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A linear operator is semisimple if it becomes diagonalizable over an algebraic closure; equivalently, its minimal polynomial has no repeated root over that closure. If an invertible operator $A$ has finite order $r$ prime to the characteristic of a field, its minimal polynomial divides $X^r-1$. The derivative $r{X}^{r-1}$ is coprime to this polynomial, so $A$ is semisimple. Its invariant irreducible summands over a finite field become multiplication by scalars in suitable finite field extensions.

For an algebraic element $\alpha$ over a field $F$ of characteristic $p>0$, suppose first that $\alpha$ is a separable algebraic element. Its minimal polynomial over $E=F({\alpha }^p)$ divides its original separable polynomial and also $T^p-{\alpha }^p=(T-\alpha {)}^p$. It is therefore separable with only one distinct root, so has degree one and $\alpha \in E$. Conversely, if the irreducible minimal polynomial $m$ is inseparable, its formal derivative vanishes and $m(T)=g(T^p)$ for some $g$ of degree $\mathrm{deg}(m)/p$. Thus $[F({\alpha }^p):F]\le \mathrm{deg}(m)/p<[F(\alpha ):F]$, proving the converse.

An algebraic element $x$ over $K$ is separable when its minimal polynomial over $K$ has no repeated root in a splitting field. For an irreducible polynomial $m_x$, this is equivalent to $m_x^{\prime }\ne 0$, because then $\gcd (m_x,m_x^{\prime })=1$.

The identity $(BA{)}^{j+1}=B(AB{)}^{j}A$ gives $m_{BA}(t)\mid t{m}_{AB}(t)$. If $AB$ is diagonalizable, its minimal polynomial has simple roots. Hence all nonzero Jordan blocks of $BA$ have size one and its zero blocks have size at most two. Squaring kills these nilpotent blocks, making both squared products diagonalizable. Their eigenvalue multisets agree by Characteristic polynomials of AB and BA, so the squared products are similar matrices.

A simple field extension is a field extension generated by one element: $F(\alpha )$ is the smallest subfield of an ambient field containing $F$ and $\alpha$. Its elements are rational expressions in $\alpha$ with coefficients in $F$ and nonzero denominators. If $\alpha$ is algebraic, its minimal polynomial $m_{\alpha }$ identifies $F(\alpha )\cong F[t]/(m_{\alpha })$, so $[F(\alpha ):F]=\mathrm{deg}{m}_{\alpha }$. The classes of $1,t,\dots ,t^{d-1}$ give a basis when $d=\mathrm{deg}{m}_{\alpha }$.

In particular, ${\alpha }^2\in F$ but $\alpha \notin F$ implies $[F(\alpha ):F]=2$. Conversely, in characteristic different from two, any degree-two extension is of this form: for $\beta \notin F$, express ${\beta }^2=a+b\beta$ in the basis $1,\beta$, then $\alpha =\beta -b/2$ has ${\alpha }^2=a+b^2/4\in F$ and generates the same extension. The restriction on the characteristic makes division by two possible.

For a linear operator on a finite-dimensional vector space over a field $K$, all polynomial kernels stabilize at their first power exactly when its minimal polynomial $m$ has no repeated irreducible polynomial factor in $K[X]$. This formulation does not require those factors to be linear.

To prove it, the Bézout's identity for polynomials gives $\ker q(T)=\ker \gcd (q,m)(T)$: a greatest common divisor is a combination of $q$ and $m$, while $q$ is its multiple. If $m$ is square-free, $\gcd (p^n,m)=\gcd (p,m)$, proving the displayed equality. Conversely, if an irreducible factor $f$ occurs to exponent $e\ge 2$, its primary decomposition theorem summand has minimal polynomial $f^e$. On this summand $f(T)$ is nonzero, while $f(T{)}^e=0$. Thus the kernels of those two powers differ. An eigenbasis follows from this criterion precisely when the distinct factors of $m$ are also linear over $K$; a real rotation through a right angle has square-free $m=X^2+1$ but no real eigenvalues.

For an algebraic integer $\alpha$ generating a number field, the polynomial discriminant of its minimal polynomial $f$ equals the discriminant of elements of a number field for its power basis. A squarefree reduction modulo $p$ has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. The discriminant-index formula for an integral lattice, $\mathrm{disc}(f)=[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]{]}^2{d}_K$, proves the assertion.

// Target: arithmetic.bigb

Let $K={\mathbb{F}}_{2^d}$, let $\alpha$ generate $K^{\times }$, and let $T:K\to {\mathbb{F}}_2$ be nonzero linear with nondegenerate bilinear form $(x,y)\mapsto T(xy)$. If the minimal polynomial of $\alpha$ is $P(X)=X^d+{\sum }_{j<d}{c}_j{X}^j$, then $x_n=T({\alpha }^n)$ satisfiesso it is produced by an LFSR of length at most $d$. Its period is exactly $2^d-1$: any period $r$ would imply $T(({\alpha }^r-1)y)=0$ for every $y\in K$, hence ${\alpha }^r=1$ by nondegeneracy.