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The following should be known, but I could not find an example.

Let $\kappa$ be an uncountable cardinal. Find a model $M$ of size $\kappa$ which has $\ge\kappa$ many automorphisms, but for some $m\in M$, $(M,m)$ has only the trivial automorphism (I call this element rigid on the title. Not sure if there is a standard name for it).

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Do you want to assume the language is countable? If the language can have size $\kappa$ it is easy to construct examples. –  Eric Wofsey May 23 at 12:54

3 Answers 3

up vote 7 down vote accepted

Let $(F,+,\cdot)$ be a rigid field, and put $M=(F,+,R)$, where $R(x,y,u,v)\iff xy=uv$. Then $(M,1)$ is interdefinable with $(F,+,\cdot)$ and therefore rigid, but $M$ has $|F|$ automorphisms $x\mapsto ax$ for each $a\in F^\times$. There exist rigid fields of any infinite cardinality, see e.g. http://mathoverflow.net/a/61082/12705.

Generalizing this and Eric Wofsey’s answer, if $(A,\cdot,R_1,R_2,\dots,F_1,F_2,\dots)$ is any rigid structure such that $\cdot$ is a group operation, we can take $M=(A,\cdot',R'_1,\dots,F'_1,\dots)$, where \begin{gather} \cdot'(x,y,z)=xz^{-1}y,\\ R'_i(x_1,\dots,x_n,z)\iff R_i(x_1z^{-1},\dots,x_nz^{-1}),\\ F'_i(x_1,\dots,x_n,z)=F_i(x_1z^{-1},\dots,x_nz^{-1})z. \end{gather} Then $M$ has automorphisms $x\mapsto xa$ for each $a\in A$, but $(M,1)$ is rigid. Elementary constructions of rigid structures (e.g., graphs) of arbitrary cardinality are easy to find, and expanding the structure with a $\cdot$ if necessary preserves rigidity.

Since all answers given so far are instances of the construction above, let me also mention an example that works in a different way. Let $F$ be a rigid field, and let $M$ be the polynomial ring $F[x]$. Since $F^\times=M^\times$ is definable in $M$, every automorphism of $M$ fixes $F$, and is uniquely determined by the image of $x$. Thus, $M$ has automorphisms $f(x)\mapsto f(ax+b)$ for $a\in F^\times$, $b\in F$, whereas $(M,x)$ is rigid.

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That was an idea I had in mind, but the only examples of a rigid field I managed to think of were $\Bbb R$'s subfields. Is there an easy proof that there are rigid fields of every cardinality? –  Asaf Karagila May 23 at 13:18
    
Thanks for the edit! –  Asaf Karagila May 23 at 13:23
    
The easiest example of a rigid structure of arbitrary cardinality I can think of is a well-ordering. Is there an example that does not require Choice? –  Eric Wofsey May 23 at 14:06
    
Rigidity of well orders does not require choice, actually. –  Emil Jeřábek May 23 at 14:08
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It is consistent with ZF that there exists a set that does not carry any rigid structure in a countable language, see mathoverflow.net/questions/6262 (it is stated there for a single binary relation, but the argument for ZFA applies to any well-orderable language, and as long as one bounds its cardinality, this transfers to ZF by the Jech–Sochor embedding theorem). –  Emil Jeřábek May 23 at 14:33

Here's one example. Take any ring $R$ equipped with the following two operations: $$(x,y,z)\mapsto x+y-z$$ $$(x,y,z,w)\mapsto x+(y-z)(w-x)$$

It is easy to see that if you add a constant symbol $0$, then the entire ring structure of $R$ is definable from these two operations. However, for any $a\in R$ the map $x\mapsto x+a$ preserves these operations and thus gives an automorphism. Hence if $R$ is any ring of cardinality $\kappa$ with no nontrivial ring-automorphisms this model will have the desired properties. One such uncountable ring is $\mathbb{R}$; I don't know whether rigid rings of all infinite cardinalities exist but I suspect they do.

