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Is there a known example of a ring endomorphism $f: \mathbb{Z}[x_1, \ldots, x_n] \to \mathbb{Z}[x_1, \ldots, x_n]$ such that $f \circ f = f$ but whose image is not isomorphic to a polynomial ring?

My own interest in this has to do with better understanding the (categorical) Cauchy completions of Lawvere theories for some familiar types of algebraic objects; here we are dealing with the Lawvere theory of commutative rings. But apparently this type of problem is of interest to algebraists in the context of hard problems like the Jacobian conjecture and the cancellation problem, so there is a certain body of work out there already on related material.

From the literature I've scanned so far (articles by Costa, Shpilrain, Picavet, Gupta, and others), a lot of attention is paid to retracts of polynomial algebras over fields, but I'm having quite a hard time finding a clear statement for the case of polynomial algebras over $\mathbb{Z}$. One tantalizing lead was a statement by Picavet here: "We were motivated by an unsolved conjecture: a projective algebra of finite type over a field $A$ is a polynomial ring. An example by Costa shows that the statement is false if $A$ is not a field." I couldn't find a statement by Costa which treated every non-field $A$ (including $\mathbb{Z}$ in particular); I suspect Picavet meant that there exist non-fields $A$ for which the statement is false. An interesting example of such $A$ can be extracted from Gupta's later negative solution to the cancellation problem, as mentioned by Jeremy Rickard here at the MO discussion Is a retract of a free object free?. (Actually, Gupta's constructions more significantly show that the statement is false for any field of positive characteristic, but this work can be used to derive some non-field examples as well.)

Akhil Mathew asked essentially the same question at Math.SE here, and got pointers to literature from Mariano Suárez-Alvarez, but I'm hoping to get something more definitive now.

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  • $\begingroup$ To "squeeze water out" you might switch from the theory of commutative rings, which is merely Maltsev, to "much more Maltsev" theories. Now on one hand, with the theory of modules over a fixed ring you are asking about finitely generated projective modules, so the whole K-theory with its vector bundles falls on you. While zooming in in the other direction if you specialize to arithmetic theories, you are in the realm of e. g. projective formulas in the Intuitionistic Propositional Calculus or in various modal logics - there are many deep results there, and it is highly nontrivial. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 4, 2015 at 6:23
  • $\begingroup$ Returning to commutative rings - one approach might be this: Yoneda places the category of affine schemes (i. e. the opposite of the category of commutative rings) inside the category of functors from finitely presented rings to sets, and you might also make it more tight by restricting to subtoposes of Zariski, étale, ..., canonical sheaves. Now the guys you ask about will be injectives in affine schemes, so you might hope to understand them by looking at injectives in one of these sheaf toposes, which are fairly simpler to get: they are retracts of powers of respective subobject classifiers. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 4, 2015 at 6:30
  • $\begingroup$ @მამუკაჯიბლაძე Thanks for the suggestions. In particular for the suggestion to look at arithmetic theories; I understand PRA as initial among Lawvere theories in which the generating object is an NNO, but I'd never thought to consider its Cauchy completion. $\endgroup$
    – Todd Trimble
    Commented Oct 4, 2015 at 10:52
  • $\begingroup$ Strangely enough, although I used wrong term - "arithmetic" instead of "arithmetical" - it still turns out to be closely related (or maybe even amounting to the same thing, I don't know). Namely I meant theories with a specific Maltsev term, the one satisfying $p(x,y,y)=p(x,y,x)=p(y,y,x)=x$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 4, 2015 at 11:09
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    $\begingroup$ @მამუკაჯიბლაძე Thanks for clarifying further; even if you had said 'arithmetical' I probably would have misunderstood you. A simple example of non-free projective objects (not far from the case of commutative rings) occurs for the theory of Boolean rings, where any non-terminal finite Boolean algebra is projective. Meanwhile, one particular Mal'cev operation in PRA comes from thinking of the natural numbers under reverse ordering as forming a cartesian closed poset, and then putting $p(x, y, z) = x^{y^z} \wedge z^{y^x}$. $\endgroup$
    – Todd Trimble
    Commented Oct 4, 2015 at 12:32

1 Answer 1

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Existence of such an example follows from the same result of Asanuma that is crucial for Gupta's work, see the article

Teruo Asanuma, "Polynomial fibre rings of algebras over noetherian rings", Inventiones mathematicae 87 (1987), 101–127 (DOI link).

It follows from Corollary 5.3 of that article that if
$$ A=\mathbb{Z}[x,y,z]/(-x^{p^e}+y+y^{sp}+pz), $$ where $p$ is a prime number and $e,s$ are positive integers such that $p^e\not\mid sp$, $sp\not\mid p^e$, then $A$ is not isomorphic to a polynomial ring, but $B=A[t]$ is isomorphic to a polynomial ring in three variables. Of course, $A$ is a retract of $B$ via the evaluation at $t=0$.

EDIT: Moreover, consider $$ A_n:=\mathbb{Z}[x,y,z,x_1,\ldots,x_n]/(-x^{p^e}+y+y^{sp}+px_1\cdots x_nz), $$ then the proof of the same Corollary 5.3 (in the case $R=\mathbb{Z}[x_1,\ldots,x_n]$) can be adapted to show that $A_n[t]$ is isomorphic to the polynomial ring in $n+3$ variables, but $A_n$ is not isomorphic to the polynomial ring in $n+2$ variables.

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    $\begingroup$ Nice. This prompts follow-up questions: (a) are there examples for $n\neq 3$? clearly not for $n=0,1$, but $n=2$? also it's not a priori clear for $n\ge 4$ (it doesn't follow from the example, precisely since adding indeterminates makes the ring free). (b) Is there an example that is not stably free (in the sense that $A$ is isomorphic to a retract of a polynomial ring, but no $A[x_1,\dots,x_m]$ is isomorphic to a polynomial ring). $\endgroup$
    – YCor
    Commented Apr 19, 2020 at 9:07
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    $\begingroup$ @YCor For $n=2$ and R a UFD, a retract of $R[x,y]$ is a polynomial ring, see Remark after Theorem 3.5 in D.L.Costa, Retracts of polynomial rings, Journal of Algebra, 44:2 (1977), 492-502 (sciencedirect.com/science/article/pii/0021869377901971). $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 19, 2020 at 10:08
  • $\begingroup$ @YCor I now added a modification covering the case $n\ge 4$ also. The only remaining question is (b), and at a first glance it seems very hard. $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 19, 2020 at 12:28

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