Problem 41 - A Computational Introduction to Number Theory and Algebra

Question

Let \(R\) be a ring, and let \(I\) and \(J\) be ideals of \(R\). With the ringtheoretic product as defined in Exercise \(7.2,\) show that: (a) \(I J\) is an ideal; (b) if \(I\) and \(J\) are principal ideals, with \(I=a R\) and \(J=b R,\) then \(I J=a b R\), and so is also a principal ideal; (c) \(I J \subseteq I \cap J\) (d) if \(I+J=R,\) then \(I J=I \cap J\).

Step-by-Step Solution

Answer

To summarize, we have demonstrated various properties concerning the product of ideals in a ring. We showed that the product of two ideals is an ideal, we proved that if both ideals are principal, their product is a principal ideal. Furthermore, we established the inclusion \(IJ \subseteq I \cap J\) and equality \(IJ = I \cap J\) under the condition \(I + J = R\). These properties of the product of ideals are useful in the study of rings and their ideals, as they help us understand how different ideals interact with each other and the structure of the ring.

Step 1: (a) Proving that \(IJ\) is an ideal

Let \(I\) and \(J\) be ideals of a ring \(R\), and let \(IJ\) denote their ring-theoretic product. To show that \(IJ\) is an ideal, we must verify the following two conditions: 1. It is closed under subtraction. 2. It is closed under multiplying by any element from the ring. Let's first prove that \(IJ\) is closed under subtraction. Consider two elements \(x_1, x_2 \in IJ\). By definition of the ring-theoretic product, we have \(x_1 = \sum_{i=1}^n a_i b_i\) and \(x_2 = \sum_{i=1}^m c_i d_i\), where \(a_i, c_i \in I\) and \(b_i, d_i \in J\). Thus, \[x_1 - x_2 = \sum_{i=1}^n a_i b_i - \sum_{i=1}^m c_i d_i = \sum_{i=1}^n a_i b_i + \sum_{i=1}^m (-c_i) d_i,\] where \(-c_i \in I\) since \(I\) is an ideal. Since the sums individually belong to \(IJ\), their sum, which is \(x_1-x_2\), is also an element of \(IJ\). Hence, \(IJ\) is closed under subtraction. Now, let's show that \(IJ\) is closed under multiplying by any element from the ring. Consider an element \(x \in IJ\) and an element \(r \in R\). Then, we have \(x = \sum_{i=1}^n a_i b_i\), where \(a_i \in I\) and \(b_i \in J\). Thus, \[r \cdot x = r \cdot \sum_{i=1}^n a_i b_i = \sum_{i=1}^n (r \cdot a_i) b_i,\] where \(r \cdot a_i \in I\) since \(I\) is an ideal. Therefore, the sum is an element of \(IJ\), so \(IJ\) is closed under multiplying by any element from the ring. Since \(IJ\) is closed under subtraction and multiplication by any element from the ring, it is an ideal.

Step 2: (b) Proof for principal ideals

Let \(I\) and \(J\) be principal ideals of a ring \(R\), with \(I=aR\) and \(J=bR\). We want to show that \(IJ = abR\) and is a principal ideal. Let \(x \in IJ\), then by the definition of the product of ideals, we have \[x = \sum_{i=1}^n a_i b_i,\] where \(a_i \in I\) and \(b_i \in J\). Since \(I=aR\) and \(J=bR\), we can rewrite the \(a_i\)'s and \(b_i\)'s as \(a_i = ar_i\) and \(b_i = bs_i\) for some \(r_i, s_i \in R\). Then, \[x = \sum_{i=1}^n (ar_i)(bs_i) = ab \sum_{i=1}^n r_is_i,\] which is an element of \(abR\). Hence, \(IJ \subseteq abR\). Conversely, let \(y \in abR\), so \(y = abz\) for some \(z \in R\). Clearly, \(y = a(bz)\) and \(a \in I, bz \in J\), so \(y\) is of the form of an element in \(IJ\). Therefore, \(abR \subseteq IJ\). Since \(IJ \subseteq abR\) and \(abR \subseteq IJ\), it follows that \(IJ = abR\), so the product of principal ideals is also a principal ideal.

