Problem 11 - A Computational Introduction to Number Theory and Algebra

Question

Let \(E\) be an extension field of a field \(F,\) and let \(\alpha_{1}, \ldots, \alpha_{n} \in E\) be algebraic over \(F\). Show that the ring \(F\left[\alpha_{1}, \ldots, \alpha_{n}\right]\) (see Example 7.45 ) is in fact a field, and that \(F\left[\alpha_{1}, \ldots, \alpha_{n}\right]\) is a finite (and hence algebraic) extension of \(F\).

Step-by-Step Solution

Answer

Question: Prove that if \(F\) is a field and \(\alpha_1, \ldots, \alpha_n \in E\) are algebraic over \(F\), then the smallest field containing \(F\) and all \(\alpha_i\) is also finite and algebraic over \(F\). Answer: The smallest field containing \(F\) and all \(\alpha_i\) can be represented as \(F[\alpha_1,\ldots,\alpha_n]\), which is a ring. To show that it is a field, we need to prove that it is closed under addition and multiplication, and that there are additive and multiplicative inverses for each element. After proving these properties, we need to show that it is finite and algebraic over \(F\). The ring is closed under addition and multiplication due to the structure of the elements in it, and additive and multiplicative inverses exist for each element since the extension is generated by algebraic elements. To demonstrate that it is a finite and algebraic extension, we can show that the dimension of the vector space \(F[\alpha_1,\ldots,\alpha_n]\) over \(F\) is finite and it is generated by algebraic elements over \(F\). Therefore, the smallest field containing \(F\) and all \(\alpha_i\) is finite and algebraic over \(F\).

Step 1: Define the ring \(F[\alpha_1,\ldots,\alpha_n]\)

Given a field \(F\) and algebraic elements \(\alpha_1, \ldots, \alpha_n \in E\), we define the ring \(F[\alpha_1,\ldots,\alpha_n]\) as the smallest ring containing \(F\) and all \(\alpha_i\). More concretely, we can represent \(F[\alpha_1,\ldots,\alpha_n]\) as a set of all finite sums and products of \(\alpha_i\) elements with coefficients in \(F\).

Step 2: Prove closed under addition and multiplication

Let \(a, b \in F[\alpha_1,\ldots,\alpha_n]\), then by definition, each of them can be written as finite sums of products of the form: $$a = \sum c_{i_1,\ldots,i_n}\alpha_1^{i_1}\cdots\alpha_n^{i_n} \quad\text{and}\quad b = \sum d_{i_1,\ldots,i_n}\alpha_1^{i_1}\cdots\alpha_n^{i_n},$$ where \(c_{i_1,\ldots,i_n}, d_{i_1,\ldots,i_n} \in F\). Now, we compute the sum and product of \(a\) and \(b\). Adding \(a\) and \(b\), we get: $$a + b = \sum (c_{i_1,\ldots,i_n} + d_{i_1,\ldots,i_n})\alpha_1^{i_1}\cdots\alpha_n^{i_n}.$$ By definition, \(c_{i_1,\ldots,i_n} + d_{i_1,\ldots,i_n} \in F\), so the sum of \(a\) and \(b\) is in the ring \(F[\alpha_1,\ldots,\alpha_n]\). Multiplying \(a\) and \(b\), we get: $$a\cdot b = \sum\sum c_{i_1,\ldots,i_n}d_{j_1,\ldots,j_n}(\alpha_1^{i_1+j_1}\cdots\alpha_n^{i_n+j_n}).$$ By definition, \(c_{i_1,\ldots,i_n}d_{j_1,\ldots,j_n} \in F,\) and the exponents are integer summations so the product of \(a\) and \(b\) is in the ring \(F[\alpha_1,\ldots,\alpha_n]\).

