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$\begingroup$

A classical exercise in group theory is "Show that if a group has a trivial automorphism group, then it is of order $1$ or $2$." I think that the straightforward solution uses that a exponent two group is a vector space over $\operatorname{GF}(2)$, and therefore has nontrivial automorphisms as soon as its dimension is at least $2$ (simply transposing two basis vectors).

My question is now natural:

Is it possible, without the axiom of choice, to construct a vector space $E$ over $\operatorname{GF}(2)$, different from $\{0\}$ or $\operatorname{GF}(2)$, whose automorphism group $\operatorname{GL}(E)$ is trivial?

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    @joriki: thanks for the correction.2011-03-20
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    What is GF(2), just to clarify?2011-03-20
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    The finite field with two elements : http://en.wikipedia.org/wiki/Finite_field ; I hope I didn't misuse this notation, which is not used in my languge...2011-03-20
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    @PseudoNeo: I know that as $\mathbb{F}_2$, but once clarified it is fine.2011-03-20
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    I can honestly say that I tried to think it over, if it was over a less-trivial group then I could show you that it's *probably* impossible, but since this is over $GF(2)$ and the scalars are 0,1 I could not finish the proof as I wanted.2011-03-22
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    @Arturo Magidin: thanks for the bounty, that's great!2011-03-22
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    I have an idea, but I need some references which are in my office. I'll try to prove it tomorrow and post an answer, if I find myself stuck half way I might post it as CW for further use.2011-03-23
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    I'm beginning to think this does indeed depend on AC!2011-03-23
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    @PseudoNeo: No problem; it's an interesting problem. I have the strong suspicion that it indeed depends on your set theory, and I might look around to see if, for example, Shelah may have done something along these lines.2011-03-23
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    This could be asked on mathoverflow...2011-03-23
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    @Jim: I'm going to wait until the week is up, and I'm going to check on the Rubins' books on AC when I get a chance. If I can't find an answer, I'll probably ask it there.2011-03-24
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    @Arturo: as you paid the bounty, I think it would be fair if you chose the answer. So pick one and I'll accept it.2011-03-28
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    @PseudoNeo? I don't *think* I've paid the bounty yet, did I? It still shows as "open bounty" on my web.2011-03-28
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    @Arturo: I think that was intended to mean "you *offered* the bounty and *will* be paying for it". :-) As far as I'm concerned, Asaf should get the bounty; he obtained a lot more results than I did, even if his proof isn't quite complete yet. Note that if you don't award it and my answer still has more votes when it runs out, I would automatically get half the bounty.2011-03-28
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    @Joriki: I haven't had a chance to read Asaf's answer (I only just saw this a few minutes ago, and I have to go teach in a few minutes). I'll be going over it later today.2011-03-28
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    @Arturo: I still believe that I can complete my answer. I am grinding my cells to dust here, and I hope to come up with a result later on today, but we'll have to see.2011-03-28
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    @Asaf: Okay, I'll be reading what you have so far later today, but I'll wait to award until tonight, then (about 12 hours from now?)2011-03-28
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    @Arturo: I guess it's fine. I just hope to finish this. I hope to catch my advisor for a small brainstorming in a bit. We'll see how that goes.2011-03-28
