ℚ(∛5) is an Algebraic Field Extension of ℚ of Degree 3 (so it's a finite degree extension too!)

Let's learn about the field extension ℚ(∛5) = ℚ(5^(1/3))! First, it is the smallest subfield of the field of complex numbers ℂ (or the field of real numbers ℝ) containing all of the field of rational numbers ℚ and ∛5 = 5^(1/3). Second, because it is a zero (root) of f(x) = x^3 - 5 ∈ ℚ[x], which is irreducible over ℚ by Eisenstein's criterion with p = 5, and deg(f(x))=3, we can say that ℚ(∛5) = ℚ(5^(1/3)) = {c2*5^(2/3) + c1*5^(1/3)+c0*1 | c0, c1, c2 ∈ ℚ}. This means that ℚ(∛5) = ℚ(5^(1/3)) is a 3-dimensional vector space over ℚ (with basis {1,5^(1/3),5^(2/3)}). By definition, we say the degrees of ℚ(∛5) = ℚ(5^(1/3)) over ℚ is 3, and we write [ℚ(∛5):ℚ]=3. This is called a finite extension, and it implies that ℚ(∛5) is an algebraic extension of ℚ. In contrast, ℚ(π) is not an algebraic extension of ℚ because π is not an algebraic number over ℚ. It is a transcendental number and the extension is a transcendental field extension.

#AbstractAlgebra #FieldTheory #FieldExtension #ExtensionField

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