# Isomorphism of fields via the forgetful functor

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 $$\pi$$?

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 $$1$$).
• 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! – Alex Kruckman Apr 29 '19 at 16:56