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. I already asked this question (mostly out of curiosity) on MSE last year, but no answer was found.

Some comments however suggest that the existence might involve the axiom of choice, or that requiring $F$ to preserve limits and filtered colimits could be useful.

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    $\begingroup$ Can we solve the question in any interesting algebraic categories? (Replace $R[X]$ with freely generated objects of a suitable kind.) $\endgroup$ Commented Jun 2, 2020 at 13:23
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    $\begingroup$ To elaborate on the comment by @AndrejBauer: How about the free pointed set functor (the functor which just disjointly adds a singleton to a set)? This should be the simplest possible case. Does it have a square root? $\endgroup$
    – Gro-Tsen
    Commented Jun 2, 2020 at 13:48
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    $\begingroup$ (Actually, to start with, I don't even know whether the identity functor on sets has a nontrivial square root.) $\endgroup$
    – Gro-Tsen
    Commented Jun 2, 2020 at 13:51
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    $\begingroup$ @Gro-Tsen If such a functor $F$ existed in the category of sets, we would have functoriality maps $S_n \to S_{|F ([n])|} \to S_{n+1}$ compatible with the natural embedding $S_n$ to $S_{n+1}$, where $[n]$ is a set with $n$ elements This contradicts $F([n])$ having fewer than $n$ elements by counting automorphisms, it contradicts $F([n])$ having more than $n$ elements as then any map from its symmetric group to $S_n$ would have image of size at most $2$ (except in a few exceptional cases), and $F(F([n]))= [n+1]$ contradicts $F([n])$ having exactly $n$ elements. $\endgroup$
    – Will Sawin
    Commented Jun 2, 2020 at 19:38
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    $\begingroup$ Because $F^2$ is faithful and conservative, the same goes for $F$. With a little work, this implies that $F$ preserves all (co)limits that $F^2$ does, e.g. finite limits, filtered colimits, and monomorphisms and epimorphisms. (The map $F(\lim A_i) \to \lim F(A_i)$ becomes a split monomorphism after applying $F$ and similarly a split epimorphism after applying $F^2$.) Then $F$ reflects such colimits as well. $\endgroup$ Commented Jun 2, 2020 at 22:33

1 Answer 1


I don't know how to answer the question as stated, but I believe that by strengthening the question a bit we can see that there does not exist a "reasonable" notion of "polyonomial ring in one-half variable" (unless perhaps we start thinking about enlarging our category or something). That is, let me address the following:

Modified question: Let $k$ be a commutative ring. Does there exist a functor $F: CAlg_k \to CAlg_k$ and a natural transformation $\iota: Id \Rightarrow F$ such that $F(F(A))$ is naturally isomorphic to $A[x]$ and such that the composite $A \xrightarrow{\iota_A} F(A) \xrightarrow{\iota_{F(A)}} F(F(A)) = A[x]$ is identified under this isomorphism with the usual map $A \to A[x]$?

It seems to me that this is an extremely minimial, reasonable additional property to ask of a construction deserving the name "polynomial algebra in one half variable". It would be nice to eventually dispense with it, but I hope to shed some light on the original question by adding the assumption of the existence of $\iota$.

Answer to Modified Question: No, there is no such functor $F$ and natural transformation $\iota$, at least when $k$ is a field. For if there were, then since the composite $F(k) \to k[x] \to F(k)[x]$ has a retract given by evaluation at zero, it would be the case that $F(k)$ is a retract of $k[x]$. There are no retracts of $k[x]$ besides $F(k) = k$ or $F(k) = k[x]$, neither of which is compatible with $F^2(k) \cong k[x]$ and $F^4(k) \cong k[x,y]$.

EDIT: I was suddenly seized with doubt, so here's a proof of the fact I just used:

Fact: Let $k$ be a field. Let $k \subseteq R \subseteq k[x]$ be a $k$-algebra which is a retract of the $k$-algebra $k[x]$. Then either $R = k$ or $R = k[x]$.

Proof: Let $\pi: k[x] \to R$ be the retraction map, and let $p = \pi(x) \in R \subseteq k[x]$ be the image of $x \in k[x]$ under $\pi$. Since $k[x]$ is a PID, the kernel of $\pi$ is a principal ideal, of the form $(f(x))$ for some $f(x) \in k[x]$. Thus we have $f(p) = 0$. By the following lemma, this implies either that $p \in k$ (in which case $R = k$) or else that $f(x) = 0$ (in which case $R = k[x]$).

Lemma: Let $k$ be a commutative ring, and let $p(x) \in k[x]$ be a polynomial. Suppose that the leading coefficient of $p(x)$ is a non-zero-divisor in $k$. Suppose that $p(x)$ is algebraic over $k$, i.e. that $f(p(x)) = 0$ for some $f(y) \in k[y] \setminus \{0\}$. Then $p(x) \in k$ is a degree-zero polynomial.

Proof: Suppose for contradiction that $\deg p(x) \geq 1$. Then the leading coefficient of $f(y)$ is $a b^n$ where $a$ is the leading coefficient of $f(y)$ and $b$ is the leading coefficient of $p(x)$ (and $n = \deg f$). This leading coefficient vanishes by hypothesis, so that $b$ is a zero-divisor in $k$, contrary to hypothesis. Therefore $\deg p(x) = 0$ as claimed.

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    $\begingroup$ The answer applies in more generality than I have stated it, but I'm not quite sure of the maximal generality it applies to, so I have not aimed for maximal generality. $\endgroup$
    – Tim Campion
    Commented Apr 24, 2021 at 1:58
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    $\begingroup$ For one thing, the answer applies to any nonzero ring $k$ because if $k$ is not a field, then we can always quotient at a maximal ideal to obtain a field $k'$; the argument then applies by looking at the $k$-algebras $k',F(k'), k'[x]$ rather than $k,F(k),k[x]$. There could in principle be loopholes, though -- for instance, this doesn't tell us whether the category of faithful $k$-algebras admits such a functor and natural transformation, unless we have a bit more information about retracts of $k[x]$ for general commutative rings $k$ (which may exist; I'm just ignorant of it). $\endgroup$
    – Tim Campion
    Commented Apr 24, 2021 at 13:52
  • $\begingroup$ Thanks very much for your nice answer. I think the modified question is very natural as well, but I was also wondering what happens if we "enlarge our category". $$ $$ I am not sure what is the good way to formulate this, something like: is there a category $\mathcal D$ and functors $E : \mathcal C \to \mathcal D, F' : \mathcal D \to \mathcal D$ such that $F'^2 \circ E \cong E \circ P$, where $P : \mathcal C \to \mathcal C$ the functor $R \mapsto R[X]$? $\endgroup$
    – Watson
    Commented Apr 24, 2021 at 15:10
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    $\begingroup$ @Watson That makes some sense. I'm pretty sure that by abstract nonsense there exists a universal example of a category receiving a functor from $CAlg_k$ and admitting a square root of $A \mapsto A[x]$; as you suggest, the first thing to ask is probably whether or not this universal category is the zero category. There should likewise be a universal such $k$-linear category, a universal such cocomplete $k$-linear category, etc. $\endgroup$
    – Tim Campion
    Commented Apr 24, 2021 at 15:35
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    $\begingroup$ @Watson At any rate, I think the "enlarge the category" version of the question sounds interesting. It's beyond the scope of the current question, but it could be an interesting research question, or perhaps it could fit as another MO question if it could be stated precisely enough. $\endgroup$
    – Tim Campion
    Commented Apr 26, 2021 at 21:37

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