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Every finite cyclic extension has a normal basis
Statement
Let be a finite Galois extension whose Galois group is cyclic, say of order . Then has a normal basis (Normal bases of a finite Galois extension): there is for which
is an ordered -basis of (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).
Facts & Assumptions
Given: A finite Galois extension with cyclic of order ; by [L6] , and is an -linear endomorphism of the -vector space , written when regarded as such. Evaluation of a polynomial at is that of Polynomial evaluation at an endomorphism: .
Let be a group and a field. Every finite family of distinct group homomorphisms is linearly independent over as a family of functions (Dedekind's linear independence theorem for distinct characters).
For every endomorphism of a finite-dimensional -vector space, is a nonzero ideal with a unique monic generator , and if and only if (The annihilator ideal is nonzero and has a unique monic generator; if and only if , The annihilator set ; once existence is proved, its unique monic generator is the minimal polynomial).
For the polynomial is monic of degree ( is monic of degree ; for its coefficient is and its constant coefficient is , while ); is defined as for any ordered basis and is independent of it (The basis-independent characteristic polynomial of an endomorphism of a finite-dimensional space, including in dimension zero).
for every endomorphism of a finite-dimensional vector space (The minimal polynomial divides the characteristic polynomial, ).
An endomorphism of a finite-dimensional vector space has a cyclic vector if and only if (A cyclic vector exists exactly when the minimal and characteristic polynomials agree); is a cyclic vector when is all of (Cyclic subspaces, cyclic vectors, and vector annihilators).
For a finite Galois extension with one has (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and , The degree of a finite field extension).
With the unique monic generator of (The vector annihilator is the unique monic generator of and divides the minimal polynomial) and , the list is an ordered basis of (A vector annihilator gives a power basis and its companion matrix).
Proof
, because has order in ; so the polynomial lies in and by [L2]. In particular .
The maps are pairwise distinct elements of , and their restrictions to are pairwise distinct group homomorphisms , since two field automorphisms of agreeing on agree on .
If with satisfies , then is the zero function on , hence on , so [L1] applied to the family of step 1.2 forces every to be ; thus no nonzero polynomial of degree less than annihilates , and .
Combining steps 1.1 and 2.1, , and since is monic and divides the monic of the same degree, .
By [L3] the polynomial is monic of degree , and by [L4]; two monic polynomials of the same degree, one dividing the other, are equal, so .
By [L5] there is a cyclic vector for , that is .
Let . By [L7] the list is an ordered basis of , which has dimension , so and is an ordered -basis of .
Since , that list is exactly the family of conjugates of , so it is a normal basis.
Remarks
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Why the minimal polynomial is forced to be . The divisibility is cheap; the content is the lower bound on its degree, and that is exactly Dedekind's independence of characters. Without it the minimal polynomial could be a proper divisor of and no cyclic vector would be available.
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The hypothesis is on the group, not on the base field. No finiteness or infiniteness of is used, so this proof also covers cyclic extensions of infinite fields, for which Every finite Galois extension of an infinite field has a normal basis gives a second and quite different argument.
Depends on
- Normal bases of a finite Galois extension
- Dedekind's linear independence theorem for distinct characters
- The annihilator set $\operatorname{Ann}(T)=\{p\in F[x]:p(T)=0\}$; once existence is proved, its unique monic generator $\mu_T$ is the minimal polynomial
- The annihilator ideal is nonzero and has a unique monic generator; $p(T)=0$ if and only if $\mu_T\mid p$
- The minimal polynomial divides the characteristic polynomial, $\mu_T\mid\chi_T$
- A cyclic vector exists exactly when the minimal and characteristic polynomials agree
- Cyclic subspaces, cyclic vectors, and vector annihilators
- The vector annihilator is the unique monic generator of $\operatorname{Ann}_T(v)$ and divides the minimal polynomial
- A vector annihilator gives a power basis and its companion matrix
- The basis-independent characteristic polynomial $\chi_T$ of an endomorphism of a finite-dimensional space, including $\chi_T=1$ in dimension zero
- $\chi_A(x)$ is monic of degree $n$; for $n\geq1$ its $x^{n-1}$ coefficient is $-\operatorname{tr}(A)$ and its constant coefficient is $(-1)^n\det(A)$, while $\chi_{0\times0}=1$
- Polynomial evaluation at an endomorphism: $p(T)=\sum_k a_kT^k$
- Finite Galois extensions and $\operatorname{Gal}(K/F)$
- Equivalent characterizations of a finite Galois extension
- Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis
- The degree $[K:F]=\dim_F K$ of a finite field extension
Used by
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Sources
- P. L. Clark, Field Theory (course notes/monograph), Proposition 8.22 (standard reference, not scraped)
- K. Conrad, Linear Independence of Characters (expository blurb), Theorem 3.7 (standard reference, not scraped)