Problem 66 - A Computational Introduction to Number Theory and Algebra

Question

Let \(\mathbb{Q}^{(m)}\) be the subring of \(\mathbb{Q}\) defined in Example \(7.26 .\) Let us define the map \(\rho: \mathbb{Q}^{(m)} \rightarrow \mathbb{Z}_{m}\) as follows. For \(a / b \in \mathbb{Q}\) with \(b\) relatively prime to \(m, \rho(a / b):=[a]_{m}\left([b]_{m}\right)^{-1}\). Show that \(\rho\) is unambiguously defined, and is a surjective ring homomorphism. Also, describe the kernel of \(\rho\).

Step-by-Step Solution

Answer

In this exercise, we found that the map \(\rho : \mathbb{Q}^{(m)} \rightarrow \mathbb{Z}_m\) is well-defined, meaning it produces a unique result for each input. We also showed that \(\rho\) is a surjective ring homomorphism, satisfying both the addition and multiplication properties and covering the whole range of \(\mathbb{Z}_m\). Finally, we described the kernel of \(\rho\) as the set of elements in \(\mathbb{Q}^{(m)}\) whose numerator is a multiple of \(m\).

Step 1: Show that \(\rho\) is unambiguously defined

Let \(a_1/b_1\) and \(a_2/b_2\) be two elements in \(\mathbb{Q}^{(m)}\) with both \(b_1\) and \(b_2\) relatively prime to \(m\). Suppose \(a_1/b_1 = a_2/b_2\). Then, we can say that \(a_1b_2 = a_2b_1\). To show that \(\rho\) is unambiguously defined, we need to prove that \(\rho(a_1/b_1) = \rho(a_2/b_2)\): \(\rho(a_1/b_1) = [a_1]_{m}([b_1]_{m})^{-1} = [a_2]_{m}([b_2]_{m})^{-1} = \rho(a_2/b_2)\) Since \(a_1b_2 = a_2b_1\), we can say that \([a_1b_2]_m = [a_2b_1]_m\). We also have the fact that the inverse of \(([b_1]_m)_m\) is unique. Therefore, it follows that \(\rho(a_1/b_1) = \rho(a_2/b_2)\) meaning \(\rho\) is unambiguously defined.

Step 2: Show that \(\rho\) is a surjective ring homomorphism

To show that \(\rho\) is a surjective ring homomorphism, we need to prove the following statements: 1. \(\rho\) is a ring homomorphism: We need to show that \(\rho (a + b) = \rho (a) + \rho (b)\) and \(\rho (ab) = \rho (a) \rho (b)\). Let \(a/b, c/d \in \mathbb{Q}^{(m)}\). Then, \(\rho((a/b) + (c/d)) = \rho((ad + bc) / (bd)) = [(ad + bc)]_m ([bd]_m)^{-1} = [ad]_m + [bc]_m = \rho(a/b) + \rho(c/d)\) And for the multiplication, \(\rho((a/b)(c/d)) = \rho((ac)/(bd)) = [ac]_m ([bd]_m)^{-1} = ([a]_m [b]_m^{-1})([c]_m [d]_m^{-1}) = \rho(a/b) \rho(c/d)\) 2. \(\rho\) is surjective: To show that \(\rho\) is surjective, we need to show that for every \(r \in \mathbb{Z}_m\), there exists an element \(q \in \mathbb{Q}^{(m)}\) such that \(\rho(q) = r\). Let \(r \in \mathbb{Z}_m\), then we can choose \(q = r/1 \in \mathbb{Q}^{(m)}\) since \(1\) is relatively prime to \(m\). Then, we have \(\rho (r/1) = [r]_m = r\), which means that \(\rho\) is surjective. Combining the results, we can conclude that \(\rho\) is a surjective ring homomorphism.

Step 3: Describe the kernel of \(\rho\)

The kernel of \(\rho\) is given by the set of elements in \(\mathbb{Q}^{(m)}\) that map to the identity element (0) in \(\mathbb{Z}_m\): \(\ker(\rho) = \{q \in \mathbb{Q}^{(m)} \mid \rho(q) = 0\}\) Let \(q = a/b \in \mathbb{Q}^{(m)}\) with \(b\) relatively prime to \(m\). Then, we have \(\rho(a/b) = [a]_m ([b]_m)^{-1} = 0\). This implies that \([a]_m = [0]_m\), which further implies that \(a \equiv 0 \pmod{m}\). Therefore, \(a\) is a multiple of \(m\). So, the kernel of \(\rho\) is given by: \(\ker(\rho) = \{a/b \in \mathbb{Q}^{(m)} \mid a \equiv 0 \pmod m\}\)

Key Concepts

Subring

A subring is a subset of a ring that is itself a ring under the same operations. For example:

• Given a ring \( R \), a subring \( S \) must contain the additive identity of \( R \), be closed under addition and multiplication, and include the inverses under addition.

In the given exercise, \( \mathbb{Q}^{(m)} \) is described as a subring of the rational numbers \( \mathbb{Q} \) that forms a ring with the operations of addition and multiplication defined in \( \mathbb{Q} \).
Here, the key challenge is ensuring that the fractions \( a/b \) involve denominators \( b \) that are relatively prime to \( m \). This condition allows certain properties, like invertibility in modulo \( m \), to be straightforwardly applied. This subring is essential for defining the map \( \rho \) and examining its properties such as surjectivity and kernel.

Kernel

The kernel of a ring homomorphism is the set of elements in the domain that map to the zero element in the codomain. It's crucial because it helps identify how the homomorphism behaves.

• For a homomorphism \( \rho: R \rightarrow S \), the kernel \( \ker(\rho) \) is defined as all \( r \in R \) such that \( \rho(r) = 0 \) in \( S \).

In this exercise, the kernel of \( \rho \) consists of fractions \( a/b \) in \( \mathbb{Q}^{(m)} \) where \( [a]_m = 0 \). To achieve this, \( a \) must be a multiple of \( m \).
This shows that all elements in the kernel are mapped to zero in \( \mathbb{Z}_m \), making the kernel a crucial aspect of understanding the structure of \( \rho \).

Surjective

A function is called surjective if every element in the codomain has a pre-image in the domain. Surjectivity is a key property for ring homomorphisms, indicating the function covers the entire codomain.

• For \( \rho: R \rightarrow S \), \( \rho \) is surjective if \( \forall s \in S, \exists r \in R \text{ such that } \rho(r) = s \).

In our exercise, \( \rho \) is shown to be surjective by proving that each element \( r \) in \( \mathbb{Z}_m \) has a corresponding element \( q = r/1 \) in \( \mathbb{Q}^{(m)} \).
This means for every conceivable output in the codomain, there's an input in the domain, making the function robust and comprehensive. Surjectivity helps affirm that the mapping is complete and overlaps every component of \( \mathbb{Z}_m \).