6
$\begingroup$

I'm self-studying Model Theory via Hodges' book Model Theory and got stuck in the following problem (see below for notation):

Section 2.2, Problem 9: For each of the following classes, show that it can be defined by a single sentence of $L_{\omega_1\omega}$.
(a) Unique factorization domains.
(b) Principal ideal domains.
(c) Dedekind domains.
(d) Semisimple rings.
(e) Left coherent rings.
(f) Left artinian rings.
(g) Noetherian local commutative rings.
(h) Groups $G$ such that if $H,K$ are any two isomorphic finitely generated subgroups of $G$ then $H$ is congruent to $K$ in $G$.

The problem reduces to restate the properties in each item in a suitable manner, without dealing directly with ideals and subgroups but rather only with elements of the ring in question (finitely many of them at a time). See below for more details.

My problem is:

Problem: I don't know how to solve items (b), (c), (d), (f), and (g), because I don't know how to write conditions on ideals (and chains of these) in the given language.

Some notation: The signature $L$ for rings is $\langle 0,1,+,-,\cdot\rangle$. Given an ordinal $\kappa$, we define the language $L_{\kappa\omega}$ in the usual manner, but we allow disjunctions and conjunction on sets of cardinality $<\kappa$, i.e., if $I$ is a set of formulas of $L_{\kappa\omega}$ with $|I|<\kappa$, then $\bigvee_{\phi\in I}\phi$ and $\bigwedge_{\phi\in I}\phi$ are formulas of $L_{\kappa\omega}$ (but we can only put quantifiers over finitely many variables: that's what the $\omega$ in $L_{\kappa\omega}$ means).


For example, here's how we can do (h) (I believe the author meant "conjugate" instead of "congruent"): We use variables $x_i,y_i,z_i,\ldots$. Moreover, if the variables $x_1,\ldots,x_k$ are bound, we write $\overline{x}=x_1\ldots x_k$.

Given variables $x,y_1,\ldots,y_k$, let's write: $$x\in\langle \overline{y}\rangle:\bigvee_{n<\omega}\exists z_1\ldots z_n\left[\left[\bigwedge_{1\leq i\leq n}\bigvee_{1\leq j\leq k}(z_i=y_j)\lor (z_i=y_j^{-1})\right)\land x=z_1\cdots z_n\right]$$ which can be read, in the language of groups, as "$x$ belongs to the subgroup generated by the $y_i$".

We can also write a formula for two finitely generated subgroups to be isomorphic: Consider variables $p_1,p_2,\ldots$, and all terms $t(\overline{p})$ in the language of groups which have only these variables. There are only countably many of these terms. Say we have variables $x_1\ldots x_k$ and $z_1\ldots z_k$. We write $$\langle\overline{x}\rangle\simeq\langle\overline{z}\rangle:\bigwedge_{t(\overline{p})}(t(\overline{x})=1\leftrightarrow t(\overline{z})=1),$$ where $t(\overline{x})$ means that we change each occurence of the variable $p_i$ by $x_{i\bmod k}$, and similarly for $t(\overline{z})$. The formula above states that the maps $x_i\mapsto z_i$ extends to an isomorphism of the generated subgroups.

Finally, the condition in the exercise can be expressed by $$\bigwedge_{n<\omega}\forall x_1\ldots x_n,z_1\ldots,z_n\left[\langle\overline{x}\rangle\simeq\langle\overline{z}\rangle\rightarrow\exists g\left(\bigwedge_{1\leq i\leq k}gx_ig^{-1}\in\langle\overline{z}\rangle\land g^{-1}z_ig\in\langle\overline{x}\rangle\right)\right]$$ and the conjunction of this sentence with the usual group axioms is a formula in $L_{\omega_1\omega}$ which works.


EDIT: Here are a few solutions for some of the items.

Item (b): A ring is a PID iff every finitely generated ideal is principal, or equivalently iff every $2$-generated ideal is principal (Bézout domain), and it is a UFD (item (a)). All of these can be written in $L_{\omega_1\omega}$.