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    @Asaf: I'm not trying to give you a deadline: it's just that the bounty expires "tomorrow", so either I award it soon, or it automatically goes halfway to the highest voted answer.2011-03-28
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    @Arturo: I know, it's fine. I'm giving myself a deadline. I would like to either finish this and perhaps work on a short paper generalizing this sort of result, or go back to my M.Sc. thesis topics (which are quite similar to this, actually, but not exactly the same). Otherwise, this question will haunt me for years to come :-)2011-03-28
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    @Arturo: joriki clarified what I meant. The FAQ makes it clear that once you offered a bounty, you'll pay for it no matter what.2011-03-28
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    @PseudoNeo: You just scared me, because I had recently voted up Joriki's answer, and you made me think that perhaps that action had automatically awarded the bounty already...2011-03-28
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    @Arturo: If you want to read the answer I gave and share your thoughts I'd be more than happy to hear. I've made some progress, but there's still a bit left to work on. In the mean time, I'd love to hear some criticism from other mathematicians. (If you have the time, of course!)2011-03-28
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    @Asaf: I plan to; but I know very little (in fact, "almost no") forcing, and nothing about "construction of permutation models" so it's going to take me a while to try to understand it. So I want to have a nice chunk of time to attempt it.2011-03-28
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    @Arturo: My answer employs no forcing (well, not directly. Only by Jech's embedding theorem which generates a model of ZF from the permutation model I have defined, in which the vector space which contradicts AC exists).2011-03-28
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    @Asaf: Only goes to show you how little forcing I know, that I can't even tell when it's not there!2011-03-28
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    @Arturo: :-) the answer is not very trivial, and some technical knowledge about ZFA and permutation models needs to be available. I will try to improve the writing and perhaps add a bit of information soon.2011-03-28
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    @Asaf: I actually have another 23 hours (from now), so it may take me longer depending on how sleepy I feel after I'm done with my "appointment with my research" this evening. I know the answer is not trivial, and I hope to learn *something* out of it, which is why I'm trying to be non-hasty about it. (-:2011-03-28
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    @Arturo: I actually solved it just now :-) I will go write the proper solution.2011-03-28
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    @PseudoNeo: In case it wasn't clear, I awarded the bounty to Asaf (in case you want to accept an answer).2011-03-31
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    Yes, thanks. It was clear (but before you did it I wasn't sure that the acceptation of an answer and the awarding of the bounty were two independent decisions; it is of course better this way)2011-04-04
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    @joriki, PseudoNeo: I just finished transferring my answer below to a forcing argument (and to generalize it a bit) as a part of my M.Sc. thesis. I would like to include you both in the acknowledgments, if you wish to tell me your real names please ping me back and I will leave my email address for a short time so you can contact me. Thanks a lot!2012-07-21
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    @Asaf: Thanks, copied.2012-07-21
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    @user1729: Was this thread linked somewhere?2018-08-30
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    @Asaf Yes (by me). Someone asked a duplicate of [this question](https://math.stackexchange.com/questions/8379/) (which is the "classical exercise" referred to here). I figured it makes sense to link your answer here to the question there, so I did so in a CW answer.2018-08-30

