Elementary Equivalence Is (ℤ, +) Identical To (ℤ×ℤ, +)?
Introduction
Hey guys! Today, we're diving into a fascinating topic in mathematical logic: the elementary equivalence between two algebraic structures. Specifically, we'll be exploring whether the group of integers under addition, denoted as (ℤ, +), is elementarily equivalent to the group of pairs of integers under component-wise addition, denoted as (ℤ×ℤ, +). This question touches on the core concepts of model theory, allowing us to understand how logical sentences hold true in different mathematical structures. Let's embark on this journey together and unravel the mysteries of elementary equivalence!
Background: Elementary Equivalence
Before we jump into the specifics, let's clarify what elementary equivalence means. In simple terms, two structures are elementarily equivalent if they satisfy the same first-order sentences in a given language. A first-order sentence is a statement that can be expressed using quantifiers (∀, ∃), logical connectives (∧, ∨, ¬, →), and relations and functions from the language. For instance, in the language of groups L = {+}, we can express the sentence “there exists an identity element” as: ∃x ∀y (x + y = y ∧ y + x = y).
So, if (ℤ, +) and (ℤ×ℤ, +) are elementarily equivalent, it means that any first-order sentence that holds true in (ℤ, +) also holds true in (ℤ×ℤ, +), and vice versa. This is a powerful concept because it implies that the two structures are indistinguishable from the perspective of first-order logic. But how do we actually prove or disprove such a claim? That’s what we're here to figure out!
The Structures: (ℤ, +) and (ℤ×ℤ, +)
Let's take a closer look at our two structures.
(ℤ, +)
The first structure, (ℤ, +), is the set of all integers (..., -2, -1, 0, 1, 2, ...) equipped with the usual addition operation. This is a classic example of an infinite cyclic group. Key properties of (ℤ, +) include:
- Identity Element: 0 is the identity element, meaning that for any integer a, a + 0 = 0 + a = a.
- Inverse Elements: Every integer a has an inverse, -a, such that a + (-a) = (-a) + a = 0.
- Associativity: Addition is associative, i.e., for any integers a, b, and c, (a + b) + c = a + (b + c).
- Commutativity: Addition is commutative, i.e., for any integers a and b, a + b = b + a.
- Cyclic: (ℤ, +) is a cyclic group, generated by the element 1 (or -1). This means that every integer can be obtained by repeatedly adding 1 (or -1) to itself.
(ℤ×ℤ, +)
The second structure, (ℤ×ℤ, +), is the set of all ordered pairs of integers, where addition is performed component-wise. That is, (a, b) + (c, d) = (a + c, b + d). This structure is also a group, and it's isomorphic to the direct product of two copies of (ℤ, +). Here are some key properties of (ℤ×ℤ, +):
- Identity Element: (0, 0) is the identity element.
- Inverse Elements: The inverse of (a, b) is (-a, -b).
- Associativity: Component-wise addition is associative.
- Commutativity: Component-wise addition is commutative.
- Not Cyclic: Unlike (ℤ, +), (ℤ×ℤ, +) is not a cyclic group. You can't generate the entire group by repeatedly adding a single element to itself.
The Challenge: Proving or Disproving Elementary Equivalence
Now, the big question: are these two structures elementarily equivalent? To answer this, we need to either prove that they satisfy the same first-order sentences or find a sentence that distinguishes them. If we can find a sentence that holds true in one structure but not in the other, we've successfully disproven elementary equivalence.
Let's think about some properties that might help us differentiate between the two structures. We've already noted that (ℤ, +) is cyclic, while (ℤ×ℤ, +) is not. But can we express this difference using a first-order sentence? This is the crucial step in our investigation.
Finding a Distinguishing Sentence
The key difference between (ℤ, +) and (ℤ×ℤ, +) lies in their cyclic nature. (ℤ, +) is cyclic, meaning it can be generated by a single element (either 1 or -1). In contrast, (ℤ×ℤ, +) is not cyclic; no single element can generate the entire group through repeated addition. This distinction is our ticket to finding a distinguishing sentence.
To capture the non-cyclic nature of (ℤ×ℤ, +) in a first-order sentence, we can express the idea that “there is no element x such that every element y can be obtained by repeatedly adding x to itself.” However, formalizing “repeatedly adding” in first-order logic requires a bit of cleverness, since first-order logic doesn't directly support iteration or recursion.
Expressing Non-Cyclicity in First-Order Logic
Instead of directly expressing cyclicity, we can express its negation, which is the existence of an element of infinite order that is not a generator. In (ℤ, +), every non-zero element generates the entire group, while in (ℤ×ℤ, +), no single element can generate the entire group. This is because (ℤ×ℤ, +) requires at least two generators.
Consider the following sentence, which states that “for every element x, there exists an element y such that y is not in the subgroup generated by x.” To express this in first-order logic, we need to say that for any x, there exists a y such that for every natural number n, y is not equal to n times x (where n times x means x added to itself n times).
However, first-order logic cannot directly quantify over natural numbers or express repeated addition. Instead, we can use a clever trick. We can say that for every x, there exists a y such that for every n, y is not equal to nx. This captures the essence of non-cyclicity.
The Distinguishing Sentence
Let's formalize this idea into a first-order sentence in the language L = {+}. We'll use the following notation:
- ∃: There exists
- ∀: For all
- ¬: Not
- =: Equals
- +: Addition
To express