# Is there a notion of polynomial ring in “one half variable”?

Let $$C$$ be the category of commutative rings.

Is there a functor $$F :C \to C$$ such that $$F(F(R)) \cong R[X]$$ for every commutative ring $$R$$ ?

(Here, we may assume those isomorphisms to be natural in $$R$$, if needed).

I tried to see what $$F(\mathbb Z)$$ or $$F(k)$$ (for a field $$k$$) should be, but I cannot come up with a contradiction to disprove the existence of $$F$$. On the other hand, I tried to build such an $$F$$, without success (e.g. try to consider some extension of $$\mathbb Z[X_r : r \in R] / (X_{a+b} - X_a - X_b, X_{ab} - X_a X_b)$$...).

• I would expect that if such a thing exists it is not any remotely natural construction but instead some crazy thing you can construct assuming global choice because merely being a functor is a relatively weak condition. – Eric Wofsey Dec 2 '19 at 7:49
• What if you also assume $F$ preserves limits and filtered colimits ? – uno Dec 3 '19 at 1:11
• @uno : what would be your ideas in that case? – Watson Dec 3 '19 at 7:19
• Now asked on MO: mathoverflow.net/q/361976/84923 – Watson Jun 4 at 5:53