Problem 1 - A Course in Galois Theory

Question

Suppose that \(S\) is a set and \(R\) is a ring. Let \(R^{S}\) denote the set of all mappings from \(S\) to \(R\). Show that \(R^{S}\) is a ring, under the operations defined by $$ (f+g)(s)=f(s)+g(s), \quad(f g)(s)=f(s) g(s) $$ Show that if \(S\) has more than one element then there exist non-zero elements \(f\) and \(g\) in \(R^{S}\) for which \(f g=0\).

Step-by-Step Solution

Answer

Yes, \( R^S \) is a ring, and zero divisors exist if \( S \) has more than one element.

Step 1: Verify Addition Closure

To show that \( R^{S} \) is a ring, we first prove it is closed under addition. Let \( f, g \in R^{S} \). By definition, their sum is \( (f + g)(s) = f(s) + g(s) \) for all \( s \in S \). Since \( f(s) \) and \( g(s) \) are elements of the ring \( R \), and since rings are closed under addition, \( f(s) + g(s) \) is also in \( R \). Therefore, \( f+g \) is a mapping from \( S \) to \( R \), showing closure under addition.

Step 2: Verify Additive Identity

We need to show that there exists an additive identity in \( R^S \). Define a function \( 0 \in R^S \) by \( 0(s) = 0_R \) for all \( s \in S \), where \( 0_R \) is the additive identity in \( R \). For any \( f \in R^S \), \( (f + 0)(s) = f(s) + 0_R = f(s) \), so \( 0 \) acts as an additive identity in \( R^S \).

Step 3: Verify Additive Inverses

For each \( f \in R^S \), we need another function \( -f \) such that \( f + (-f) = 0 \). Define \( (-f)(s) = -f(s) \) for each \( s \in S \), where \( -f(s) \) is the additive inverse of \( f(s) \) in \( R \). Then \( (f + (-f))(s) = f(s) + (-f(s)) = 0_R \), and thus \( f + (-f) = 0 \), showing \( R^S \) contains additive inverses.

Step 4: Verify Multiplication Closure

Next, we show closure under multiplication. Let \( f, g \in R^S \). Their product as defined is \( (fg)(s) = f(s)g(s) \) for all \( s \in S \). Since \( R \) is a ring and closed under multiplication, \( f(s)g(s) \) is in \( R \), indicating \( fg \) maps \( S \) to \( R \), demonstrating closure under multiplication.

Step 5: Show Distributive Laws Hold

We verify the distributive laws in \( R^S \). For any \( f, g, h \in R^S \), we check \((f(g+h))(s) = f(s)(g(s) + h(s)) = f(s)g(s) + f(s)h(s) = ((fg) + (fh))(s)\), showing left distributivity. Similarly, \(((f+g)h)(s) = (f(s) + g(s))h(s) = f(s)h(s) + g(s)h(s) = ((fh) + (gh))(s)\) confirms right distributivity, proving both distributive properties hold.

Step 6: Identify Zero Divisors for Multiple Element Set

Assume \( S \) has more than one element, i.e., there exist \( a, b \in S \) with \( a eq b \). Define functions \( f, g \in R^S \) by \( f(a)=1_R, f(s)=0_R \) for \( s eq a \) and \( g(b)=1_R, g(s)=0_R \) for \( s eq b \). Then, \((fg)(s) = f(s)g(s) = 0_R \) for every \( s \in S \), indicating \( fg = 0 \) in \( R^{S} \), even though \( f \) and \( g \) are non-zero.

Key Concepts

Ring Theory

Ring Theory is a fundamental branch of abstract algebra. A ring, in mathematical terms, is a set equipped with two binary operations traditionally called addition and multiplication, where:

• it is closed under these operations,
• it has an additive identity (usually denoted as 0),
• each element has an additive inverse, and
• it satisfies the associative and distributive laws.

A ring doesn't necessarily have to have a multiplicative identity or multiplicative inverses, which differentiates it from a field.

In the context of the exercise, given a set \( S \) and a ring \( R \), the set \( R^{S} \) embodies all possible functions that map elements from \( S \) to \( R \). The major task of the exercise is to affirm that \( R^{S} \) also fulfills the criteria to be a ring when defined operations are applied. This involves checking if functions' sum and product defined function-wise meet closure, identity, and inverseness according to ring theory principles.

Function Spaces

Function spaces in mathematics are often sets of functions sharing a common domain and codomain, accompanied by specific operations.

In the provided exercise, \( R^{S} \) is a function space where each function maps a set \( S \) into a ring \( R \). When we refer to function spaces, we think of concepts such as vector spaces, where:

• addition of two functions is another function, and
• function multiplication is defined appropriately as per the problem context.

The operations in \( R^{S} \) are defined pointwise, meaning each element of \( S \) is considered separately under the function and the operations.

Moreover, function spaces like \( R^{S} \) help demonstrate how abstract ideas like rings can operate over sets, allowing us to explore constructive examples (like creating zero divisors) that uphold theoretical properties, enriching our understanding in more practical contexts.

Zero Divisors

Zero Divisors in ring theory are elements \( x \) and \( y \) such that \( x eq 0 \), \( y eq 0 \), yet their product is zero, i.e., \( xy = 0 \). They indicate that the structure lacks the cancellation property, a fascinating aspect in abstract algebra.

In the exercise case, you're tasked to show that if the set \( S \) comprises multiple elements, one can construct non-zero functions \( f \) and \( g \) in \( R^{S} \) that result in zero when multiplied. This is achieved by choosing specific points in \( S \) for \( f \) and \( g \) to assume non-zero (and divergent) values, causing their product over the entire domain to become zero.

• For example, consider \( f(a) = 1_R \) while \( f(s) = 0_R \) for \( s eq a \). Likewise, define \( g \) with \( g(b) = 1_R \) and \( g(s) = 0_R \) where \( s eq b \).
• Each function is non-zero since at certain elements they reflect 1 in \( R \).

Thus, \( (fg)(s) = f(s)g(s) = 0 \) across all \( s \) in \( S \) establishes the existence of zero divisors within function spaces derived from non-trivial rings, illustrating the rich complexity beyond standard number systems.