Lagmental Vicfred

Linearized Polynomials Encode F_q-Linear Maps by Vicfred

tags: mathematics, finite fields, Frobenius, normal bases

Polynomials with q-power exponents act additively and F_q-linearly on extension fields. I want the notation, the mechanism, and the failure mode visible at the same time.

Notation

For \(q=p^r\), the Frobenius map \(F(x)=x^q\) controls extensions of \(\mathbf F_q\). Its orbits determine minimal polynomials, trace, norm, and the Galois group.

$$ L(x)=a_0x+a_1x^q+\cdots+a_mx^{q^m} $$

There are two layers here: the object \(\mathsf D\) and the law \(\mathsf C\). Writing them separately makes the direction of \(\Longrightarrow\) visible and keeps an accidental converse from slipping in.

$$ L(x+y)=L(x)+L(y),\qquad L(cx)=cL(x)\quad(c\in\mathbf F_q) $$

Stress the formula

Now evaluate one representative case. The result should agree with the structural law above, but it is obtained without assuming the conclusion.

$$ \ker L\le(\overline{\mathbf F}_q,+),\qquad\#\ker L\le q^m\quad(a_m\ne0) $$

Interpretation

The formula is reusable precisely because it says which pieces are structural and which belong only to the worked example.

$$ \begin{aligned} \mathsf{D}\;&:\quad L(x)=a_0x+a_1x^q+\cdots+a_mx^{q^m},\\[5pt] \mathsf{C}\;&:\quad L(x+y)=L(x)+L(y),\qquad L(cx)=cL(x)\quad(c\in\mathbf F_q). \end{aligned} $$

Limit of the argument

Frobenius is \(\mathbf F_q\)-linear on an extension but not generally linear over a larger coefficient field. Exponents must match the chosen base.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] L(x+y)=L(x)+L(y),\qquad L(cx)=cL(x)\quad(c\in\mathbf F_q) \end{gathered}} $$

I would use the boxed line as a reference later, while returning to the full display whenever a hypothesis becomes uncertain. That division keeps compression from becoming ambiguity.

This article was posted on Thu 19 February 2026. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.