Background and motivation

I've always been fascinated about algebraic statements independent from ZFC set theory. One such fascinating example comes from considering $\rm{Ext}^1_\mathbb{Z}(A,\mathbb{Z})$. If $A$ is free then this abelian group is trivial. Is the converse true? The converse is known as the Whitehead problem.

Now the Whitehead problem was shown independent of ZFC by Shelah. This is somewhat unsatisfying, but it can be proved assuming the axiom of constructibility ($V=L$). In fact, there are other reasonable sounding and simple statements in analysis and topology that are also independent from ZFC but become theorems once $V=L$ is assumed. Another statement is that the global (a.k.a. homological) dimension of the ring $\prod_{i=1}^\infty \mathbb{C}$ is two if and only if the continuum hypothesis holds, which is implied by adding $V=L$ again. Thus I feel warmly about the axiom of constructibility.

Of course, the dissatisfaction remains, because there are other statements that are independent from $ZFC + V=L$ (well, perhaps I should write $ZF + V=L$ to save space). Question 18058 and Question 11480 are examples.

Question, loosely stated

Now, I am curious if there are any known algebraic (see postscript) statements, reasonably naturally sounding (use judgement), that are independent from $ZF + V=L$? Or perhaps independent from $ZF + A$ where $A$ is your favourite set-theoretic axiom independent from $ZFC$? Perhaps some easy low-hanging fruit for this search would be in the area of homological dimension theory? Has anyone done work on this type of thing?

I am sure there must be some statements of some kind. In the proof of the Whitehead problem under adding $V=L$, one can first deduce some combinatorial statement that requires little set-theoretic machinery and then use it to prove the Whitehead problem. So perhaps adding other axioms, one can also deduce various combinatorial gadgets and them use them to get new algebraic statements that are independent from the original $ZFC$? I would even like to hear about statements implied by additional axioms, but whose independence is not proved. (One of Devlin's books explains this).

(Remark: Although with enough brute force, one should be able to churn out such things no matter how many new axioms one adds, I would be interested in finding enough axioms of set theory so that the remaining independent statements would be so bizarre sounding that they would be essentially be uninteresting for all of time. Presumably as one adds more and more axioms to set theory, this would happen, no?)

Since I am not a set theorist, I would appreciate answers that are understandable to someone who knows the basics of set theory (say a typical first grad course) but knows very little about forcing.


By "algebraic" I mean roughly some statement in the language of groups, rings, ideals, modules, fields, etc., somewhat natural sounding, that does not itself refer to the additional axiom (e.g. using some set whose cardinal is inaccessible, or something along these lines). Thus I am not looking for statements about real numbers or set theory, although those are interesting too.

  • $\begingroup$ I posted a question to look for Whitehead-like problems, but despite some upvotes no answers were posted. mathoverflow.net/questions/101212/… $\endgroup$
    – Asaf Karagila
    Jan 4, 2013 at 19:53
  • $\begingroup$ Your question, also interesting, looks for open problems that were proven in some definition of "countable case" and not necessarily independent or implied by stronger set theoretic axioms, so I am hoping that the different criteria and wording might prompt some answers. $\endgroup$ Jan 4, 2013 at 20:04
  • $\begingroup$ Jason, while you are technically correct, I have realized that just like the axiom of choice ensures that the passing from finite to countable is relatively smooth; combinatorial principles (which hold in $V=L$) ensure similar transitions for higher cardinals. Our questions are indeed different, but they do share a common background (in my opinion). $\endgroup$
    – Asaf Karagila
    Jan 4, 2013 at 20:49

3 Answers 3


Let me address the part of your question seeking algebraic statements independent of ZFC+V=L.

The basic situation is that in set theory our tools are not so flexible for finding statements independent of ZFC+V=L, as opposed to finding statements independent of ZFC. The main reason for this is that one cannot directly use forcing to prove that a statement is independent of ZFC+V=L, because no nontrivial forcing extension can satisfy this theory. Simply put, forcing extensions never satisfy V=L. Thus, our main tool for proving independence over ZFC does not work at all for proving independence over ZFC+V=L.

This phenomenon has led some set theorists to view the theory ZFC+V=L as "nearly complete", settling essentially every set-theoretic question. But of course, the theory isn't really complete---it cannot be by the incompleteness theorem---it just feels so complete in comparison with the ubiquitous independence phenomenon for ZFC. Nevertheless, one can find several classes of statements independent of ZFC+V=L. Let me list a few of them.

  • There are of course the consistency statements, such as Con(ZFC), which if true, are independent of ZFC+V=L.

