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This question is related to this one. In that question, it is stated that nilpotent elements of a non-commutative ring with no non-trivial ring automorphisms form an ideal. Ted asks in the comment for examples of such rings but there are no answers. I would also like to know whether there are such rings and hence this question.

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In general, you can start off with distinct commutative rings with no nontrivial automorphisms and take some kind of noncommutative product. I think the free product should work, so the free product of the reals and the algebraic reals should give an example. – George Lowther Apr 19 '12 at 23:54
...actually, the rings in the product should lack invertible elements (other than $\pm1$) for this to work. So take the free product of the real algebraic integers with the constructible real algebraic integers. Its a bit contrived I know -- there must be a more natural example. – George Lowther Apr 20 '12 at 0:04
@plm: I do not see how Artin-Wedderburn rules out Artinian examples. Can you clarify? – Qiaochu Yuan Apr 23 '12 at 20:06
As a token of appreciation, I will award this bounty to an answer of your choice (I recall you wanted to give several bounties here). – Asaf Karagila May 6 '12 at 15:03
@AsafKaragila Wow, thanks so much! Please award this bounty to George Lowther's answer. – user23211 May 6 '12 at 15:36
up vote 19 down vote accepted

I think I have one. Let $k$ be the field with $2$ elements. Let $R$ be the $k$-algebra with generators $x$, $y$ and $z$, modulo the relations $$zx=xz,\ zy=yz,\ yx=xyz.$$ It is not hard to see that monomials of the form $x^i y^j z^k$ are a basis for $R$. We will call these the standard monomials.

For any $f \neq 0$ in $R$, write $f = \sum f_{ij}(z) x^i y^j$. We will define the leading term of $f$ to be the term $f_{ij}(z) x^i y^j$ where we choose $i+j$ as large as possible, breaking ties by choosing the largest possible power of $i$.

Lemma The center of $R$ is $k[z]$.

Proof Let $Z$ be central and write $Z$ in the basis of standard monomials. Since $Zx=xZ$, there are no powers of $y$ in $Z$. Since $Zy=yZ$, there are no powers of $x$ in $Z$. $\square$.

Lemma Every automorphism of $R$ acts trivially on the center of $R$.

Proof Every automorphism of $k[z]$ is of the form $z \mapsto az+b$ for $a \in k^{\ast}$. Since $k$ has two elements, we must have $\sigma: z \mapsto z+b$. Any automorphism of $R$ descends to an automorphism of the abelianization, which is $k[x,y,z]/(xy(z-1))$. Since $z-1$ is a zero divisor in the abelianization, $z+b-1$ must be a zero divisor as well, and this forces $b$ to be zero. $\square$.

Lemma If $f$ and $g$ have leading terms $f_{ij}(z) x^i y^j$ and $g_{kl}(z) x^k y^l$, then the leading term of $fg$ is $z^{jk} f_{ij}(z) g_{kl}(z) x^{i+k} y^{j+l}$.

Proof A computation. $\square$

Now, suppose that we have an automorphism $x \mapsto X$, $y \mapsto Y$, $z \mapsto z$ of $R$. Let the leading terms of $X$ and $Y$ be $f(z) x^i y^j$ and $g(z) x^k y^l$.

Lemma The vectors $(i,j)$ and $(k,l)$ are linearly independent.

Proof We are supposed to have $YX=zXY$. Taking leading terms $$z^{il} f(z) g(z) x^{i+k} y^{j+l} = z^{jk+1} f(z) g(z) x^{i+k} y^{j+l}.$$ So $il-jk=1$ and $\det \left( \begin{smallmatrix} i & j \\ k & l \end{smallmatrix} \right)=1$. $\square$

Consider the images $z^a X^b Y^c$ of the standard monomials. Their leading terms are $z^{a+b+c} f(z)^b g(z)^c x^{bi+ck} y^{bi+cl}.$ Using the above lemma, these leading terms are all distinct. So there is no cancellation of leading terms in any sum $\sum s_{abc} z^a X^b Y^c$. So we see that every element in the image of the automorphism must have a leading term of the form $h(z) x^{bi+ck} y^{bj+cl}.$

But automorphism are surjective! So $(1,0)$ and $(0,1)$ must be positive integer combinations of $(i,j)$ and $(k,l)$. So either $(i,j) = (1,0)$ and $(k,l) = (0,1)$ or viceversa. We see that $X$ and $Y$ are of degree $1$ in $x$ and $y$. From this point, it's an easy computation.