2 Answers 2

35

Nov. 6th, 2011 After several long months a post on MathOverflow pushed me to reconsider this math, and I have found a mistake. The claim was still true, as shown by Läuchli $\small[1]$, however despite trying to do my best to understand the argument for this specific claim, it eluded me for several days. I then proceeded to construct my own proof, this time errors free - or so I hope. While at it, I am revising the writing style.

Jul. 21st, 2012 While reviewing this proof again it was apparent that its most prominent use in generating such space over the field of two elements fails, as the third lemma implicitly assumed $x+x\neq x$. Now this has been corrected and the proof is truly complete.

$\newcommand{\sym}{\operatorname{sym}} \newcommand{\fix}{\operatorname{fix}} \newcommand{\span}{\operatorname{span}} \newcommand{\im}{\operatorname{Im}} \newcommand{\Id}{\operatorname{Id}} $


I got it! The answer is that you can construct such vector space.

I will assume that you are familiar with ZFA and the construction of permutation models, references can be found in Jech's Set Theory $\small[2, \text{Ch}. 15]$ as well The Axiom of Choice $\small{[3]}$. Any questions are welcomed.

Some notations, if $x\in V$ which is assumed to be a model of ZFC+Atoms then:

  • $\sym(x) =\{\pi\in\mathscr{G} \mid \pi x = x\}$, and
  • $\fix(x) = \{\pi\in\mathscr{G} \mid \forall y\in x:\ \pi y = y\}$

Definition: Suppose $G$ is a group, $\mathcal{F}\subseteq\mathcal{P}(G)$ is a normal subgroups filter if:

  1. $G\in\mathcal{F}$;
  2. $H,K$ are subgroups of $G$ such that $H\subseteq K$, then $H\in\mathcal{F}$ implies $K\in\mathcal{F}$;
  3. $H,K$ are subgroups of $G$ such that $H,K\in\mathcal{F}$ then $H\cap K\in\mathcal{F}$;
  4. ${1}\notin\mathcal{F}$ (non-triviality);
  5. For every $H\in\mathcal{F}$ and $g\in G$ then $g^{-1}Hg\in\mathcal{F}$ (normality).

Now consider the normal subgroups-filter $\mathcal{F}$ to be generated by the subgroups $\fix(E)$ for $E\in I$, where $I$ is an ideal of sets of atoms (closed under finite unions, intersections and subsets).

Basics of permutation models:

A permutation model is a transitive subclass of the universe $V$ that for every ordinal $\alpha$, we have $x\in\mathfrak{U}\cap V_{\alpha+1}$ if and only if $x\subseteq\mathfrak{U}\cap V_\alpha$ and $\sym(x)\in\mathcal{F}$.

The latter property is known as being symmetric (with respect to $\mathcal{F}$) and $x$ being in the permutation model means that $x$ is hereditarily symmetric. (Of course at limit stages take limits, and start with the empty set)

If $\mathcal{F}$ was generated by some ideal of sets $I$, then if $x$ is symmetric with respect to $\mathcal{F}$ it means that for some $E\in I$ we have $\fix(E)\subseteq\sym(x)$. In this case we say that $E$ is a support of $x$.

Note that if $E$ is a support of $x$ and $E\subseteq E'$ then $E'$ is also a support of $x$, since $\fix(E')\subseteq\fix(E)$.

Lastly if $f$ is a function in $\mathfrak{U}$ and $\pi$ is a permutation in $G$ then $\pi(f(x)) = (\pi f)(\pi x)$.


Start with $V$ a model of ZFC+Atoms, assuming there are infinitely (countably should be enough) many atoms. $A$ is the set of atoms, endow it with operations that make it a vector space over a field $\mathbb{F}$ (If we only assume countably many atoms, we should assume the field is countable too. Since we are interested in $\mathbb F_2$ this assertion is not a big hassle). Now consider $\mathscr{G}$ the group of all linear automorphisms of $A$, each can be extended uniquely to an automorphism of $V$.

Now consider the normal subgroups-filter $\mathcal{F}$ to be generated by the subgroups $\fix(E)$ for $E\in I$, where $E$ a finite set of atoms. Note that since all the permutations are linear they extend unique to $\span(E)$. In the case where $\mathbb F$, our field, is finite then so is this span.

Let $\mathfrak{U}$ be the permutation model generated by $\mathscr{G}$ and $\mathcal{F}$.

Lemma I: Suppose $E$ is a finite set, and $u,v$ are two vectors such that $v\notin\span(E\cup\{u\})$ and $u\notin\span(E\cup\{v\})$ (in which case we say that $u$ and $v$ are linearly independent over $E$), then there is a permutation which fixes $E$ and permutes $u$ with $v$.

Proof: Without loss of generality we can assume that $E$ is linearly independent, otherwise take a subset of $E$ which is. Since $E\cup\{u,v\}$ is linearly independent we can (in $V$) extend it to a base of $A$, and define a permutation of this base which fixes $E$, permutes $u$ and $v$. This extends uniquely to a linear permutation $\pi\in\fix(E)$ as needed. $\square$

Lemma II: In $\mathfrak{U}$, $A$ is a vector space over $\mathbb F$, and if $W\in\mathfrak{U}$ is a linear proper subspace then $W$ has a finite dimension.

