Let $\mbox{Rings}$ be the category of commutative rings with $1$. 


Is there an equivalence of categories $F: \mbox{Rings} \to \mbox{Rings}$ such that
$$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$



Invariance of $\mathbb{Z}[x]$ under a self-equivalence of the category of commutative rings with 1
The category $\mbox{CRing}$ of commutative rings is *rigid*, i.e. every equivalence $\mbox{CRing} \to \mbox{CRing}$ is isomorphic to the identity, 


Proof: Let $F : \mbox{CRing} \to \mbox{CRing}$ be an equivalence. The main part is to prove that $A := F(\mathbb{Z}[x])$ is isomorphic to $\mathbb{Z}[x]$. Charles' answer shows that $A$ is a retract of a polynomial ring $\mathbb{Z}[x\_1,...,x\_n]$, in particular an integral domain. Since $\mathbb{Z}$ is the initial ring, $F$ preserves it and since $\mathbb{Z} \subseteq \mathbb{Z}[x]$ is a split monomorphism, the same is true for $\mathbb{Z} \to A$. Thus we have $\mathbb{Z} \subseteq A \subseteq \mathbb{Z}[x\_1,...,x\_n]$. Now assume that we already knew that $A$ is a UFD with $\text{tr.deg}(Q(A)/\mathbb{Q}) = 1$. Then [2], Theorem 4.1 would imply that $A \cong \mathbb{Z}[x]$. Note that the only nontrivial input in the lemmas preceding this theorem is Lüroth's theorem applied to $A \otimes \mathbb{Q}$, which is not needed here because be show below directly $Q(A) \cong \mathbb{Q}(x)$.


So let's check the two properties of $A$. Since $A$ is a retract of $\mathbb{Z}[x\_1,...,x\_n]$, which is UFD, we conclude from [1], Prop. 1.8 that also $A$ is UFD.


Now it remains to prove that $Q(A) \cong \mathbb{Q}(x)$, which Kevin Ventullo has already sketched in the comments. First note that $F$ preserves nontrivial rings since $0$ is the terminal ring. A monomorphism is the same as an injective ring homomorphism (since the forgetul functor from rings to sets is representable), thus $F$ preserves injective ring homomorphisms. Now fields are precisely the nontrivial rings with only one proper ideal, i.e. such that every homomorphism to a nontrivial ring is injective. Thus $F$ preserves fields (and also field extensions).


The universal property of the quotient field and that we already know that $F$ preserves $\mathbb{Z}$, integral domains, fields and injections implies $F(\mathbb{Q}) \cong \mathbb{Q}$ and $Q(A) = F(\mathbb{Q}(x))$. Now the extension $\mathbb{Q}(x)$ of $\mathbb{Q}$ is characterized by the property it is transcendental and every transcendental extension factors over it. Thus it remains to characterize algebraic field extensions $L/K$.


To do this, we call $L/K$ *locally finite* if for every extension $E/K$ there are only finitely many $K$-homomorphisms $L \to E$. It is clear that every finite extension is locally finite. Besides, every locally finite extension is algebraic: If not, choose a transcendence basis $B$ and extend infinitely many homomorphisms from $K(B)$ to it's algebraic closure to get a contradiction. It follows that algebraic extensions are precisely the filtered colimits of locally finite extensions.


So we have proved that $F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$. Now let $R$ be a commutative ring. The usual coring structure on $\mathbb{Z}[x]$ induces a ring structure on $\text{Hom}(\mathbb{Z}[x],R)$ which is naturally isomorphic to $R$. Every coring structure on $\mathbb{Z}[x]$ is isomorphic to the usual one ([3], Prop. 3.1). Thus, we get, naturally in $R$,


$R \cong \text{Hom}(\mathbb{Z}[x],R) \cong \text{Hom}(F(\mathbb{Z}[x]),F(R)) \cong \text{Hom}(\mathbb{Z}[x],F(R)) \cong F(R)$.


Thus $F$ is isomorphic to the identity. Actually [3] gives another, similar proof that $\mbox{CRing}$ is rigid, and more generally it classifies the automorphisms of the category of $R$-algebras, where $R$ is an integral domain. I should have checked the literature in the first place.




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[1] Douglas L. Costa, *Retracts of Polynomial Rings*, J. Algebra 44 (1977), pp. 492 - 502


[2] S. Abhyankar, P. Eakin, W. Heinzer, *On the uniqueness of the coefficient ring in a polynomial ring*, J. Algebra 23 (1970), pp. 310 - 342


[3] W. E. Clark, G. M. Bergman, *The Automorphism Class Group of the Category of Rings*, J. Algebra 24 (1973), pp. 80 - 99


