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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 $\mathbb Z[X_r : r \in R] / (X_{a+b} - X_a - X_b, X_{ab} - X_a X_b)$...).

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    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Dec 2 '19 at 7:49
  • $\begingroup$ What if you also assume $F$ preserves limits and filtered colimits ? $\endgroup$ – uno Dec 3 '19 at 1:11
  • $\begingroup$ @uno : what would be your ideas in that case? $\endgroup$ – Watson Dec 3 '19 at 7:19
  • $\begingroup$ Now asked on MO: mathoverflow.net/q/361976/84923 $\endgroup$ – Watson Jun 4 at 5:53

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