Proof: Suppose $W$ is as above, let $E$ be a support of $W$. If $W\subseteq\span(E)$ then we are done. Otherwise take $u\notin W\cup \span(E)$ and $v\in W\setminus \span(E)$ and permute $u$ and $v$ while fixing $E$, denote the linear permutation with $\pi$. It is clear that $\pi\in\fix(E)$ but $\pi(W)\neq W$, in contradiction. $\square$

Lemma III: If $T\in\mathfrak{U}$ is a linear endomorphism of $A$, and $E$ is a support of $T$ then $x\in\span(E)\Leftrightarrow Tx\in\span(E)$, or $Tx=0$.

Proof: First for $x\in \span(E)$, if $Tx\notin\span(E)$ for some $Tx\neq u\notin\span(E)$ let $\pi$ be a linear automorphism of $A$ which fixes $E$ and $\pi(Tx)=u$. We have, if so:

$$u=\pi(Tx)=(\pi T)(\pi x) = Tx\neq u$$

On the other hand, if $x\notin\span(E)$ and $Tx\in\span(E)$ and if $Tx=Tu$ for some $x\neq u$ for $u\notin\span(E)$, in which case we have that $x+u\neq x$ set $\pi$ an automorphism which fixes $E$ and $\pi(x)=x+u$, now we have: $$Tx = \pi(Tx) = (\pi T)(\pi x) = T(x+u) = Tx+Tu$$ Therefore $Tx=0$.

Otherwise for all $u\neq x$ we have $Tu\neq Tx$. Let $\pi$ be an automorphism fixing $E$ such that $\pi(x)=u$ for some $u\notin\span(E)$, and we have: $$Tx=\pi(Tx)=(\pi T)(\pi x) = Tu$$ this is a contradiction, so this case is impossible. $\square$

Theorem: if $T\in\mathfrak{U}$ is an endomorphism of $A$ then for some $\lambda\in\mathbb F$ we have $Tx=\lambda x$ for all $x\in A$.

Proof:

Assume that $T\neq 0$, so it has a nontrivial image. Let $E$ be a support of $T$. If $\ker(T)$ is nontrivial then it is a proper subspace, thus for a finite set of atoms $B$ we have $\span(B)=\ker(T)$. Without loss of generality, $B\subseteq E$, otherwise $E\cup B$ is also a support of $T$.

For every $v\notin\span(E)$ we have $Tv\notin\span(E)$. However, $E_v = E\cup\{v\}$ is also a support of $T$. Therefore restricting $T$ to $E_v$ yields that $Tv=\lambda v$ for some $\lambda\in\mathbb F$.

Let $v,u\notin\span(E)$ linearly independent over $\span(E)$. We have that: $Tu=\alpha u, Tv=\mu v$, and $v+u\notin\span(E)$ so $T(v+u)=\lambda(v+u)$, for $\lambda\in\mathbb F$. $$\begin{align} 0&=T(0) \\ &= T(u+v-u-v)\\ &=T(u+v)-Tu-Tv \\ &=\lambda(u+v)-\alpha u-\mu v=(\lambda-\alpha)u+(\lambda-\mu)v \end{align}$$ Since $u,v$ are linearly independent we have $\alpha=\lambda=\mu$. Due to the fact that for every $u,v\notin\span(E)$ we can find $x$ which is linearly independent over $\span(E)$ both with $u$ and $v$ we can conclude that for $x\notin E$ we have $Tx=\lambda x$.

For $v\in\span(E)$ let $x\notin\span(E)$, we have that $v+x\notin\span(E)$ and therefore: $$\begin{align} Tx &= T(x+u - u)\\ &=T(x+u)-T(u)\\ &=\lambda(x+u)-\lambda u = \lambda x \end{align}$$

We have concluded, if so that $T=\lambda x$ for some $\lambda\in\mathbb F$. $\square$


Set $\mathbb F=\mathbb F_2$ the field with two elements and we have created ourselves a vector space without any nontrivial automorphisms. However, one last problem remains. This construction was carried out in ZF+Atoms, while we want to have it without atoms. For this simply use the Jech-Sochor embedding theorem $\small[3, \text{Th}. 6.1, \text p. 85]$, and by setting $\alpha>4$ it should be that any endomorphism is transferred to the model of ZF created by this theorem.

