Structure of linear functionals in finite fields

Question

If $(V,\langle\rangle)$ is an inner product space and $f: V\rightarrow\mathbb R$ is a linear functional, then $f$ has representation $f(x)=\langle v,x\rangle$ for a unique $v$ of norm $1$. This is known as Reisz's representation.

My question: Is there an analogous characterization of $K$-linear functionals $f: L\rightarrow K$ where $K\subset L$ is an extension of finite fields?

Answer 1

The closest thing that comes to mind is a characterization using the trace.

Let $|K|=q$, $|L|=Q=q^n$. Then the (relative) trace mapping is defined as $$ tr_{L/K}:L\to K, x\mapsto x+x^q+x^{q^2}+\cdots+x^{q^{n-1}}=\sum_{i=0}^{n-1}x^{q^i}. $$ It is easy to show that $tr_{L/K}$ is A) $K$-linear and B) surjective. It follows that

• To each $z\in L$, the mapping $f_z:L\to K$, $f_z(x)=tr_{L/K}(zx)$ is $K$-linear.
• If $z\neq0$ then multiplication by $z$ is a surjective $K$-linear transformation $L\to L$, hence $f_z$ is also surjective and therefore non-zero.
• Consequently the $K$-vector space of functions $f_z$ has dimension $n=\dim Hom_K(L,K)$.
• By rank-nullity all the functionals $f:L\to K$ are of the form $f=f_z$ for some $z$.

Summary: Instead of $\langle z,x\rangle$ we can use $tr_{L/K}(zx)$.

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A few closing remarks:

• There is no concept of a norm (in the sense of functional analysis) here.
• We could actually use any function $L\to K$ satisfying A and B instead of $tr_{L/K}$ for the argument to go through. The trace is usually used simply because it is kind of canonical and has that nice description in terms of the powers of the Frobenius.
• This argument is often encountered when constructing additive characters of finite fields, when we precompose the exponential mapping $\chi:\Bbb{F}_p\to\Bbb{C}^*, \chi(x)=e^{2\pi i x/p}$ by $\Bbb{F}_p$-linear functionals $K\to\Bbb{F}_p$ to get all the additive characters of a finite field $K$. Characters are a common tool for mapping counting problems from a finite field to $\Bbb{C}$ and its subsets.