Field extensions: adjoining a number without losing arithmetic
Distinguish algebraic and transcendental elements, construct a generated field, and calculate inside a quadratic extension.
On this page
The question for today
What you’ll understand
- Describe a field extension as an inclusion with compatible operations.
- Construct the field generated by one algebraic element.
- Separate algebraicity of individual elements from finiteness of an extension.
Solving an equation can require a larger arithmetic universe. The rational numbers contain neither root of x² − 5, although the equation uses only rational coefficients. Adjoining one root is more disciplined than declaring a new symbol: the enlarged system must support addition, multiplication, and division by every nonzero element. A field extension records this enlargement while keeping the original field visible. The base field matters because the same number can satisfy different algebraic constraints over different coefficient fields.
An extension remembers which numbers were already available
Write L/F when F is a subfield of L and both fields use the same addition, multiplication, zero, and identity. This notation does not mean a quotient. Rational numbers sit inside real numbers, and real numbers inside complex numbers. A more focused example is the smallest subfield of the real numbers containing both the rationals and √5. The word smallest means intersection of every subfield containing those ingredients, so it specifies a unique generated field.
For an element α in some larger field, F(α) allows all rational expressions in α with coefficients in F whose denominators are nonzero. By contrast, F[α] initially allows only polynomial expressions. They coincide when α is algebraic, but not for a transcendental element t: polynomial expressions in t cannot include 1/t. Keep these two constructions separate until a relation proves that the needed inverses can be reduced to polynomials.
A polynomial relation makes powers reusable
An element is algebraic over F if some nonzero polynomial with coefficients in F vanishes at it. The word nonzero excludes the polynomial that vanishes at everything. Since (√5)² = 5, every higher power of √5 reduces to a rational multiple of either 1 or √5. Sums and products therefore remain in the set of expressions a + b√5. To establish a field, division still needs an argument; closure under multiplication alone is insufficient.
Multiply a + b√5 by its conjugate a − b√5. Their product is a² − 5b², which is rational. If this product were zero with rational a and b not both zero, either b would be zero and force a = 0, or a/b would be a rational square root of 5. Neither is possible. The conjugate therefore gives an inverse for every nonzero expression. This exact calculation explains why adjoining this algebraic element requires only two rational coordinates.
Algebraic does not mean finite, and it is always relative
An element can be transcendental over one field and algebraic over another. The real number π is transcendental over the rationals, but belongs to the coefficient field Q(π), so there it satisfies X − π. Algebraicity is a relation between an element and a base field, not an absolute label attached to a written symbol. An extension is algebraic when every element in it is algebraic over the specified base.
A finite-dimensional field extension is algebraic: the powers 1, α, α², and so on must eventually be linearly dependent over the base field. The converse needs a finite generation hypothesis. The field of all algebraic numbers is algebraic over Q but has no finite dimension over Q, because it contains roots of irreducible polynomials with arbitrarily large degree. A later lesson will turn the first linear dependence among powers into a precise measure of how much adjoining α costs.
Work it through
Divide in Q(√5)
Compute (2 + √5)/(1 + √5) and express the answer using rational coefficients of 1 and √5.
The denominator is nonzero because √5 is positive. Multiply numerator and denominator by 1 − √5, the conjugate of the denominator.
The numerator becomes 2 − 2√5 + √5 − 5 = −3 − √5. The denominator becomes 1 − 5 = −4.
Divide both coefficients by −4 to obtain 3/4 + √5/4. Multiplying this answer by 1 + √5 returns 2 + √5, which checks the calculation in the original field.
The quotient is (3 + √5)/4, an element of Q(√5) with two rational coordinates.
The living notebook
Explore the model
Interactive lab · vectors
Coordinates need a chosen scalar field
The lab controls the angle and magnitude of one real two-dimensional vector and displays its two projections. Predict which projection changes sign when the angle crosses 90°, then test it. Use this to distinguish an object from its coordinates; Q(√5) also has two coordinates over Q, but this real-vector display does not perform field multiplication or test rational independence.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA coordinate picture helps only after the basis and allowed scalars have been specified.
Pause & reason
Check your understanding
Your turn
Try it yourself
Show that 1/(2 + √3) belongs to Q(√3), and explain why this does not place √3 in Q.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Multiply by 2 − √3 and keep track of the coefficient of √3.
Compare with the solution
The product (2 + √3)(2 − √3) is 4 − 3 = 1, so the inverse is 2 − √3. Its coefficients are rational, which places it in Q(√3). Membership in the extension permits a nonzero √3 coefficient; membership in Q does not. Irrationality of √3 ensures the two coordinate representations cannot coincide.
Keep the idea
Take it with you
- A field extension is an inclusion together with a specified base field.
- An algebraic relation reduces powers and can make inverses polynomial expressions.
- An algebraic extension need not be finite-dimensional.
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