  • The existence of certain large cardinals, such as inaccessible, Mahlo, hyper-Mahlo, weakly compact, indescribable, or unfoldable cardinals, if consistent with ZFC, is independent of ZFC+V=L. The reason is that if there are such cardinals, then they exist also in L, and truncating the universe shows the consistency of the non-existence of such cardinals. Perhaps some of these large cardinals can be cast in purely algebraic terms, and I expect this may be the most promising kind of example for you.

  • The existence of transitive models of any given extension of ZFC, such as ZFC + `there is a supercompact cardinal', if consistent with ZFC, is independent of ZFC+V=L. The reason, as I explain in a recent paper on the axiom of constructibility, is that $V$ and $L$ have transitive models of the same theories. So if there is such a transitive model inside a model of ZFC, then there is such a transitive model inside a model of ZFC+V=L, and meanwhile, it is consistent with V=L that there are no transitive models of ZFC at all.

  • The number of ordinals $\alpha$ such that $L_\alpha\models\text{ZFC}$ is highly independent (provided that it is consistent to be large, which is essentially a mild large cardinal assumption). The reason is that if there are, say, $\omega^2+\omega\cdot 5+7$ many such ordinals $\alpha$, then by chopping off at the $\omega^2+\omega\cdot 4+26^{th}$ such ordinal, the number of such $\alpha$ drops accordingly.

Unfortunately, it seems that few of these kinds of independent statements have an essentially algebraic nature. I think the best examples will arise simply by stating the large cardinal axioms in algebraic-like terms, such as they are.

  • $\begingroup$ Thank you for the detailed answer. Since forcing extensions cannot satisfy $ZFC + V=L$, what about considering some theory that does not satisfy $V=L$. For instance, in some models, nonfree abelian groups $A$ with $\mathrm{Ext}^1(A,\mathbb{Z})$ do exist, so would it be easier to find further algebraic statements in such models that are independent? $\endgroup$ Jan 4, 2013 at 20:43
  • $\begingroup$ (I meant to say $\mathrm{Ext}^1(A,\mathbb{Z}) = 0$) $\endgroup$ Jan 4, 2013 at 20:44
  • $\begingroup$ Jason, yes, probably the right question here is simply to ask for the best examples of algebraic statements independent of ZFC. We don't have so many examples. You can peruse mathoverflow.net/questions/1924/…. $\endgroup$ Jan 4, 2013 at 20:48
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    $\begingroup$ Hi Hamkins, those propositions in your answer are all increasing the consistency strength. Are there some proposition conservative on arithmetic sentences, like CH, independent with V=L? $\endgroup$ Sep 17, 2021 at 5:01
  • $\begingroup$ Yes, I believe one can get instances like that from the universal Sigma1 definable sequence. $\endgroup$ Sep 17, 2021 at 14:34

One can find examples of algebraic statements that are independent of ZFC + $V = L$ by considering "absolute" versions of standard algebraic propositions. This happens for example when one seeks large strongly non-isomorphic (i.e. non-$L_{\infty \omega}$-equivalent) or absolutely rigid families of structures, absolutely indecomposable abelian groups and modules, or absolute $E$-rings of arbitrarily big cardinality. Roughly, up to the first $\omega$-Erdos cardinal $\kappa(\omega)$, in general such families exist, but above $\kappa(\omega)$, they do not.

The beauty of $\kappa(\omega)$ is that if it exists, then it exists in $L$, so that the existence of absolutely rigid systems in every cardinal is independent of ZFC + $V = L$.

In loose terms, the absolute version of a property $P$ is obtained by requiring that any structure $M$ retains $P$ in any generic extension. This can be recast as in algebraic terms using back-and-forth systems or infinitary equivalence.

Questions about the existence of this type of strongly non-isomorphic structures were first asked by Nadel and pursued more recently by Eklof, Fuchs, Goebel, Herden and Shelah.


We can use a theorem of Eklof and Mekler to get a statement about abelian groups that is independent of $ZFC + V=L$ under the additional assumption that it is consistent that weakly compact cardinals exist.

For an uncountable cardinal $\kappa$, we say an abelian group is $\kappa$-free iff every subgroup of cardinality $< \kappa$ is free. Clearly a free group is $\kappa$-free for all $\kappa$.

Consider the following statement:

For every uncountable regular cardinal $\kappa$, there exists an abelian group $A$ of cardinality $\kappa$ that is $\kappa$-free but not free.

If there exists a weakly compact cardinal $\kappa$, then every $\kappa$-free abelian group of cardinality $\kappa$ is free, so the above statement is false. Since weakly compact cardinals retain their weak compactness in L, this means that the falsity of the above statement is consistent with $V=L$.

On the other hand, if there are no weakly compact cardinals, then Eklof and Mekler show in their Main Theorem that $V=L$ implies the existence of a $\kappa$-free but not free group of cardinality $\kappa$ for every uncountable regular $\kappa$.


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