Question

Let \(E\) be an extension of \(F\). Show that the set of all elements of \(E\) that are algebraic over \(F\) is a subfield of \(E\) containing \(F\).

Short answer

Question: Show that the set of all algebraic elements in an extension \(E\) over a base field \(F\) forms a subfield of \(E\) containing \(F\). Answer: To prove this, we have shown that the set of algebraic elements is closed under addition, subtraction, multiplication, and division (excluding division by zero), contains the identity elements 0 and 1, and every element in \(F\) is algebraic over \(F\). Thus, the set of all elements of \(E\) that are algebraic over \(F\) is a subfield of \(E\) containing \(F\).

Step-by-step solution

Prove that the set of algebraic elements contains additive inverses

Let \(e_1, e_2 \in E\) be algebraic over \(F\). We have to show that \(e_1 - e_2\) is also algebraic over \(F\). Since \(e_1\) and \(e_2\) are algebraic, we know there exist polynomials \(p(x)\) and \(q(x)\) in \(F[x]\) such that \(p(e_1) = 0\) and \(q(e_2) = 0\). Now consider the polynomial \(h(x) = p(x)q(x)\), which is also in \(F[x]\). We can write \(h(e_1 - e_2) = p(e_1)q(e_1) - p(e_2)q(e_2)\). Since \(p(e_1) = 0\) and \(q(e_2) = 0\), we have \(h(e_1 - e_2) = 0\), which means that \(e_1 - e_2\) is algebraic over \(F\). Thus, the set of algebraic elements is closed under subtraction.

Prove that the set of algebraic elements is closed under addition

Now we have to show that the set of algebraic elements is closed under addition. From Step 1, we know that \(e_1 - (-e_2)\) is algebraic. But we also know that \(-e_2\) is algebraic, since \((-1)e_2\) is algebraic (\((-1)\) belongs to \(F\)). So \(e_1 + e_2\) is also algebraic.

Prove that the set of algebraic elements is closed under multiplication

Let \(e_1\) and \(e_2\) be algebraic elements over \(F\). We need to show that \(e_1e_2\) is also algebraic over \(F\). Since \(e_1\) and \(e_2\) are algebraic, there exist polynomials \(p(x)\) and \(q(x)\) in \(F[x]\) such that \(p(e_1) = 0\) and \(q(e_2) = 0\). Now consider the polynomial \(h(x) = p(x) \cdot q(x)\), which is in \(F[x]\). Now, \(p(e_1 e_2) \cdot q(e_1 e_2) = p(e_1)q(e_1)e_2^2 = 0\), which means that \(e_1e_2\) is algebraic over \(F\). Thus, the set of algebraic elements is closed under multiplication.

Prove that the set of algebraic elements is closed under division (excluding division by zero)

Let \(e_1\) and \(e_2\) be algebraic elements over \(F\), with \(e_2 \neq 0\). We have to show that \(\frac{e_1}{e_2}\) is also algebraic over \(F\). Since both \(e_1\) and \(e_2\) are algebraic, there exist polynomials \(p(x)\) and \(q(x)\) in \(F[x]\) such that \(p(e_1) = 0\) and \(q(e_2) = 0\). We can now create the polynomial \(h(x) = p(x)q(1/x) = \sum_{i,j} c_{ij}x^i(1/x)^j\). Since all coefficients \(c_{ij}\) are in \(F\), \(h(x)\) is in \(F[x]\). Now, we have \(h(\frac{e_1}{e_2}) = p(\frac{e_1}{e_2})q(\frac{1}{\frac{e_1}{e_2}}) = p(\frac{e_1}{e_2})q(\frac{e_2}{e_1})\). We know that \(p(e_1) = 0\), so \(p(\frac{e_1}{e_2})q(\frac{e_2}{e_1}) = 0\), which implies that \(\frac{e_1}{e_2}\) is algebraic over \(F\). Thus, the set of algebraic elements is closed under division, excluding division by zero.

Prove that the set of algebraic elements contains the identity elements 0 and 1

To show that the set of algebraic elements contains the identity elements 0 and 1, we have to show that there exist polynomials in \(F[x]\) such that \(p(0) = 0\) and \(q(1) = 0\). Consider the polynomial \(p(x) = x\), which is in \(F[x]\). We have \(p(0) = 0\), so 0 is indeed algebraic. Now consider the polynomial \(q(x) = x-1\), which is also in \(F[x]\). We have \(q(1) = 0\), so 1 is algebraic as well. Thus, the set of algebraic elements contains the identity elements 0 and 1.

Prove that every element in \(F\) is algebraic over \(F\)

Let \(f \in F\). We have to show that \(f\) is algebraic over \(F\). Consider the polynomial \(p(x) = x - f\), which is in \(F[x]\). We have \(p(f) = f - f = 0\), so \(f\) is algebraic over \(F\). Thus, the set of algebraic elements contains \(F\). By showing that the set of algebraic elements is closed under addition, subtraction, multiplication, and division (excluding division by zero), contains the identity elements 0 and 1, and every element in F is algebraic over F, we have proven that the set of all elements of \(E\) that are algebraic over \(F\) is a subfield of \(E\) containing \(F\).

Key concepts

Field Theory

In field theory, fields are mathematical structures where you can add, subtract, multiply, and divide any two elements, provided the divisor is not zero. Fields are central to algebra and number theory.
Each field extension is a larger field that contains a smaller field. These extensions are particularly useful for studying solutions to polynomial equations.
For any field extension, specific conditions must be met for it to be considered a subfield, which often involves ensuring closure under operations like addition and multiplication.

Subfields

A subfield is a subset of a field that is also a field itself. To determine if a subset of a field is a subfield, it must satisfy several properties:

• It should include the same zero and one elements of the larger field.
• It must be closed under addition, subtraction, and multiplication.
• It must be closed under division, except for division by zero.

In the context of algebraic extensions, the set of elements that are algebraic over a given field forms a subfield. This means these elements comply with the rules of field operations within the extension.

Polynomials

Polynomials are expressions consisting of variables and coefficients. They are fundamental to understanding algebraic extensions. They allow us to define what it means for an element to be algebraic over a field.
A polynomial with coefficients in a field can define the "roots" that satisfy it. If an element satisfies a polynomial equation with coefficients in the base field, it indicates that the element is algebraic over that field.

• For example, if a polynomial equation like \(p(x) = x^2 - 2\) has roots that are elements of an extension field, those elements are algebraic over the base field.

By studying polynomials, we can learn about the relationships between different fields and uncover deeper algebraic structures.

Algebraic Elements

Algebraic elements over a field \(F\) are elements in a larger field (or extension), \(E\), that satisfy a polynomial equation with coefficients in \(F\). To be algebraic means there exists a non-zero polynomial such that substituting the element into the polynomial yields zero.
These elements help define the structure of field extensions. For example, if \(e\) in \(E\) satisfies a polynomial \(p(x)\) in \(F[x]\), then \(e\) is algebraic over \(F\). They form subfields under certain operations, demonstrating arithmetic closure within an extension. Understanding algebraic elements is key to grasping how larger fields can be structured and explored.