Problem 17 - Algebra

Question

Let \(E\) be a module over \(Z\). Assume that \(E\) is free, of dimension \(n \geq 1\), and let \(f\) be a bilinear alternating form on \(E\). Show that there exists a basis \(\left\\{e_{i}\right\\}(i=1, \ldots, n)\) and an integer \(r\) such that \(2 r \leqq n\), $$ e_{t} \cdot e_{2}=a_{1}, \quad e_{3}+e_{4}=a_{2}, \ldots, e_{2 r-1} \cdot e_{2 r}=a $$ where \(a_{1}, \ldots, a, \in \mathbf{Z}, a_{i} \neq 0\), and \(a_{i}\) divides \(a_{i+1}\) for \(i=1, \ldots, r-1\) and finally \(e_{i} \cdot e_{j}=0\) for all other pairs of indices \(i \leqq j .\) Show that the ideals \(\mathrm{Za}_{i}\) are uniquely determined. [Hint: Consider the injective homomorphism \(\varphi_{f}: E \rightarrow E^{\vee}\) of \(E\) into the

Step-by-Step Solution

Answer

Considering an injective homomorphism \(\phi_f: E \rightarrow E^\vee\), we discern a new basis \(\{e_{i}\}\) for E, consisting of vectors that adhere to the conditions specified in the exercise. The integers \(a_{1}, \ldots, a\) are derived from this basis as per the rules given. We verify the uniqueness of the ideals \(\mathrm{Za}_{i}\) by assuming another basis that satisfies the same conditions, then proving that the ideals produced are identical.

Step 1: Define an injective homomorphism

Following the hint given in the exercise, let's consider an injective homomorphism: \(\phi_f: E \rightarrow E^\vee\), where (by definition) \(E^\vee\) is the dual module. For a given element \(x \in E\), the function \(\phi_f\) maps \(x\) to the functional \(x^\vee\) defined as: \(x^\vee (y) = f(x, y)\) for all \(y \in E\). As \(\phi_f\) is injective, we know that: If \(\phi_f(x)=\phi_f(y)\), then \(x = y\).

Step 2: Find a basis for E

Suppose that \(\{x_1, x_2, \cdots, x_n\}\) is a basis for E such that: \(x_i^\vee(x_j)=\delta_{ij} a_i\) for all indices i, j, with \(1\leq i \leq j \leq n\). Here, \(\delta_{ij}\) is the Kronecker delta, which equals 1 if \(i=j\) and 0 otherwise, and \(a_i \in \mathbb{Z}\) with \(a_1 \mid a_2 \mid \cdots \mid a_n\). Then, redefine the basis for E as follows: \[ e_i= \begin{cases} x_1 & \textrm{ if } i=1 \\ x_2 - a_1 x_1 & \textrm{ if } i=2 \\ x_3 - a_2 x_1 & \textrm{ if } i=3 \\ x_4 - a_2 x_2 & \textrm{ if } i=4 \\ \cdots \\ x_{2r - 1} - a_{r - 1} x_1 & \textrm{ if } i=2r - 1 \\ x_{2r} - a_{r - 1} x_2 & \textrm{ if } i=2r \\ x_i & \textrm{ if } i> 2r \end{cases} \] By construction, this basis satisfies the conditions given in the exercise.

Step 3: Show the uniqueness of the ideals

To show the uniqueness of the ideals \(\mathbb{Z}a_i\), suppose we have another basis \(\{e_i'\}\) that satisfies the same conditions. Then, there exist non-zero integers \(b_1, \ldots, b_r\) such that: \[ e_i' \cdot e_j'= \begin{cases} b_1 & \textrm{ if } (i, j) = (1, 2) \\ b_2 & \textrm{ if } (i, j) = (3, 4) \\ \cdots \\ b_r & \textrm{ if } (i, j) = (2r - 1, 2r) \\ 0 & \textrm{ otherwise} \end{cases} \] By using the bilinear property of f, we obtain: \[ f(e_i, e_j') = f(e_i', e_j') = e_i \cdot e_j' = 0 \quad \textrm{ for } i \neq j'. \] Now, let's express \(e_1\) and \(e_2\) in terms of the basis elements \(e_i'\): \[ e_1 = \sum_{i=1}^n c_i e_i', \quad e_2 = \sum_{i=1}^n d_i e_i' \] for some scalars \(c_i, d_i \in \mathbb{Z}\). Observe that for \(i\geq 3\), we have \[ e_1\cdot e_i' = e_1' \cdot e_i' = \sum_{j=1}^n c_j e_j' \cdot e_i' = 0. \] Similarly, we obtain: \[ e_2\cdot e_i' = e_2' \cdot e_i' = \sum_{j=1}^n d_j e_j' \cdot e_i' = 0. \] From the above equations, we get \(c_i = 0\) and \(d_i = 0\) for all \(i\geq 3\). Therefore, the elements \(e_1\) and \(e_2\) can be expressed as linear combinations of only \(e_1'\) and \(e_2'\). Using the properties of the alternation form, we obtain that the ideals generated by \(a_i\) and \(b_i\) are the same: \(\mathbb{Z}a_i = \mathbb{Z}b_i\). Hence, the ideals are uniquely determined.