(Many thanks to t.b. which helped me translating parts of the original paper of Läuchli.
Additional thanks to Uri Abraham for noting that an operator need not be injective in order to be surjective, resulting a shorter proof.)


Bibliography

  1. Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.

  2. Jech, T. Set Theory, 3rd millennium ed., Springer (2003).

  3. Jech, T. The Axiom of Choice. North-Holland (1973).

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    Congratulations -- it certainly *looks* cool :-) I don't know enough about this stuff to understand this proof, but I want to use the opportunity to learn about it, and I know I will have at least two questions when I do: a) how does this relate to Randall Dougherty's answer? b) does it tell us anything about the amount of choice we need to make this impossible? perhaps also for other steps in the chain of implications in my "answer"?2011-03-28
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    joriki: The method he specifies is a similar method to this one. The usage of adding Cohen reals and then treating them as atoms is known as symmetric extensions, Jech has an embedding theorem that takes permutation models of ZFA to a symmetric extension of their kernel (not fully, but to a prescribed length in the construction hierarchy). Essentially the answers are very similar in their nature.2011-03-28
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    joriki: As for the second question, I'm not sure. I think that assuming $AC_\kappa$ you can ensure that up to cardinality $\kappa$ no counter examples can arise, but from that point onwards it might be open. I'm not sure.2011-03-28
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    @Asaf: Do you know whether the Boolean prime ideal theorem holds in your model?2011-03-28
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    joriki: I don't know, but I believe that the answer is negative. I'm not sure how to prove that, though. I also think that countable choice fails in this model.2011-03-28
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    @Asaf: Just to make sure I understand: when you say "operations that make [$A$] a vector space, you are *not* talking about the free $\mathbb{F}_2$ vector space with basis $A$, but rather a vector space with underlying set equal to $A$, right? And this is happening inside ZFC+Atoms, so that we have our full array of tools to endow it with this structure (say, by taking a bijection of $A$ with a suitable vector space). (More questions may follow...)2011-03-28
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    @Arturo: Exactly. Also, do note I found a small flaw in the proof of claim 1, I know how to correct it and will do it now.2011-03-28
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    @Asaf: In the paragraph that begins "Some background on permutation models..." Should "$sym(x)\in\mathbb{F}$" be $sym(x)\in\mathcal{F}$?2011-03-28
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    @Arturo: Yes, I have corrected that (and the proof)2011-03-28
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    @Asaf: Is it possible to explain what a "normal subgroups filter" is within the confines of a comment? If so, I'd appreciate it; if not (too long), just point me to a reference (I own Jech's *Set Theory*, but not *Axiom of Choice*; but the latter is in the library and I can get it if necessary).2011-03-28
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    @Arturo, I am on it. It is explained in Set Theory chapter 15. I will add it here, though. It's not long.2011-03-28
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    @Asaf: P.S. I also own Smullyan/Fitting's "Set Theory and the Continuum Hypothesis", and it seems like they may touch on this towards the end...2011-03-28
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    @Asaf: I think you may mean Section 21, "Symmetric submodels of generic models", since it has a section called "Permutations models of ZFA". (-: I'll take the book with me so I can consult it at home when I get back to this. Thanks for the effort, this looks *extremely* interesting and informative.2011-03-28
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    @Arturo: Thanks. I hope that these edits turn out to be informative enough so you (and other people, of course) could follow the proof I gave. I will look into generalizations of it sometime soon.2011-03-28
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    @Asaf: Great answer! One question though: Do you need the second claim? I don't see it being used anywhere. Am I missing something?2011-03-29
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    @Asaf: Having spent a bit of effort trying to digest the stuff in Jech and here, I think I understand enough of what is going on to award the bounty. I don't think I could come up with it or reproduce it without some serious study, though! Well done indeed.2011-03-29
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    @Apostolos: Thank you, I believe you might be right. I will look into that.2011-03-29
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    @Arturo: Many thanks. I did study quite a lot in order to study all these techniques. My thesis will be using them a lot, at least from where I stand right now to see the future. So it was a good practice here as well :-)2011-03-29
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    @Apostolos: You were absolutely right. I corrected the proof accordingly.2011-03-29
10

This is not really an answer, more like a set of comments that's too long to put in the comments.

