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Let $L=K(\sqrt{a_1},\sqrt{a_2})$ with independent square classes and characteristic different from two, and let $G$ be generated by the automorphism negating both square roots. The two orbit polynomials are $t^2-a_1$ and $t^2-a_2$, so their coefficients generate only $K$, whereas

Multiplication by a generator of ${\mathbb{F}}_{q^n}^{\times }$ on the ${\mathbb{F}}_q$-space ${\mathbb{F}}_{q^n}$ gives a cyclic subgroup regular on nonzero vectors. Its normalizer contains the field Frobenius automorphism. In degree three over ${\mathbb{F}}_2$ the full normalizer has order 21.

For an automorphism $\sigma$ of a left Noetherian ring $A$, the skew polynomial ring of an automorphism $A[t;\sigma ]$ is left Noetherian. For a left ideal $L$, the normalized leading-coefficient ideals $I_n=\{{\sigma }^{-n}(a_n):{\sum }_{j=0}^n{a}_j{t}^j\in L\}$ form an ascending chain of left ideals of $A$. Choose finite generating lifts through the stabilization degree. Every element of $L$ has its leading coefficient cancelled by a suitable left multiple of one of these lifts, and induction on degree proves finite generation. Passing to the skew Laurent polynomial ring preserves the property by clearing negative powers. The right-sided statement follows by applying the argument to the opposite ring.

Let $A$ have no zero divisors and $\sigma$ be an automorphism. For nonzero Laurent polynomials $f=\sum a_i{x}^i$ and $g=\sum b_j{x}^j$ with greatest exponents $m,n$, the coefficient of $x^{m+n}$ in $fg$ is $a_m{\sigma }^m(b_n)$. No other exponent pair contributes to that highest degree. Both factors are nonzero, so their product is nonzero, proving the noncommutative domain property in every characteristic.

For an automorphism $\sigma$ of a unital ring $A$, this ring consists of finite sums ${\sum }_{j\ge 0}{a}_j{t}^j$, with multiplication determined by $ta=\sigma (a)t$. Both left and right coefficient normal forms are available because $\sigma$ is invertible. The noncommutative Hilbert basis theorem implies that it is a right Noetherian ring when $A$ is, and a left Noetherian ring when $A$ is. Invertibility of $\sigma$ matters in this two-sided formulation.

A linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

A topological abelian group $A$ with a continuous action of a topological group $G$ by group automorphisms. Continuity means joint continuity of $G\times A\to A$. For a discrete coefficient group this is equivalent to every element of $A$ having an open stabilizer.

Suppose a group of field automorphisms acts transitively on elements $y_i$, and each $y_i$ has a unique $p$th root $x_i$ in a stable extension. If every automorphism extends, it sends $x_i$ to the unique root above its image of $y_i$, so the extended group acts transitively on the $x_i$.

A formal associative deformation is trivial if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}(\mathrm{mod}t)$ fixing the identity transports the deformed multiplication to the original one. This is equivalence by a formal change of coordinates, stronger than merely an abstract algebra isomorphism between unspecified middle terms.

A determinant-preserving unitary bundle automorphism covering the identity on the base. For an $SU(2)$ bundle its values are in the fiberwise special unitary group. We use $A^u=u^{-1}Au+u^{-1}du$. Its curvature transforms by $F_{A^u}=u^{-1}{F}_{A}u$. The automorphism is a change of unitary frame, not a change of the base point.

// Destination: geometry-and-topology.bigb

A universal homogeneous model realizes every type over a small parameter set. Embed a small model containing a realization into the universal model, then use homogeneity to move the embedded parameters back to the original ones.