The following article https://tinyurl.com/yydxzxe3 says that

"For the case of fields, given a field F and an isomorphism of sets U(F) → S, there is a unique field whose underlying set is S and which is isomorphic to F as a field via the given function"

The U above is the forgetful functor which we know for fields doesn't have a left adjoint but this could still be possible as we need a weaker condition (i.e. all bijections give us isomorphisms not all morphisms ).

But I don't get how. Say $\mathbb{F}_{2^n} = \mathbb{F}_{2}[x]/p(x) $ is a fixed finite field and I have a permutation $\pi$ on $2^n$ elements. How do I construct $f(x)$ such that $ \mathbb{F}_{2}[x]/f(x) \cong \mathbb{F}_{2^n}$ via $\pi$?

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The observation here is just that we can "transport structure" along any isomorphism.

Suppose $F$ is a field and $\pi\colon U(F)\to S$ is a bijection of sets. We define operations $+$ and $\times$ on $S$ in such a way that $(S;+,\times)$ is a field and $\pi\colon F\to (S;+,\times)$ is a field isomorphism. For any $a,b\in S$, define: \begin{align*} a+b &= \pi(\pi^{-1}(a) + \pi^{-1}(b))\\ a\times b &= \pi(\pi^{-1}(a)\times \pi^{-1}(b)).\end{align*} Note that on the right hand sides of the above equations, $+$ and $\times$ are the field operations in $F$.

In your example of a permutation $\pi$ of the set $U(\mathbb{F}_{2^n}) = \mathbb{F}_2[x]/p(x)$, this doesn't amount to finding a new quotient of $\mathbb{F}_2[x]$, we just get totally new field structure on the set $\mathbb{F}_2[x]/p(x)$ which has nothing to do with the original one, or with the ring structure on $\mathbb{F}_2[x]$ (e.g. in most cases it will have a different $0$ and a different $1$).

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  • $\begingroup$ Exactly, but since we know that all finite fields are of the above type we should be able to find a new quotient, right? The motivation for the question is that a bijection in F_2^n is given by 2^n images but by converting it to a field isomorphism I just need to specify images of the n generators. This would be helpful in creating smaller computational circuits (given that I somehow precompute the f) $\endgroup$ – Tushant Mittal Apr 29 '19 at 16:54
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    $\begingroup$ All finite fields are of the above type up to isomorphism. The new field is isomorphic to $\mathbb{F}_{2^n}$ by construction, so it's isomorphic to the same quotient of $\mathbb{F}_2[x]$, not a new one! $\endgroup$ – Alex Kruckman Apr 29 '19 at 16:56
  • $\begingroup$ Ah. I get it now. The isomorphism, equality problem again. That's exactly what the article is about and I still made that error! $\endgroup$ – Tushant Mittal Apr 29 '19 at 16:56

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