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Nice! I did think of $\mathbb{Z}\langle X,Y\rangle/(XY-3YX)$, but that has nontrivial automorphisms taking $X$ to $-X$ and taking $Y$ to $-Y$. Your example uses the field of 2 elements to force the maps $X\mapsto-X,Y\mapsto-Y$ to be trivial and replaces $3$ by an indeterminate $z$, which seems to be a very neat trick to make this idea work. – George Lowther Apr 26 '12 at 22:12
@GeorgeLowther Our minds run on similar tracks, then. I tried $yx=2xy$ right before this one. – David Speyer Apr 26 '12 at 22:20
Thank you very much for the example! – user23211 Apr 27 '12 at 10:41

Apparently such rings do exist.

After trying to construct one without much success, I did some googling and found the following article:

  • Maxson, C. J. 1979. Rigid rings. Proc. Edinburgh Math. Soc., 21(2): 95–101.

In it, the author uses the following definition: a ring $R$ (with non-zero multiplication, not necessarily possesing a multiplicative identity) is said to be rigid if $R$ admits no endomorphisms other than $0_R$ and $\operatorname{id}_R$. The author notes that he knows no examples of non-commutative rigid rings.

However, according to

a non-commutative rigid ring was constructed in the article

  • Friger, M. D. 1986. About Rigid Torsion-Free Rings. Siberian Math. J., 3: 217–219,

which, unfortunately, I have not been able to locate.

Rigidity seems to be a stronger condition than the one OP wants, so there may still be some hope of a simple example. I shall conclude this list of sources with some of my own (probably pretty naive, since I'm not in any way an expert on this) thoughts.

While thinking about the problem I spent quite some time trying to confirm or refute $\mathbb{Z}\langle X,Y\rangle/(X^3-3,Y^3-5)$ as an example. I had no success though. My intuition was that a ring with wanted properties should behave similarly to $\mathbb Z$ in some way, but that "generators" could be distinguished in some way. "Adjoining third roots of $3$ and $5$" that do not commute with each other seemed like a good idea. The problem with a square root $x$ is that the negative and positive one seems to be basically undistinguishable, thus probably yielding an automorphism of the form $x\mapsto -x$. Third roots don't seem to have this problem though.

Anyway, I didn't see a good way to prove anything about this ring, so I went to the quaternions $\mathbb H$. I noticed that $a=3^{\frac13}(\cos\frac{2\pi}3+i\sin\frac{2\pi}3)$ and $b=5^{\frac13}(\cos\frac{2\pi}3+j\sin\frac{2\pi}3)$ are third roots of $3$ and $5$ in $\mathbb H$. Furthermore, they don't commute with each other. So, I thought I should try my luck and decided to observe the smallest subring $R$ of $\mathbb H$ that contains $1,a,b$. However, after a lot of thought and some computer-assisted computations, it turned out $a^2+a+1$ is a non-central invertible element, thus yielding an automorphism of this ring:

$$\begin{array}{ll} \Phi: R\to R \\ \Phi(x)=8(a^2+a+1)x((abab^2a-ab^2aba)^2+506)^2(a-1) \end{array}$$

To see why this is an automorphism, note that $(abab^2a-ab^2aba)^2+506=-\frac14$.

Anyway, this still doesn't seem to rule out $\mathbb Z\langle X,Y\rangle/(X^3-3,Y^3-5)$ as a possible example, so if anyone sees how to prove or refute that this is an example, I'd be very happy to know.

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Thanks a lot for the thoughts and for the sources! Apparently the Friger paper is available in a library in my city. I'll try to check it out tomorrow. – user23211 Apr 26 '12 at 17:40
I managed to work out an example of a noncommutative ring with no trivial automorphisms (although rather complicated, and not rigid) which I was about to post. Having seen this answer though, I'm thinking about $\mathbb{Z}\langle X,Y\rangle/(X^3-3,Y^3-5)$. If that is rigid, then it is a much simpler example. – George Lowther Apr 26 '12 at 19:10
@GeorgeLowther Please do post your example though. There can't be too many of them! – user23211 Apr 26 '12 at 21:17
I have Friger's paper in front of me but it's in Russian so I have trouble understanding it. I'm going to make a copy so if someone would be willing to read it and post some abstract here, please contact me. I'm also trying to read the paper but I may well fail. – user23211 Apr 27 '12 at 10:33

Here's a construction of a non-commutative ring with no non-trivial automorphisms. It is not rigid though (rigid=no nontrivial endomorphisms. See Dejan's answer).

First, define sets $S_1,S_2\subset\mathbb{N}$, where $S_1$ is the set of square-free numbers whose prime factors are equal to 1 mod 4 and $S_2$ is the set of square-free numbers equal to 3 mod 4. (The precise choice of $S_1,S_2$ won't really matter).

Now, for $i=1,2$, define $G_i$ to be the set of nonnegative rational numbers of the form $p/q$ for $p$ a nonnegative integer and $q\in S_i$. These sets are closed under addition, so are commutative monoids. