Normal Extensions and Field Automorphisms
Understand normality as closure under conjugates, and distinguish embeddings from automorphisms.
On this page
The question for today
What you’ll understand
- Test normality using irreducible polynomials and splitting fields
- Determine the allowed images of field generators
- Explain why normality and separability are separate hypotheses
An algebraic generator comes with other roots of its minimal polynomial. Choosing one root constructs a field, but it does not guarantee that the field contains its companions. Normality closes this gap. It says that a field extension cannot contain one root of an irreducible polynomial over the base while omitting another. This is the condition that turns embeddings into internal symmetries, provided we keep track of which base field must remain fixed.
Containing a whole family of conjugates
An algebraic extension L/K is normal when every irreducible polynomial in K[x] with a root in L splits completely in L. The quantifier concerns irreducible polynomials over K; demanding that every polynomial over K split would be a much stronger claim. For example, Q(sqrt(2)) is normal over Q because it is the splitting field of x squared minus two. It certainly does not contain roots of every rational polynomial.
A finite extension is normal precisely when it is a splitting field of some polynomial over its base. For the forward direction, choose finitely many algebraic generators and multiply their minimal polynomials: normality supplies every conjugate, while the chosen generators already generate the field. Conversely, a base-field embedding permutes the roots of a splitting polynomial, so it preserves the whole field they generate. This embedding viewpoint is often easier to use than testing infinitely many polynomials.
Images are constrained by polynomial relations
A K-automorphism of L is a bijective field homomorphism from L to itself that fixes each element of K. Its value on a generator must be another root of the generator's minimal polynomial. This is a necessary condition, but several generators can satisfy additional relations, so their images cannot always be chosen independently. The safe procedure is to present the field by generators and relations, then verify that each proposed assignment respects every relation.
The real field Q(cube root of two) illustrates the missing distinction. Its generator has three distinct complex conjugates, yet only one is real. Any Q-automorphism must therefore fix the generator, so the automorphism group is trivial even though the extension has degree three. Its other Q-embeddings land in different subfields of the complex numbers. Adjoining a primitive cube root of unity supplies the missing roots and produces a larger normal extension.
Normal does not mean separable
A finite Galois extension is both normal and separable. These requirements do different jobs: separability ensures enough distinct embeddings, and normality keeps their images inside one field. Together they imply that the automorphism group has size equal to the extension degree. Characteristic zero makes separability automatic, which is why it can disappear from rational-number examples without disappearing from the theorem itself.
In characteristic p, the extension F_p(t)(t to the power one over p) is normal because x to the p minus t has only one distinct root and already splits in that extension. It is inseparable of degree p, and its base-fixing automorphism group is trivial. Thus a count of symmetries is also a diagnostic: too few automorphisms may signal missing conjugates, repeated conjugates, or both. Identify which hypothesis failed before applying a Galois theorem.
Work it through
Four symmetries of a biquadratic extension
Let L = Q(sqrt(2), sqrt(5)). Determine its degree and all Q-automorphisms, checking that the two sign choices really are independent.
If sqrt(5) were a + b sqrt(2) with rational a and b, squaring would give 5 = a squared + 2b squared + 2ab sqrt(2). Irrationality forces ab = 0; neither a squared = 5 nor b squared = 5/2 has a rational solution. Therefore sqrt(5) is not in Q(sqrt(2)), and the tower degree is 2 times 2 = 4.
The field is the splitting field of (x squared minus 2)(x squared minus 5), so it is normal; characteristic zero gives separability. Every automorphism sends sqrt(2) to either sign of sqrt(2), and sqrt(5) to either sign of sqrt(5). The four assignments preserve the defining relations and extend by substitution on the basis 1, sqrt(2), sqrt(5), sqrt(10).
Composing two sign changes multiplies their signs independently. Each nonidentity map has order two, so the group is C2 times C2. The map changing both signs fixes sqrt(10), whereas the maps changing exactly one sign negate it. This checks the action on a product rather than counting assignments alone.
The extension has degree four and exactly four automorphisms. Its internal symmetries capture every embedding because the extension is normal and separable.
The living notebook
Explore the model
Interactive lab · quantum
An invariant can forget a transformation
Set the mixing angle to 90 degrees and vary the relative phase. Predict that both basis probabilities stay one half while the Bloch coordinates change. This is a limited analogy for information retained by an invariant: the two-level state is not a field extension or an automorphism computation.
- Probability |0⟩
- 71.1%
- Probability |1⟩
- 28.9%
- Bloch (x, y, z)
- 0.78, 0.45, 0.42
A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. 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 forAn unchanged observation need not mean that the underlying object stayed fixed; field automorphisms must instead preserve every field operation and the chosen base.
Pause & reason
Check your understanding
Your turn
Try it yourself
Determine whether Q(sqrt(3), i) is normal over Q, find its degree, and count its Q-automorphisms.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use a splitting polynomial, then check whether i already belongs to the real field Q(sqrt(3)).
Compare with the solution
It is the splitting field of (x squared minus 3)(x squared plus 1), so it is normal and separable. The real quadratic field Q(sqrt(3)) cannot contain i, so adjoining i contributes another factor of two and the degree is four. The independent assignments sqrt(3) to either sign of sqrt(3) and i to either sign of i give four automorphisms. Complex conjugation is the one that fixes sqrt(3) and negates i.
Keep the idea
Take it with you
- Normality retains all conjugates of every irreducible polynomial that acquires a root
- An embedding can land in another field even when no nontrivial internal automorphism exists
- Finite Galois extensions require both normality and separability
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