Step 3: Prove the existence of additive and multiplicative inverses

Since \(F[\alpha_1,\ldots,\alpha_n]\) contains \(F\), all elements in \(F\) have trivial additive inverses. Now let's find additive inverses for the remaining elements. Given \(a \in F[\alpha_1,\ldots,\alpha_n]\) as defined before, its additive inverse is: $$-a = \sum (-c_{i_1,\ldots,i_n})\alpha_1^{i_1}\cdots\alpha_n^{i_n}.$$ By definition, \(-c_{i_1,\ldots,i_n} \in F\), so the additive inverse of \(a\) is in the ring \(F[\alpha_1,\ldots,\alpha_n]\). Now, consider a nonzero element \(a \in F[\alpha_1,\ldots,\alpha_n]\). Since each \(\alpha_i\) is algebraic over \(F\), there exists a minimal polynomial \(p_i(x)\) such that \(p_i(\alpha_i)=0\) and it has coefficients in \(F\). Let \(p(x)\) be the polynomial resulting from taking the product of all the minimal polynomials of \(\alpha_i\). Then, \(p(\alpha_i)=0\) for all \(i\), and \(p(x)\) has coefficients in \(F\). Now consider the polynomial \(q(x)=xp(x)-1\). Since \(p(x)\) has coefficients in \(F\), so does \(q(x)\). Now, by applying the division algorithm, we have: $$q(x) = a(x)p(x) + b(x)$$ for some polynomials \(a(x), b(x) \in F[x]\). Evaluating both sides at \(a\), we get: $$0 = a(a)p(a) + b(a).$$ Thus, \(b(a)\) is the multiplicative inverse of \(a\), and since \(b(x) \in F[x]\), we have \(b(a) \in F[\alpha_1,\ldots,\alpha_n]\).

Step 4: Prove it is a finite and algebraic extension

To show that it is a finite extension, we will show that the dimension of \(F[\alpha_1,\ldots,\alpha_n]\) as a vector space over \(F\) is finite. Consider the basis of the vector space \(B = \{\alpha_1^{i_1}\cdots\alpha_n^{i_n} | 0\leq i_j

Key Concepts

Algebraic Extensions

An algebraic extension of a field is a type of extension wherein every element of the extension field is algebraic over the base field. That's a fancy way of saying each element satisfies a polynomial equation with coefficients from the base field. In the given exercise, each \( \alpha_i \) is described as algebraic over the field \( F \). This means there exists some polynomial \( p(x) \) with coefficients in \( F \) such that \( p(\alpha_i) = 0 \). This condition is what makes the elements "algebraic" rather than "transcendental" or non-algebraic.
To further understand, think of elements like \( \alpha_i \) as having a specific dependency on the simpler base field \( F \). Their existence within the extension relies on their relationship with \( F \) via this polynomial equation. Such extensions take a raw mathematical structure \( F \) and expand it into something richer, using these algebraic relationships.
Algebraic extensions are central in fields like number theory or algebraic geometry, offering a neat way to handle solutions to polynomial equations that wouldn't have fit into the original field \( F \).

Finite Extensions

When we say that a field extension is finite, we are discussing the concept of having a finite number of generators that span the extension relative to the base field. In layman's terms, it means you can count how many ways the extended field differs from the base field using simple arithmetic operations. In our exercise, \( F[\alpha_1, \ldots, \alpha_n] \) is claimed to be a finite extension of \( F \).
To illustrate, using algebraic elements \( \alpha_i \), you can create many combinations using sums and products, but the number of "linearly independent" combinations (or the basis of this extension as a vector space over \( F \)) is finite. Think of it like a set of LEGOs: you have a finite number of pieces (the \( \alpha_i \) elements), and only so many ways you can build with them within this expanded "mathematical space."
Finite extensions are significant because they often signify manageability and compactness in exploring solutions to polynomial equations, closing the gap between algebra and geometric interpretations in mathematics.

Minimal Polynomials

Minimal polynomials play a critical role in field extensions, especially in algebraic extensions, because they provide the simplest polynomial with coefficients in the base field that a given algebraic element satisfies. For an algebraic element like \( \alpha \), its minimal polynomial over a field \( F \) is defined as the smallest degree monic polynomial in \( F[x] \) with \( \alpha \) as a root.
In our exercise solution, understanding minimal polynomials helps us explore inverses in the extension field \( F[\alpha_1, \ldots, \alpha_n] \). The minimal polynomial governs the algebraic dependencies and ensures every element \( \alpha_i \) fits succinctly into the structure of \( F[\alpha_1, \ldots, \alpha_n] \) without redundancy. Knowing \( p(\alpha_i) = 0 \) aids in constructing and understanding this field extension by guiding how these elements "sit" with respect to \( F \).
Minimal polynomials also have practical and theoretical applications, serving as vital tools in coding theory, encryption algorithms, and simplifying complex polynomial interrelations in mathematics.

