct of Rings

Closure Under Addition

In ring theory, one of the key properties that define an ideal is its closure under addition. This means that if you take any two elements from the ideal and add them together, the result will still be an element within the same ideal. This concept acts like a guarantee that the operations within the ideal remain consistent.

For example, let's look at two elements, $(a_1, b_1)$ and $(a_2, b_2)$, that belong to a product of ideals $I_1 \times I_2$. If you add them together as shown: \[(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2)\]

Each component $a_1 + a_2$ and $b_1 + b_2$ continues to remain in their respective ideals $I_1$ and $I_2$. Thus, the whole new pair $(a_1 + a_2, b_1 + b_2)$ also lands snugly within $I_1 \times I_2$. Therefore, we conclude that any product of ideals, such as $I_1 \times I_2$, meets the criterion of closure under addition.

Closure Under Multiplication

Another essential property of an ideal in ring theory is being closed under multiplication. Closure under multiplication refers to the condition where multiplying any element of the ideal by another element from the ring still results in a product that belongs to the ideal.

Consider how this works with a pair from the ideal $(a, b)$ belonging to $I_1 \times I_2$ and another from the product ring $(r, s)$ from $R_1 \times R_2$. Their multiplication is carried out like this:
\[(a, b) \cdot (r, s) = (ar, bs)\]

For this to belong to $I_1 \times I_2$, both $ar$ and $bs$ should lie in their respective ideals $I_1$ and $I_2$. Since $I_1$ and $I_2$ are already known to be ideals, each ensures that products like $ar$ and $bs$ are also within $I_1$ and $I_2$. This proves that $I_1 \times I_2$ thoroughly maintains the closure under multiplication property.

Product of Rings

The concept of the product of rings can seem quite abstract, but it plays a crucial role in ring theory. The idea is that you take two rings, say $R_1$ and $R_2$, and combine them in a way that both retain their original operations but cooperate in tandem.

In the context of the problem, when we build a product of rings $R_1 \times R_2$, we also create a product of ideals $I_1 \times I_2$. Here, the operations within the ideals reflect merely as those of the original rings but within pairs. Each ring component is operating within its lane, but the results combined with the other create a new structure—a greater product ring.

In terms of addition: Components from each respective ideal add to stay within the bounds of their ideal.
For multiplication: A component from the ideal multiplied by any element in the ring still lies within the ideal.

These properties make the product of rings not just a simple combination of elements or operations. They define a robust structure that harmonizes the mathematical properties of both rings throughout their operation.

Problem 29

Problem 31

Other exercises in this chapter

Problem 28

Let $I$ be an ideal of $R$, and $S$ a subring of $R$. Show that $I \cap S$ is an ideal of $S$

View solution

Problem 29

Let $I$ be an ideal of $R,$ and $S$ a subring of $R .$ Show that $I+S$ is a subring of $R,$ and that $I$ is an ideal of $I+S$.

View solution

Problem 31

Write down the multiplication table for $\mathbb{Z}_{2}[X] /\left(X^{2}+X\right)$. Is this a field?

View solution

Problem 32

Let $I$ be an ideal of a ring $R,$ and let $x$ and $y$ be elements of $R$ with $x \equiv y(\bmod I)$. Let $g \in R[X]$. Show that \(g(x) \equiv g(

View solution