Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.
$$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$

share|improve this question
4  
$F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ already implies $F \cong \text{id}$. Proof: If $|R|$ is the set underlying $R$, then canonically $|R| \cong Hom(\mathbb{Z}[x],R)$. Apply $F$ on the right and get $|R| \cong |F(R)|$. This map is a ring homomorphism: To see this, use naturality to reduce to the case $R = \mathbb{Z}[x]$ or $R = \mathbb{Z}[x,y]$. The first works by definiton, and the second by the description of $\mathbb{Z}[x,y]$ as the coproduct of $\mathbb{Z}[x]$ and $\mathbb{Z}[y]$. –  Martin Brandenburg Feb 19 '11 at 1:35
2  
Well, since mono=injective in rings, you know you can pick out the subcategory of fields in a categorical manner; these are the precisely the rings such that all morphisms out are monomorphisms. Hence, you can make sense of domains (they admit monos into fields), hence the field of fractions of a domain, since this is a universal property. $Q$ is the field of fractions of the initial object, and $Q(x)$ is a universal transcendental extension of $Q$ (finite algebraic extensions are those which have only finitely many morphisms into another field; arbitrary algebraic extensions are colimits... –  Kevin Ventullo Feb 19 '11 at 4:04
2  
of finite extensions). So $F(\mathbb{Z}[x])$ is a domain whose field of fractions is $Q(x)$. Moreover, it admits morphisms into any field, so its intersection with $Q$ is just $Z$. –  Kevin Ventullo Feb 19 '11 at 4:06
1  
Well, it's not quite a solution since I haven't ruled out, say $Z[2x,x^2]$, so I didn't post it as an answer. Universal is probably the wrong word to use, but basically $Q(x)$ is the only transcendental extension of $Q$ such that any other transcendental extension can be factored through it (though not uniquely). –  Kevin Ventullo Feb 19 '11 at 10:09
1  
I just want to remark that I've proven, among several other things of which I don't know if they are useful, that the whole thing works if we replace (rings) to ($k$-algebras) for an algebraically closed field $k$. –  Martin Brandenburg Feb 19 '11 at 15:19
show 5 more comments

3 Answers

up vote 18 down vote accepted

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

share|improve this answer
    
Hi Martin, Can you please explain more why $|R| \cong |F(R)|$ is a ring homomorphism? Also does it depend on the choice of an isomoprhism $\mathbb{Z}[x]\to F(\mathbb{Z}[x])$? –  Nico Bellic Feb 21 '11 at 23:35
    
I think my proof was incomplete. See the edit. –  Martin Brandenburg Feb 22 '11 at 10:39
    
Hi Martin, This is probably a stupid question, but looking at Lang didn't yield an answer. How do we know that there is no other purely transcendental, non-isomorphic to $\mathbb{Q}(x)$, extension of $\mathbb{Q}$ that also embeds inside every other transcendental extension of $\mathbb{Q}$? –  Nico Bellic Feb 22 '11 at 22:11
    
We don't need "purely transcendental", just "transcendental". Assume $K$ is a trancendental extension which embeds into any other. In particular it embeds into $\mathbb{Q}(x)$. In particular, it is purely transcendental and has transcendence degree at most $1$, but it cannot be $0$. Thus $K \cong \mathbb{Q}(x)$. –  Martin Brandenburg Feb 23 '11 at 16:29
    
I've just realized that I worked with a wrong definition of "purely transcendental". So, the last argument probably has to be replaced by Lüroth's theorem. –  Martin Brandenburg May 20 '11 at 9:21
add comment

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.

share|improve this answer
    
Hi Charles. Your plan of attack looks very nice. I think that everything works, possibly except from the last paragraph. I had that intuition too, but I looked it up in MathSciNet, and there's some literature about retracts of polynomial rings... –  Fernando Muro Feb 21 '11 at 7:01
add comment

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 $\text{Alg}(R)$.

share|improve this answer
    
In your lemma it would be more clear to state the assumption that $R,S$ are commutative rings. –  Peter Samuelson May 20 '11 at 18:10
    
This is nice! I just want to point out that (as you probably know) the category of $R$-modules can be recovered as the category of abelian group objects in $R$-algebras. (This works more generally for algebras over an operad.) –  Akhil Mathew Nov 28 '12 at 13:44
    
Yes this is essentially what happens in my proof. The group-operation is $\alpha$. –  Martin Brandenburg Nov 28 '12 at 14:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.