Is there a notion of polynomial ring in "one half variable"?

Question

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.

Interesting question! Have you tried throwing plethystic algebra (as in the Borger/Wieland paper) at it? — darij grinberg, Jun 02 '20 at 12:17
Can we solve the question in any interesting algebraic categories? (Replace $R[X]$ with freely generated objects of a suitable kind.) — Andrej Bauer, Jun 02 '20 at 13:23
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? — Gro-Tsen, Jun 02 '20 at 13:48
(Actually, to start with, I don't even know whether the identity functor on sets has a nontrivial square root.) — Gro-Tsen, Jun 02 '20 at 13:51
In any category with duality, for example, finite-dimensional vector spaces, the identity functor has duality as a square-root. — Mark Wildon, Jun 02 '20 at 17:25
@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. — Will Sawin, Jun 02 '20 at 19:38
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. — R. van Dobben de Bruyn, Jun 02 '20 at 22:33
@darijgrinberg I only looked at the first theorem of the paper, but it concerns functors that are both left and right adjoints. I don't think $R \to R[X]$ is such a functor because it only commutes with finite limits and filtered colimits (as R. van Dobben de Bruyn notes). Do you have an idea to get around this? (This also shows that uno's idea on mathstackexchange of checking that $F$ commutes with all limits can't work.) — Will Sawin, Jun 02 '20 at 23:16
Following @R.vanDobbendeBruyn 's reasoning, $F$ also preserves and reflects sifted colimits (and in particular, $F$ preserves surjections). — Tim Campion, Jun 24 '20 at 15:25
In fact, $F$ preserves all connected colimits, including pushouts and coequalizers. This means that $F$ lifts to a left adjoint functor $CRing \to CRing_{F(\mathbb Z)/}$ to the coslice category. — Tim Campion, Jun 24 '20 at 15:50
@TimCampion oh of course. While it's not true that $(A_1 \otimes_R A_2)[x] = A_1[x] \otimes_R A_2[x]$, it is true (and more relevant) that $(A_1 \otimes_R A_2)[x] = A_1[x] \otimes_{R[x]} A_2[x]$. I missed that somehow. — R. van Dobben de Bruyn, Jun 24 '20 at 21:23
@TimCampion but I'm unable to reproduce how you get surjections as a type of sifted colimits (or in fact any type of colimit). Could you say a bit more? — R. van Dobben de Bruyn, Jun 25 '20 at 01:27
@R.vanDobbendeBruyn Unless I'm mistaken, surjections are the same as regular epimorphisms. A regular epi is the coequalizer of it kernel pair. This is a reflexive coequalizer -- a sifted colimit. — Tim Campion, Jun 25 '20 at 02:55
@Gro-Tsen How about the following for a (non-canonical) square root of the identity on $Set$: let $V$ denote the universe of sets, and for each $X\in V$ choose a bijection $b_{XY}:X\to Y$ such that $b_{XY}^{-1}=b_{YX}$. Define a functor $F:Set\to Set$ by $F(X)=b_{XY}(X)=Y$ and $F(f:X\to Y)=b_{XW}\circ f\circ b_{XZ}^{-1}:Z\to W$. Then $F$ is functorial unless I'm mistaken and $F\circ F=1_{Set}$. — Alec Rhea, Jul 15 '20 at 14:06
@AlecRhea : it seems that in his comment above, Will Sawin proved that such a functor cannot exist (but I haven't read his argument in detail). Maybe your $F$, if it works, is actually isomorphic to the identity functor? — Watson, Jul 16 '20 at 06:15
Ah, it definitely is isomorphic to the identity, and this is what Gro-Tsen must have meant by trivial. (I assumed the trivial square root was the identity itself, my bad) — Alec Rhea, Jul 16 '20 at 06:43
@Watson Actually, Will's comment was about the free pointed set functor, not the identity functor. But anyway, if $F$ is a square root of the identity functor (on sets), then it is an equivalence and must send singletons to singletons. Then there is a natural isomorphism from the identity to $F$, which on a set $X$ is the map $X\to FX$ that sends the element in the image of a map $p:\ast\to X$ to the element in the image of $Fp:F\ast\to FX$. — Jeremy Rickard, Apr 24 '21 at 07:09
@მამუკაჯიბლაძე Hmm... I wonder what the analog of a spin structure is here... — Tim Campion, Apr 24 '21 at 15:42
@ZachTeitler I don't understand your suggestion. How do you define the complement of a set in a functorial way? I don't even understand how it's defined on objects, unless your sets are all subsets of one "universal set"... — Tim Campion, Apr 24 '21 at 15:43

Tim Campion · Answer 1 · 2021-04-24T14:24:16.463

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.

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. — Tim Campion, Apr 24 '21 at 01:58
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). — Tim Campion, Apr 24 '21 at 13:52
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 $R \mapsto R[X]$? — Watson, Apr 24 '21 at 15:10
Of course, we also require that $E$ is somehow "injective", e.g. faithful, or fully faithful -- I don't really know what conditions we should add to avoid silly counterexamples. — Watson, Apr 24 '21 at 15:10
@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. — Tim Campion, Apr 24 '21 at 15:35
@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. — Tim Campion, Apr 26 '21 at 21:37
Dear @Tim Campion, thank you very much for your interest, and your nice comment. I am hesitating (but not quite reluctant) to ask this new question here on MO, because I am not sure how to state it precisely enough, and also because it could be a research project on its own (?), and finally I am not quite aware of basic notions in category theory to make any progress (even though ideas come could from anywhere, e.g. some square-root $\sqrt{P} : \mathcal D \to \mathcal D$ of $P : R \mapsto R[X]$ could maybe involve graphs, who knows...). — Watson, May 01 '21 at 16:58
On the other hand, it could be nice to ask it just to "advertise" this nice (open?) problem. Square roots of functors haven't been studied very much (at least I did not find any literature on this precise topic). If you are interested and have some time, then feel free to ask it yourself. I'd be happy to read it ;-) — Watson, May 01 '21 at 16:58

Is there a notion of polynomial ring in "one half variable"?

1 Answers1