Question

Consider the ring \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) of functions \(f: \mathbb{R} \rightarrow \mathbb{R},\) with addition and multiplication defined point-wise. (a) Show that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain, and that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\) consists of those functions that never vanish. (b) Let \(a, b \in \operatorname{Map}(\mathbb{R}, \mathbb{R}) .\) Show that if \(a \mid b\) and \(b \mid a,\) then \(a r=b\) for some \(r \in \operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\) (c) Let \(\mathcal{C}\) be the subset of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) of continuous functions. Show that \(\mathcal{C}\) is a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R}),\) and that all functions in \(\mathcal{C}^{*}\) are either everywhere positive or everywhere negative. (d) Find elements \(a, b \in \mathcal{C},\) such that in the ring \(\mathcal{C},\) we have \(a \mid b\) and \(b \mid a,\) yet there is no \(r \in \mathcal{C}^{*}\) such that \(a r=b\).

Short answer

**Question:** Consider the set \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) of functions \(f: \mathbb{R}\rightarrow \mathbb{R}\). (a) Show that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain and find the invertible elements. (b) Show that if \(a \mid b\) and \(b \mid a\), then \(a = br\) for some invertible element \(r\). (c) Show that \(\mathcal{C}\), the set of continuous functions, is a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) and that all functions in \(\mathcal{C}^{*}\) are either everywhere positive or everywhere negative. (d) Find continuous functions \(a\) and \(b\) such that \(a \mid b\) and \(b \mid a\), but there is no invertible \(r\) in \(\mathcal{C}^{*}\) with \(a = br\). **Short Answer:** Directly showing that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain can be done using functions \(f(x)=x\) and \(g(x)=x-1\), which multiply to a zero function at two points. The invertible functions are those that never vanish. When \(a \mid b\) and \(b \mid a\), there exists an invertible element \(r\), such that \(a = br\). \(\mathcal{C}\), the set of continuous functions, is a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\), and all its invertible functions are either everywhere positive or everywhere negative. The continuous functions \(a(x)=x^2\) and \(b(x)=x^4\) satisfy \(a \mid b\) and \(b \mid a\), but \(a\) is not equal to \(br\) for any invertible function \(r\) in \(\mathcal{C}^{*}\) as \(r(x) = \frac{1}{x^2}\) is not continuous at \(x=0\).

Step-by-step solution

Showing \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain

Recall that an integral domain is a commutative ring with unity (a multiplicative identity) where the product of two non-zero elements is also non-zero. We can show that \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain by finding two nonzero functions \(f\) and \(g\) such that their product \(fg\) is zero. Consider the functions \(f(x) = x\) and \(g(x) = x-1\). Note that both \(f(x)\) and \(g(x)\) are nonzero functions. However, their product is \(fg(x) = x(x-1)\). Observe that \(fg(0) = fg(1) = 0\). Since the product is zero at least at two points, \(fg(x)\) is a zero function, and hence \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) is not an integral domain.

Finding invertible elements of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\)

To find the invertible elements (those functions that never vanish) of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\), we need to find the functions \(f(x)\) for which there exists another function \(g(x)\) such that the product \(fg(x) = 1\) for all \(x \in \mathbb{R}\). If \(f(x)\) is invertible, then it must never be zero since its product with \(g(x)\) is always 1. So, a function is invertible if and only if it never vanishes. Hence, the set of invertible elements is \(\operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\). (b)

Showing that if \(a | b\) and \(b | a,\) then \(a = br\) for some \(r \in \operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\)

If \(a | b\), then there exists a function \(r_1\) such that \(b = a r_1\). If \(b | a\), then there exists a function \(r_2\) such that \(a = b r_2\). Combining these, we get: \(a = b r_2 = (ar_1)r_2\). Now, we can cancel out the \(a\) term since both sides are non-zero: \(1 = r_1 r_2\). This shows that there exists an invertible element \(r = r_1 \in \operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\) such that \(a = br\) (c)

Showing that \(\mathcal{C}\) is a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\)

In order to show that \(\mathcal{C}\) is a subring of \(\operatorname{Map}(\mathbb{R},\mathbb{R})\), we need to show that if \(f,g \in \mathcal{C}\), then \((f+g)\) and \((f \cdot g)\) are also in \(\mathcal{C}\). First, let \(f,g\) be continuous functions. The sum and product of continuous functions are also continuous (by elementary real analysis). Thus, \(\mathcal{C}\) is closed under addition and multiplication and hence forms a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\).