Step 3: (c) Inclusion \(IJ \subseteq I \cap J\)

Let \(x \in IJ\), then by definition of the product of ideals, we have \[x = \sum_{i=1}^n a_i b_i,\] where \(a_i \in I\) and \(b_i \in J\). Each term \(a_i b_i\) is an element of \(I\) (since \(I\) is an ideal) and an element of \(J\) (since \(J\) is an ideal). Letting \(m = \sum_{i=1}^n a_i b_i\), we have that \(m \in I \cap J\). Therefore, \(x \in I \cap J\), and we have the desired inclusion: \(IJ \subseteq I \cap J\).

Step 4: (d) Equality \(IJ = I \cap J\) under the condition \(I + J = R\)

We've already shown that \(IJ \subseteq I \cap J\) in part (c). To prove the reverse inclusion under the given condition, let \(x \in I \cap J\). Since \(I + J = R\), there exists \(i \in I\) and \(j \in J\) such that \(i + j = 1\). Thus, we have \[x = x(i+j) = xi + xj,\] where \(xi \in I\) (since \(I\) is an ideal), and \(xj \in J\) (since \(J\) is an ideal). Therefore, \(x = xi + xj\) is a sum of terms of the form \(ab\), with \(a \in I\) and \(b \in J\), and \(x \in IJ\). This gives us the desired inclusion: \(I \cap J \subseteq IJ\). Since both inclusions hold, \(IJ = I \cap J\) when \(I + J = R\).

Key Concepts

Ideal

In ring theory, an ideal is a special subset of a ring that allows the construction of quotient rings, which are crucial for understanding the structure of rings. Given a ring \(R\), a subset \(I\) is called an ideal if it satisfies the following two properties:

• **Closure under addition**: For any elements \(a, b \in I\), the sum \(a - b\) is also in \(I\). This property indicates that the set of ideals remains closed under the operation of subtraction.
• **Closure under ring multiplication**: For any element \(a \in I\) and any \(r \in R\), the product \(r \, a\) is in \(I\). This means multiplying an ideal element by any ring element still gives an element in the ideal.

These conditions ensure that ideals can interact with other elements of the ring in a controlled manner, forming a foundation for other algebraic constructions. Ideals also help in classifying homomorphisms, which are functions preserving the structure between algebraic objects.

Principal Ideals

A principal ideal is a special type of ideal in a ring, generated by a single element. If \(R\) is a ring and \(a\) is an element of \(R\), the principal ideal generated by \(a\) is denoted \(aR\) and is defined as:

• The set \(\{ar \mid r \in R\}\).

This means every element in the principal ideal is obtained by multiplying the generator \(a\) with various elements from the ring \(R\).
Principal ideals are straightforward but powerful. They allow for simplifications in understanding more complex ring structures. For instance, in the ring of integers \(\mathbb{Z}\), any ideal is principal and can be generated by a single integer.Considerations of principal ideals become particularly interesting in rings where not all ideals are principal, like in the ring of polynomials or number fields. Understanding how principal ideals work and relate to each other helps solve equations and factor polynomials or integers.

Ring-theoretic Product

The ring-theoretic product of two ideals \(I\) and \(J\) in a ring \(R\) is a way to combine these ideals into a new ideal. This product, denoted \(IJ\), is defined as:

• The set of all finite sums of products of the form \(ab\), where \(a \in I\) and \(b \in J\).

To formally express this, an element \(x \in IJ\) is written as \(x = \sum_{i=1}^{n} a_i b_i\) for \(a_i \in I\) and \(b_i \in J\). This definition originates from the idea that elements of \(IJ\) can be built by summing over all possible products one can form by taking one member from \(I\) and one from \(J\).
This product has the following important properties:

• It constructs an ideal that is always contained within the intersection \(I \cap J\).
• When \(I\) and \(J\) are principal, the product \(IJ\) is also principal, generated by the product of the generators of \(I\) and \(J\).

The ring-theoretic product is a versatile tool in algebra for structuring rings and ideals, as well as in proofs involving ring homomorphisms and factorizations.