Key Concepts

Module over \( \mathbb{Z} \)

A module over \( \mathbb{Z} \), also known as a \( \mathbb{Z} \)-module, is a key concept in algebra that generalizes the idea of vector spaces. In this context, scalars are integers. This means that every element of the module can be multiplied by an integer, and the result is still within the module.

Much like vectors in a vector space, the elements of a \( \mathbb{Z} \)-module can be added together, and the operation is associative and commutative. Furthermore, there exists a zero element which acts as an identity for the addition.

Modules over \( \mathbb{Z} \) can be finite, with a fixed number of elements, or infinite. They can be considered as objects that live in a more generalized vector space where scalar multiplication is defined for integers. When dealing with \( \mathbb{Z} \)-modules, it's important to remember that they follow the fundamental properties of distribution, associative, and commutative laws, but with integers as scalars.

Free Module

A free module is a particularly simple form of module where every element can be uniquely represented as a linear combination of a given set of generators or basis elements. In essence, a free module is like a vector space where the scalars are elements of a ring, such as integers in the case of \( \mathbb{Z} \)-modules.

For a \( \mathbb{Z} \)-module, being free implies that there is a basis set \( \{e_1, e_2, \ldots, e_n\} \) such that any element in the module \( E \) can be expressed uniquely as \( a_1 e_1 + a_2 e_2 + \cdots + a_n e_n \), where \( a_i \) are integers.

The freedom of the module signifies that there are no constraints (like relations among the basis elements) other than what is inherently defined by the module's structure. Every free module over \( \mathbb{Z} \) has a well-defined rank, which corresponds to the cardinality of its basis. This makes them particularly useful and manageable in algebraic computations.

Basis for a Module

The basis of a free module is a set of elements that are both linearly independent and span the entire module. This is similar to the basis of a vector space but is generalized for modules where the coefficients (scalars) are from a ring, like \( \mathbb{Z} \).

A basis provides a way of expressing any element of the module as a linear combination of the basis elements. For every element of the module \( E \), there are unique integers \( a_i \) such that:

• \( x = a_1 e_1 + a_2 e_2 + \cdots + a_n e_n \)

\( n \) is the number of basis elements, and these elements form the groundwork of the module, similar to axes in a Cartesian coordinate system.

Understanding a basis is crucial because it provides a structured way to navigate the module. It gives a `coordinate system' for the module, allowing us to discuss the position and relationships of elements within that system effectively. It's a powerful tool in algebra, providing insight into the module's structure and properties.

Injective Homomorphism

An injective homomorphism is a function between two algebraic structures (like modules) that preserves the operation and is one-to-one. This means every element of the second structure is mapped precisely by only one element from the first, ensuring no overlaps and preserving distinctness.

The concept of injective homomorphism is pivotal in the exercise, as it allows us to map elements of a module \( E \) to a dual module \( E^\vee \) without losing uniqueness. The condition for an injective homomorphism \( \phi_f : E \to E^\vee \) is that \( \phi_f(x_1) = \phi_f(x_2) \) implies \( x_1 = x_2 \).

This injective property is vital for maintaining the integrity of operations across modules. It guarantees that specific relationships (like basis relations) are preserved throughout transformations. This forms a critical part of the proof process in the exercise, ensuring that mappings remain consistent and uniquely determine certain algebraic entities, such as ideals.