Item (d): first recall given a ring $R$, the ideal generated by some $x\in R$ is $(x)=\left\{ax+xb+cxd:a,b,c,d\in R\right\}$, and $R$ is simple iff it is the ideal generated by any nonzero element. Thus a semisimple ring is one for which there are $n$ and $x_1,\ldots,x_n$, with $(x_i)\cap (x_j)=0$ if $i\neq j$, and for which $(x_i)=(x)$ for any $x\in (x_i)$. These can all be written in $L_{\omega_1\omega}$.

Item (e): A ring $R$ is left coherent iff every finitely generated ideal is finitely presented. We can restate this as follows: For every $n$ and for every $a_1,\ldots,a_n$, there exist $m$ and terms $t_1(\overline{x}^1,\overline{p}),\ldots,t_m(\overline{x}^m,\overline{p})$, of the form $t_i(\overline{x}^i,\overline{p})=x^i_1p_1+\cdots+x^i_np_n$ (where $x^i_j$ and $p_j$ are variables) in the language of rings, and there exist $r^i_j\in R$, $1\leq i\leq m$, $1\leq j\leq n$, for which $t_i(\overline{r}^i,\overline{a})=0$ for all $i$, and such that for every $s_1,\ldots,s_n$, if $s_1a_1+\cdots+s_na_n=0$ then there are $q_1,\ldots,q_m$ such that for each $i$ $s_i=\sum_jq_jr_i^j$.

This means that the relations $t_i(\overline{r},\overline{p})=0$, which make sense for left $R$-modules, are satisfies by the generators $a_i$, the map from the free module generated by $p_1,\ldots,p_n$, mapping $p_i\mapsto a_i$, has precisely the module generated by $t_i(\overline{r},\overline{p})$ as its kernel.

$\endgroup$
  • $\begingroup$ What does it mean for two subgroups to be congruent? $\endgroup$ – bof Oct 9 '16 at 0:59
  • $\begingroup$ @bof I am assuming that is a misprint, and that the author meant "conjugate" instead. $\endgroup$ – Questioner Oct 9 '16 at 17:54
3
$\begingroup$

You are right - infinitary logic does not allow you to directly quantify over ideals, chains of ideals, etc. So you need to find ways around this. For instance, re: (b), note that an integral domain is a PID iff it satisfies "Every finitely generated ideal is principal" (EDIT: This is of course nonsense, you also need the ascending chain condition on principal ideals, but that's no additional problem; see the comment below). This sentence can be written in infinitary logic. To see this, note that e.g. "Every 2-element-generated ideal is principal" can be expressed in standard first-order logic as $$\forall x\forall y\exists z(\forall w[\exists a, b(w=ax+by)\implies \exists u(w=uz)]),$$ and so it's not hard to see from this that "every finitely generated ideal is principal" is a countable conjunction of first-order sentences.

In general, if you can reduce from arbitrary ideals to finitely generated ideals, you'll be in the realm of infinitary logic.

$\endgroup$
  • $\begingroup$ That was my idea. However it is not true that $R$ is a PID iff every finitely generated ideal is principal (math.stackexchange.com/a/218516/160185). We also need the acc for principal ideals, which can be written in the language, so item (b) can be dealt with. However this does not work for the other items (as far as I know). Any suggestions? $\endgroup$ – Questioner Oct 8 '16 at 22:51
  • $\begingroup$ @Questioner (Dangit re: PID! I even knew that, I was just being sloppy! Fixed.) Hm, you have a good point - I thought that the other parts wouldn't be any harder, but I can't figure them out either. $\endgroup$ – Noah Schweber Oct 8 '16 at 23:03
2
$\begingroup$

I believe that the bulk of this exercise is taken from Theorem 10 of Definablity problems for modules and rings, by Gabriel Sabbagh and Paul Eklof (JSL 1971). I will copy the statement of their theorem here.

