Building number systems by adjoining roots -- from the rationals to algebraic number fields
A field extension is built by starting with a base field (like the rationals Q) and adjoining new elements -- typically roots of polynomials. The resulting larger field inherits all the arithmetic of Q, plus the new element and everything you can build from it. Understanding these extensions is the foundation of Galois theory.
In this lesson, do arithmetic in field extensions, build extension towers, and discover how minimal polynomials control the structure of each extension.
Elements of Q(sqrt(d)) look like a + b*sqrt(d). Addition is straightforward. Multiplication uses the key relation: sqrt(d) * sqrt(d) = d. Try multiplying (1 + sqrt(2))(1 - sqrt(2)) to see how the irrational parts cancel.
Compute in the field extension Q(sqrt(2)). Elements have the form a + b*sqrt(2) where a, b are rational.
Key insight: The relation sqrt(d)^2 = d is exactly the minimal polynomial x^2 - d = 0 at work. Every element of Q(sqrt(d)) can be written uniquely as a + b*sqrt(d), making {1, sqrt(d)} a basis for Q(sqrt(d)) as a vector space over Q. The dimension of this vector space is the degree of the extension.
Build a tower of field extensions by successively adjoining elements. The tower law says that degrees multiply: [E:F] = [E:K] * [K:F]. Watch the total degree grow as each new element is adjoined.
Build a tower of field extensions starting from Q. Click elements to adjoin them and watch the tower grow.
Key insight: Adjoining sqrt(2) to Q gives degree 2. Adjoining sqrt(3) on top gives another factor of 2 (since sqrt(3) is not in Q(sqrt(2))), for total degree 4. But adjoining sqrt(6) on top of Q(sqrt(2), sqrt(3)) gives degree 1 -- it is already there as sqrt(2) * sqrt(3). This is why the tower law is multiplicative, not additive.
Every algebraic number has a unique minimal polynomial over Q: the lowest-degree monic polynomial with rational coefficients that it satisfies. The degree of this polynomial equals the degree of the field extension. Its other roots are the conjugates of the original element.
Select an algebraic element to see its minimal polynomial over Q, its degree, and all conjugate roots plotted on the complex plane.
Minimal polynomial over Q
x^2 - 2
Degree
2
Conjugate roots (2 total)
Why these conjugates?
sqrt(2) satisfies x^2 - 2 = 0, which is irreducible over Q by Eisenstein at p=2. The conjugate is -sqrt(2), obtained by the automorphism sqrt(2) -> -sqrt(2).
Key insight: The conjugates of sqrt(2) + sqrt(3) are +/-sqrt(2) +/- sqrt(3) (four values), giving a minimal polynomial of degree 4. This is because any automorphism of Q(sqrt(2), sqrt(3))/Q can independently flip the signs of sqrt(2) and sqrt(3), producing four distinct conjugates.