What is q-linear embedding?

Question

Let h be a non-negative integer and f: $F_{q^k}$->$F_{q^{k+h}}$ be a q-linear embedding. Then $a_tf(x^{q^t})+a_{t-1}f(x^{q^{t-1}})+...+a_0f(x)$ is a q-linear mapping from $F_{q^k}$ to $F_{q^{k+h}}$.

Questions:

1. What is q-linear embedding? I know in the field extension, embedding a field A to B means extend A by adjoining elements to B?
2. How to understand the q-linear mapping?

Answer 1

The elements here seem to be:

• A (fixed) $\Bbb{F}_q$-linear mapping $f:\Bbb{F}_{q^k}\to\Bbb{F}_{q^{k+h}}$. This is just a linear transformation from a $k$-dimensional space to a $(k+h)$-dimensional space. Observe that in general $\Bbb{F}_{q^k}$ is not a subfield of $\Bbb{F}_{q^{k+h}}$, so there is no natural embedding available. We simply fix some linear transformation with a trivial kernel (to make it injective). The mapping $f$ is not an embedding of fields in the sense that it does not preservec products, it is an embedding of vector spaces such that the vector spaces happen to be fields. That may sound like splitting hairs, but we cannot guarantee that this would be an embedding of fields in the natural sense, and that confuses many readers of your question.
• The Frobenius automorphism $\phi:x\mapsto x^q$ of the domain $\Bbb{F}_{q^k}$. It is also $\Bbb{F}_q$-linear. Hence so are its compositions with itself $\phi\circ\cdots\circ\phi:x\mapsto x^{q^i}$, $i$ some natural number. And the compositions of iterates of $\phi$ and $f$, $x\mapsto f(x^{q^i})$. These latter mappings have $\Bbb{F}_{q^{k+h}}$ as their domain.
• The mappings $y\mapsto ay$ from $\Bbb{F}_{q^{k+h}}$ to itself. Here $a$ is a constant from the field $\Bbb{F}_{q^{k+h}}$, and this is just multiplication in that field. The distributivity law then gives immediately that these mappings are linear over the small field $\Bbb{F}_q$.
• Consequently a mapping of the form $x\mapsto af(x^{q^i})$ is an $\Bbb{F}_q$-linear transformation from $\Bbb{F}_{q^k}$ to $\Bbb{F}_{q^{k+h}}$ as a composition of such.

A mapping of the form $$x\mapsto \sum_{i=0}^ta_i f(x^{q^i})$$ is thus $\Bbb{F}_q$-linear as a sum of such.

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As a smallest example consider the case $q=2$, $k=2$, $h=1$. The domain is the field $\Bbb{F}_4$ that is a vector space over $\Bbb{F}_2$ with basis $\{1,\alpha\}$, where $\alpha$ satisfies the equation $\alpha^2=\alpha+1$ (which pretty much defines the multiplication in $\Bbb{F}_4$). The codomain is the field $\Bbb{F}_8$ which we can think of as a 3-dimensional vector space over $\Bbb{F}_2$ with a basis $\{1,\beta,\beta^2\}$, with the multiplication determined by $\beta^3=\beta+1$.

In this toy example we can use (for example) $$f:\Bbb{F}_4\to\Bbb{F}_8, a_0+a_1\alpha\mapsto a_0+a_1\beta^2$$ as an $\Bbb{F}_2$-linear "embedding". After all, this obviously has a trivial kernel, so it is an embedding of vector spaces. The linear transformations your source seems to be describing are sums of mappings like $$ x\mapsto \beta f(x^2). $$ With the chosen $f$ we would get $$ \begin{aligned} a_0+a_1\alpha&\mapsto \beta f(a_0+a_1\alpha^2)\\ &=\beta f(a_0+a_1(1+\alpha))\\ &=\beta f([a_0+a_1]+a_1\alpha)\\ &=\beta([a_0+a_1]f(1)+a_1 f(\alpha))\\ &=\beta(a_0+a_1)+\beta^3a_1\\ &=\beta(a_0+a_1)+(1+\beta)a_1\\ &=a_1+a_0\beta, \end{aligned} $$ making it plain that this mapping is $\Bbb{F}_2$-linear.