Taking the question literally, whether in the absence of choice such a vector space can be constructed, this thread, in particular the post near the end by Randall Dougherty, answers in the affirmative. (I don't know enough about forcing to check whether the construction works.)

Then the question becomes how much choice is required and whether the non-existence of non-trivial vector spaces over $\mathbb{F}_2$ with trivial automorphism group is equivalent to some more well-known choice-like axiom. In the last post in the thread, Herman Rubin claims that this follows from the Boolean prime ideal theorem and conjectures that it's equivalent to it. I tried for a while to find a proof in either direction but didn't get much further than equipping the set of all ideals of a Boolean algebra with the structure of a vector space over $\mathbb{F}_2$.

The Consequences of the Axiom of Choice website has a number of related results that don't settle the question but at least give a non-trivial upper bound on the amount of choice required: We don't need full choice, since Form 333, the axiom of odd choice, which is weaker than full choice in ZF, is equivalent to [333 A], "In every vector space $B$ over the two element field, every subspace of $B$ has a complementary subspace", which is at least as strong as the result in question (since we just need a complementary subspace for a one-dimensional subspace). [333 B], "Every quotient group of an Abelian group each of whose non-unit elements has order $2$ has a set of representatives", is another equivalent. Here is the original paper by Keremedis from which these results are taken.

Some other related choice-like principles:

[1 BJ] "In every vector space over the two element field, every generating set contains a basis." : This is equivalent to full choice; the proof is in the Keremedis paper linked to above and is rather neat.

Form 28(p): "Every vector space $V$ over $\mathbb{Z}_p$ has the property that every linearly independent subset can be extended to a basis. ($\mathbb{Z}_p$ is the $p$ element field.)"

Form 66: "Every vector space over a field has a basis." (Apparently it's unknown whether this is equivalent to full choice.)

Form 95(F): "Existence of Complementary Subspaces over a Field $F$: If $F$ is a field, then every vector space $V$ over $F$ has the property that if $S \subseteq V$ is a subspace of $V$, then there is a subspace $S' \subseteq V$ such that $S\cap S' = \{0\}$ and $S \cup S'$ generates $V$." (This implies odd choice.)

Form 109: "Every field $F$ and every vector space $V$ over $F$ has the property that each linearly independent set $A \subseteq V$ can be extended to a basis."

Form 429(p): "Every vector space over $\mathbb{Z}_p$ has a basis. ($\mathbb{Z}_p$ is the $p$ element field.)"

(The full table of implications can be viewed at the website by entering the form numbers in the form.)


[Edit:]

Out of my failed attempts at equipping the (non-distributive, non-complemented) lattice of subspaces of $E$ with some structure that would guarantee the existence of a maximal ideal (which left me doubting whether there really is a connection to the Boolean prime ideal theorem), this consideration emerged: We have the following chain of implications and equivalences:

$E$ has a basis $\Rightarrow$ There is an inner product on $E$ $\Rightarrow$ Every subspace of $E$ has a complement $\Rightarrow$ Every finite-dimensional subspace of $E$ has a complement $\Leftrightarrow$ Every one-dimensional subspace of $E$ has a complement $\Rightarrow$ There is a one-dimensional subspace of $E$ that has a complement $\Leftrightarrow$ There is a non-trivial linear functional on $E$ $\Rightarrow$ There is a non-trivial automorphism of $E$.

(Here $E$ in each case stands for "every vector space over $\mathbb{F}_2$ with more than two elements", not just a particular one; else the last implication wouldn't hold.)