(By the way, if you're wondering how I came up with those two operations, this is just the ring analogue of a heap. However, you can't just use heaps, because any group with more than two elements has nontrivial automorphisms.)

Edit: As mentioned in Emil's answer, it turns out that there are rigid fields of any infinite cardinality.

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Using forcing we can always blow up the continuum to be $\geq\kappa$, and then find such ring! :-) –  Asaf Karagila May 23 at 13:22

I claim that one can modify a counterexample with $\kappa$ non-logical symbols to obtain a counterexample with only countably many non-logical symbols. It is easy to construct structures with many relation symbols or function symbols and with many automorphisms, but where if we fix an element, then the new structure would be rigid (as was noted by Eric Wofsey). For example, if $F$ is a field and $f_{a}:F\rightarrow F$ is the mapping defined by $f_{a}(x)=ax$, then $(F,(f_{a})_{a\in A})$

Suppose that $(X,(R_{\alpha,n})_{\alpha<\kappa,n\in\mathbb{N}})$ is a structure where each $R_{\alpha,n}$ is an $n$-ary relation. Suppose now that $0\in X$ and $(X,0,(R_{\alpha,n})_{\alpha<\kappa,n\in\mathbb{N}})$ has trivial automorphism group. Then for each $n$, let $R_{n}$ be the $n$-ary relation on $X\times\kappa$ where $((x_{1},\alpha_{1}),...,(x_{n},\alpha_{n}))\in R_{n}$ if and only if $\alpha_{1}=...=\alpha_{n}$ and $(x_{1},...,x_{n})\in R_{\alpha_{1},n}$. Let $\leq$ be the partial ordering on $X\times\kappa$ where $(x,\alpha)\leq(y,\beta)$ if and only if $\alpha\leq\beta$.

Suppose $f:X\rightarrow X$ is an automorphism of $(X,(R_{\alpha,n})_{\alpha,n})$. Then define $\phi:X\times\kappa\rightarrow X\times\kappa$ by letting $\phi(x,\alpha)=(f(x),\alpha).$ Then $\phi$ is an automorphism of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$.

Now suppose that $\phi:X\times\kappa\rightarrow X\times\kappa$ is an automorphism of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$. Then it is easy to see that if $\phi(x,\alpha)=(y,\beta)$, then $\alpha=\beta$. Therefore for each $\alpha<\kappa$, there is some $f_{\alpha}:X\rightarrow X$ such that $\phi(x,\alpha)=(f_{\alpha}(x),\alpha)$. However, if $\alpha<\beta$, then $(x,\alpha)\leq(x,\beta)$, so $(f_{\alpha}(x),\alpha)=\phi(x,\alpha)\leq\phi(x,\beta)=(f_{\beta}(x),\beta)$. Therefore, we have $f_{\alpha}(x)=f_{\beta}(x)$. We conclude that there is some $f:X\rightarrow X$ with $\phi(x,\alpha)=(f(x),\alpha)$. Furthermore, the automorphism $f$ fixes $0$ if and only if the automorphism $\phi$ fixes $(0,0)$.

$(x_{1},...,x_{n})\in R_{\alpha,n}$ if and only if $((x_{1},\alpha),...,(x_{n},\alpha))\in R_{\alpha}$ if and only if $((f(x_{1}),\alpha),...,(f(x_{n}),\alpha))=(\phi(x_{1},\alpha),...,\phi(x_{n},\alpha))\in R_{\alpha}$ if and only if $(f(x_{1}),...,f(x_{n}))\in R_{\alpha}$. Therefore, the function $f$ is an automorphism of $(X,(R_{\alpha,n})_{\alpha,n})$. We therefore have the automorphism group of $(X,(R_{\alpha,n})_{\alpha,n})$ be in a one-to-one correspondence with the automorphism group of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$. Furthermore, since only one automorphism of $X$ fixes $0$, only the trivial automorphism of $(X\times\kappa,(R_{n})_{n\in\mathbb{N}},\leq)$ fixes $(0,0)$.

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