Previous

Problem 10

Next

Problem 12

Other exercises in this chapter

Problem 9

Show that if the factorization of \(f\) over \(F[X]\) into monic irreducibles is \(f=f_{1}^{e_{1}} \cdots f_{r}^{e_{r}},\) and if \(\alpha=[h]_{f} \in F[X] /(f)

View solution

Problem 10

In the field \(E\) in Example 16.16, find all the elements of degree 2 over \(\mathbb{Z}_{2}\)

View solution

Problem 12

Consider the real numbers \(\sqrt{2}\) and \(\sqrt[3]{2}\). Show that \((\mathbb{Q}[\sqrt{2}, \sqrt[3]{2}]: \mathbb{Q})=(\mathbb{Q}[\sqrt{2}+\sqrt[3]{2}]: \math

View solution

Problem 13

Consider the real numbers \(\sqrt{2}\) and \(\sqrt{3}\). Show that \((\mathbb{Q}[\sqrt{2}, \sqrt{3}]: \mathbb{Q})=(\mathbb{Q}[\sqrt{2}+\sqrt{3}]: \mathbb{Q})=4\

View solution

Key Concepts

Algebraic Extensions

An algebraic extension of a field is a type of extension wherein every element of the extension field is algebraic over the base field. That's a fancy way of saying each element satisfies a polynomial equation with coefficients from the base field. In the given exercise, each \( \alpha_i \) is described as algebraic over the field \( F \). This means there exists some polynomial \( p(x) \) with coefficients in \( F \) such that \( p(\alpha_i) = 0 \). This condition is what makes the elements "algebraic" rather than "transcendental" or non-algebraic.
To further understand, think of elements like \( \alpha_i \) as having a specific dependency on the simpler base field \( F \). Their existence within the extension relies on their relationship with \( F \) via this polynomial equation. Such extensions take a raw mathematical structure \( F \) and expand it into something richer, using these algebraic relationships.
Algebraic extensions are central in fields like number theory or algebraic geometry, offering a neat way to handle solutions to polynomial equations that wouldn't have fit into the original field \( F \).

Finite Extensions

When we say that a field extension is finite, we are discussing the concept of having a finite number of generators that span the extension relative to the base field. In layman's terms, it means you can count how many ways the extended field differs from the base field using simple arithmetic operations. In our exercise, \( F[\alpha_1, \ldots, \alpha_n] \) is claimed to be a finite extension of \( F \).
To illustrate, using algebraic elements \( \alpha_i \), you can create many combinations using sums and products, but the number of "linearly independent" combinations (or the basis of this extension as a vector space over \( F \)) is finite. Think of it like a set of LEGOs: you have a finite number of pieces (the \( \alpha_i \) elements), and only so many ways you can build with them within this expanded "mathematical space."
Finite extensions are significant because they often signify manageability and compactness in exploring solutions to polynomial equations, closing the gap between algebra and geometric interpretations in mathematics.

Minimal Polynomials

Minimal polynomials play a critical role in field extensions, especially in algebraic extensions, because they provide the simplest polynomial with coefficients in the base field that a given algebraic element satisfies. For an algebraic element like \( \alpha \), its minimal polynomial over a field \( F \) is defined as the smallest degree monic polynomial in \( F[x] \) with \( \alpha \) as a root.
In our exercise solution, understanding minimal polynomials helps us explore inverses in the extension field \( F[\alpha_1, \ldots, \alpha_n] \). The minimal polynomial governs the algebraic dependencies and ensures every element \( \alpha_i \) fits succinctly into the structure of \( F[\alpha_1, \ldots, \alpha_n] \) without redundancy. Knowing \( p(\alpha_i) = 0 \) aids in constructing and understanding this field extension by guiding how these elements "sit" with respect to \( F \).
Minimal polynomials also have practical and theoretical applications, serving as vital tools in coding theory, encryption algorithms, and simplifying complex polynomial interrelations in mathematics.