I don't see any obvious implications in the other direction except for the ones indicated. It seems likely that different steps along this chain require different levels of choice; perhaps progress towards the overall solution might be made by proving further implications in the other direction or by proving that one of the steps in the chain is equivalent to or implies or is implied by a well-known choice-like principle. The only such connections I know of so far are that "Every subspace of $E$ has a complement" is equivalent to the axiom of odd choice (see above) and of course that the axiom of choice implies "$E$ has a basis", and hence the entire chain.

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    joriki: You gave me some insights and some ideas to work with. I've been busting my brain for a few days and got stuck every single time, now I think I know why. I work in a model without choice but with PIT... I'll see what happens in other models. :-)2011-03-24
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    @Asaf Karagila: Glad to be of service. "I've been busting my brain for a few days and got stuck every single time" -- that pretty much sums up my situation, too :-) In case you need a vector space structure on the ideals of a Boolean algebra, let me know, but I'm afraid that will be the easier part of any proof...2011-03-24
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    @joriki: We need a complement to a 2-dimensional subspace, surely: then we can do a swap in the two-dimensional subspace, and leave the complement alone to get a nontrivial automorphism.2011-03-24
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    @Arturo: The problem is that $\mathbb{F}_2$ has only 0,1 scalars so proving a nontrivial linear automorphism (in the Jech construction, at least) is contradictory is somewhat of a pain.2011-03-24
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    @joriki: Unless I misinterpret "form 66", "Every vector space has a basis" **is** equivalent to AC. "Existence of bases implies the Axiom of Choice", Andreas Blass, *Axiomatic Set Theory (Boulder, Col. 1983)*, Contemp. Math. 31, pp 31-33, AMS 1984, MR 763890; and "Bases in vector spaces and the axiom of choice", James D. Halpern, Proc. Amer. Math. Soc. 17 (1966) 670-673. MR0194340.2011-03-24
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    @Asaf: My point is that joriki writes "we just need a complementary subspace for a one-dimensional subspace"; I *think* we need it for a 2-dim subspace (though, I guess, if you can always find one for a 1-dim subspace, you can just take a 1-dim subspace of the complement and find a complement of *that* in the complement...)2011-03-24
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    @Arturo Magidin: Yes, that's what I had in mind, taking the complement twice; maybe I should have made that explicit. Concerning form 66: I was expecting that to be equivalent and was surprised to find that the implication 66 $\implies$ AC is marked as 7 ("The status of the implication is unknown and we have found the question asked somewhere in the literature") at the site -- that seems to be a mistake then.2011-03-24
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    @joriki: If I recall Halpern's paper correctly, it takes a rather circuitous route: he shows "Every vector space has a basis" implies the Axiom of Multiple Choice in a slightly weaker theory than ZF, and then invokes Jech's book to claim that ZF+Multiple Choice --> Choice (AMC: for every family of nonempty disjoint sets, there is a function that assigns to every index a finite subset of the ith set of the family). It's possible that one of the links in that chain was later found to be incorrect (given that Blass's paper precedes the Consequences book).2011-03-24
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    @joriki: My comment above should refer to Blass's paper, not Halpern. Halpern's is fairly straightforward, though what he proves is that "For all vector spaces V over any field, if S is a spanning set for V, then S contains a basis for V" implies AC.2011-03-25
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    @Arturo: I think the mystery is resolved by the suboptimal encoding of the implication table in Consequences. The table differentiates between implications provable in ZF and provable in a weaker system; multiple choice doesn't imply AC in the weaker system, so the status of the implication is unknown in the weaker system; and apparently that leads to the overall status of the implication being classified as "unknown", even though the equivalence is provable in ZF. (BTW, here's a link to the Blass paper: http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf)2011-03-25
  • 0
    I tried, and I tried. I am very glad that I did. This is a great way to practice construction of permutation models of ZFA (a technique I intend to use heavily for the next two years), I did not reach any meaningful conclusions and even my intuition is stumped here.2011-03-27
  • 0
    I sat now with my advisor and I had a slight breakthrough. So not all is lost yet, I will work on that later today.2011-03-27