Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1. - MathOverflow

Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1.

2011-02-19T00:02:37Z

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

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

Answer by Charles Rezk for Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1.

2011-02-19T04:12:18Z

Here's a proposal for a proof, though I can't finish it.

A morphism in a category is a regular epimorphism if it coequalizes some pair of morphisms.
An object $P$ in a category is projective if $\mathrm{Hom}(P,X)\to \mathrm{Hom}(P,Y)$ is surjective for every regular epimorphism $X\to Y$.
An object $K$ in a category is compact if $\mathrm{Hom}(K,-)$ takes filtered colimits to filtered colimits of sets.
An object $X$ in a category is irreducible if the only retracts of $X$ are itself and the initial object.

A self-equivalence of a category will preserve the these conditions.

Claim: $\mathbb{Z}[x]$ is (up to isomorphism) the unique irreducible compact projective in commutative rings.

What I do know (I think):

Regular epis in $\mathrm{Rings}$ are exactly the surjective ring homomorphisms.
Compact objects in $\mathrm{Rings}$ are exactly the finitely presented rings.

Since polynomial rings are certainly projective, this means that the compact projective objects in $\mathrm{Rings}$ are precisely the retracts of the $\mathbb{Z}[x_1,\dots,x_n]$s.

My intuition is that retracts of a polynomial ring are always isomorphic to a polynomial ring. In which case $\mathbb{Z}[x]$ would be the only irreducible compact projective. But I could be completely wrong about that.

Answer by Martin Brandenburg for Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1.

2011-02-21T15:35:19Z

The category $\mbox{Ring}$ is rigid, i.e. every equivalence $\mbox{Ring} \to \mbox{Ring}$ is isomorphic to the identity,

Proof: Let $F : \mbox{Ring} \to \mbox{Ring}$ 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 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 $\mbox{Ring}$ 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.

[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

Answer by Martin Brandenburg for Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1.

2011-05-20T08:32:17Z

Here is a new, short proof, that the category of rings is rigid:

Lemma: Let $R,S$ be rings, such that $\text{Alg}(R), \text{Alg}(S)$ are equivalent as categories. Then $R,S$ are isomorphic.

Proof: $R$ is the initial object of $\text{Alg}(R)$ and the comma category $\text{Alg}(R) / R$ is (via unitalization) equivalent the category $\text{Alg}'(R)$ of $R$-algebras which are not assumed to be unital. Now $0$ is a zero object of $\text{Alg}'(R)$, thus null morphisms are defined. A morphism $\alpha : A \times A \to A$ in $\text{Alg}'(R)$ with $\alpha (\text{id}_A,0) = \text{id}_A, \alpha (0,\text{id}_A) = \text{id}_A$ exists if and only if $(x,y) \mapsto x+y$ is multiplicative, i.e. the multiplication on $A$ is trivial, that is $A$ is just a $R$-module. Thus we have reconstructed $\text{Mod}(R)$ categorically from $\text{Alg}(R)$. Finally, we can reconstruct $R$ as the center of $\text{Mod}(R)$. qed

Theorem: Every self-equivalence of $\text{Ring}$ is isomorphic to the identity.

Proof: Let $F : \text{Ring} \to \text{Ring}$ be an equivalence. Now $\text{Alg}(R)$ is just the comma category $R \backslash \text{Ring}$, so we get an equivalence $\text{Alg}(R) \cong \text{Alg}(F(R))$. The lemma implies $F(R) \cong R$, in particular, $F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ as rings. But then we get also an isomorphism of corings and then $F \cong \text{id}$, as explained in my previous proof. qed

Remark, however, that this proof is not applicable for automorphisms of $\text{Alg}(R)$.