Characterizing functions in \(\mathcal{C}^{*}\)

A continuous function \(f(x)\) is in \(\mathcal{C}^{*}\) if and only if it never vanishes, as we found for \(\operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\). By the intermediate value theorem applied to continuous functions, if \(f(x)\) changes sign, then it must vanish at some point in its domain. Therefore, the functions in \(\mathcal{C}^{*}\) must be either everywhere positive or everywhere negative. (d)

Finding elements \(a\) and \(b\) in \(\mathcal{C}\), such that \(a|b\), \(b|a\), but \(a \ne br\) for any \(r \in \mathcal{C}^{*}\)

Let \(a(x) = x^2\) and \(b(x) = x^4\) be elements in \(\mathcal{C}\). Observe that both \(a\) and \(b\) are continuous functions since they are powers of continuous functions. We can see that \(a|x^2\), so \(a|a^2\), and thus \(a|b\). Similarly, \(b|x^4\), so \(b|b^2\), and thus \(b|a\). Both \(a\) and \(b\) divide each other, but there does not exist an \(r \in \mathcal{C}^{*}\) such that \(a=br\). To see this, suppose such function \(r(x)\) exists such that \(a(x) = br(x)\). Then we would have \(x^2 = x^4 r(x)\). For all \(x \ne 0\), we would have \(r(x) = \frac{1}{x^2}\). However, \(r(x)\) is not continuous at \(x=0\), and thus is not in \(\mathcal{C}^{*}\). As a result, no such \(r\) exists.

Key concepts

Commutative Ring with Unity

In the realm of abstract algebra, a commutative ring with unity is an algebraic structure that includes a set equipped with two binary operations: addition and multiplication. For a ring to qualify as a commutative ring with unity, it must satisfy several properties. First, it must be commutative, meaning that the order in which two elements are multiplied does not affect the product; mathematically, for all elements a and b in the ring, it holds that ab = ba.

Secondly, there must be a unity (also known as a multiplicative identity), an element that, when multiplied by any element in the ring, leaves that element unchanged. In symbols, for an element 1 in the ring and any element a, it holds that 1a = a1 = a.

It is essential to understand that while the commutative ring \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\) in our exercise indeed has a unity (the constant function 1), it fails to be an integral domain, another refined type of ring, because it does not satisfy the additional requirement that the product of any two non-zero elements be non-zero. We identify this failure through the existence of non-zero functions whose product at certain points is zero, thus violating a core condition for an integral domain.

Invertible Elements of a Ring

In the context of ring theory, invertible elements or units of a ring are those elements that have a multiplicative inverse within the ring. Specifically, for an element a to be invertible, there must exist another element b within the same ring such that ab = ba = 1, where 1 is the unity of the ring.

In the case of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\), the set of invertible elements, denoted as \(\operatorname{Map}(\mathbb{R}, \mathbb{R})^{*}\), is comprised of those functions that never equal zero at any point in their domain. The absence of zero values ensures that an inverse function exists, whereby the product of the two functions results in the constant function 1 across the entire domain of real numbers. This property indicates that in the case of function rings, finding invertibles equates to finding functions that do not have any roots in their domain.

Subring

The notion of a subring is pivotal in understanding the hierarchical structure within ring theory. A subring is essentially a subset of a ring that is, in itself, a ring with respect to the same operations of addition and multiplication defined on the original ring. For a subset S of a ring R to qualify as a subring, it must contain the unity of R, be closed under the ring operations (meaning that the sum and product of any elements in S still belong to S), and have additive inverses for all its elements within S.

In our exercise, the set \(\mathcal{C}\) of continuous functions is shown to be a subring of the larger ring \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\). This is because the sum and product of continuous functions remain continuous, hence \(\mathcal{C}\) is closed under these operations, and the additive inverses of functions in \(\mathcal{C}\) are the negations of these functions, which are also continuous. Demonstrating that a subset forms a subring helps constrain our focus to a more manageable and specific part of a ring, often for the purpose of deeper analysis or simplification.

Continuous Functions in Ring Theory

In the intersection of ring theory and real analysis, the study of continuous functions within rings has special significance. A function f from the real numbers \(\mathbb{R}\) to itself is said to be continuous if, intuitively, small changes in the input lead to small changes in the output, without sudden jumps or breaks. In the language of rings, a set of continuous functions can form a ring themselves if they satisfy the required ring operations.

As encountered in our exercise, the subset \(\mathcal{C}\) of continuous functions is indeed a subring of \(\operatorname{Map}(\mathbb{R}, \mathbb{R})\). Within this subring, the invertible elements, which are those functions without any real roots, possess an intriguing property – they must maintain the same sign over their entire domain. This characteristic is derived from the intermediate value theorem of real analysis, which declares that if a continuous function changes sign, it must cross zero at some point, thereby invalidating its invertibility. The exploration of continuous functions in ring theory not only brings a geometric flavor to the rather algebraic domain but also bridges these two mathematical realms by applying analytic concepts to algebraic structures.