October 22nd
Today I learned that the trace pairing over finite fields gives all possible $\FF_q$-linear maps $\FF_{q^r}\to\FF_q,$ from here . We begin by noting that the the Galois group\[\op{Gal}(\FF_{q^r}/\FF_q)=\langle\sigma_q:\alpha\mapsto\alpha^q\rangle.\]It follows that the trace looks like\[\op{Tr}(\alpha)=\sum_{\sigma\in\op{Gal}(\FF_{q^r}/\FF_q)}\sigma(\alpha)=\sum_{k=0}^{r-1}\alpha^{q^k}.\]This doesn't simpliyfy further in any nice way. As usual, we can check that $\op{Tr}:\FF_{q^r}\to\FF_q$ by noting that the Latin square property of groups implies that the above sum is held constant by any automorphism $\sigma.$ We could also say more explicitly that $\FF_q$ is the set of elements satisfying $x^q=x,$ which we can check manually because\[\op{Tr}(\alpha)^q=\left(\sum_{k=0}^{r-1}\alpha^{q^k}\right)^q=\sum_{k=0}^{r-1}\alpha^{q^{k+1}}=\op{Tr}(\alpha),\]where the last equality holds because $\alpha^{q^r}=\alpha.$
Again as usual, $\op{Tr}$ provides an $\FF_q$-linear map from $\FF_{q^r}\to\FF_q.$ Both of these can be seen from the definition using automorphisms or by manually expanding, so we won't show that here. What is more interesting is the proof that the image is all of $\FF_q.$ Indeed, the dimension of the image is at most $1,$ so we just have to show that the image is at least dimension $1.$ But $\op{Tr}(\alpha)=0$ if and only if $\alpha$ is a root of\[\op{Tr}(x)=\sum_{k=0}^{r-1}x^{q^k},\]a polynomial of degree $q^{r-1} \lt q^r.$ It follows (by Lagrangs) that $\op{Tr}$ has at most $q^{r-1}$ roots, so it is nonzero somewhere, so its image has at least dimension $1.$ This is what we wanted.
Because we have a well-behaved linear map, we can use it to define the inner product\[\langle x,y\rangle\longmapsto\op{Tr}(xy),\]which we can quickly check to be bilinear. This induces quite a few linear maps $\langle\gamma,\bullet\rangle,$ or $x\mapsto\op{Tr}(\gamma x).$ How many? We claim that all linear maps $\FF_{q^r}\to\FF_q$ have this form. For an upper bound, note that tracking where basis elements go tells us that there are $q^r$ total linear maps. It remains to show that the trace pairing induces this many maps.
To be explicit, our pairing induces mappings\[\gamma\longmapsto(x\mapsto\op{Tr}(\gamma x)).\]Now, this works for any $\gamma\in\FF_{q^r},$ so there are $q^r$ total maps induced here. So it is enough to show that these are unique; i.e., that this is injective. Indeed, if $\gamma$ and $\gamma'$ induce the same mappings so that we want to show $\gamma=\gamma'.$ Taking the difference implies that\[x\mapsto\op{Tr}((\gamma-\gamma')x)=0\]is identically $0.$ If $\gamma-\gamma'$ is nonzero, then we can set $x=(\gamma-\gamma')^{-1}\alpha$ where $\op{Tr}(\alpha)\ne0,$ which exists by the above discussion. So because $\op{Tr}((\gamma-\gamma')x)$ really is identically $0,$ we must have $\gamma-\gamma'=0,$ implying that $\gamma=\gamma'.$ This is what we wanted.