We remind the reader that all notions are at left; e.g. artinian means left artinian.

THEOREM 10. If W is any one of the following classes of rings, then W is definable in $L_{\omega_1,\omega}$:
W is the class of all

(a) unique factorization domains,
(b) principal ideal domains,
(c) Dedekind domains,
(d) semisimple rings,
(e) artinian rings,
(f) semihereditary rings,
(g) Prufer rings,
(h) valuation rings,
(i) coherent rings,
(j) semiprimary rings,
(k) semiperfect rings.

The classes introduced in (f), (g), and (h) are elementary.

You can find the required $L_{\omega_1,\omega}$-sentences in their paper, but I will copy their sentence defining Dedekind domains.

(c) $\Lambda$ is a Dedekind domain $\Leftrightarrow$ every finitely generated ideal $A$ is uniquely a product of finitely generated prime ideals: $A = \prod_P P^{e_P(A)}$; and for any two finitely generated ideals $A, B$, $A\subseteq B$ if and only if $e_P(A)\geq e_P(B)$ for all finitely generated primes $P$.


I have skimmed what is written in the problem statement and in the other answers on this page, and I think that the answers for (a), (b), (e) and (h) are either correct or at least on the right track. I think that the answer given for (d) in the original post is incorrect. The word semisimple does not mean "finite product of simple rings", but rather it means that $R$, as a left module, is a finite sum of simple submodules. There are simple rings that are not semisimple. In any case, see the Sabbagh-Eklof paper for (d).

What remains is (g): Noetherian local commutative rings. You will find on page 644 of the Sabbagh-Eklof paper the fact that the class of left Noetherian rings is not definable in $L_{\infty,\omega}$. You will also find on that page the statement that the authors do not know if the class of Noetherian commutative rings is definable in $L_{\infty,\omega}$. So either this implied problem has been solved, or else the hypotheses of being a local ring is essential to the answer of (g).

Of course, the class of commutative local rings is easy to define in $L_{\omega,\omega}$: simply say that $\forall x\forall y(xy=yx)$ and that the sum of two nonunits is a nonunit. I'm not sure what you have to add to this to carve out exactly the Noetherian ones. It is not hard to say in $L_{\omega_1,\omega}$ that the maximal ideal $\mathfrak m$ is finitely generated, and that $\cap_{n=1}^{\infty} \mathfrak m^n = (0)$. These things must be true if the commutative local ring is Noetherian, but they are not enough to force the ring to be Noetherian.

$\endgroup$
  • 2
    $\begingroup$ An anecdote: A few years ago, part (g) was assigned on a problem set for the graduate model theory course at Berkeley. The professor, having taken the problem from Hodges, hadn't thought about the solution before assigning it. Lots of people (including those, like me, who weren't in the class) discussed the problem over the following week, and no one found a solution. I do hope someone comes up with the answer (or a proof that this class is not definable...) $\endgroup$ – Alex Kruckman Oct 19 '16 at 14:11
1
$\begingroup$

I outline the solution to b). This should give you some ideas for how to do the others, (plus I dont remember all the definitions at the moment). Note that a PID is a unique factorization domain (UFD). The property of being a UFD is $L_{\omega_1,\omega}$. This is a), I just noticed. A UFD has the property that there is no infinite decreasing sequence of divisors, $\cdots a_2|a_1|a_0$. Equivalently there is no infinite sequence of increasing principal ideals $(b_0)\subseteq (b_1)\subseteq (b_2)\subseteq \cdots$. The consequence of this is that if $R$ is a UFD and every finitely generated ideal is principal then $R$ is a PID. Thus the problem reduces to expressing that every finitely generated ideal is principal. For example $$\forall a,b \exists c [(a,b)=(c)]$$ I think you can see that $(a,b)=(c)$ is first order, so you just take a conjunction of similar such sentences.

For c) every ideal in a Dedekind domain is finitely generated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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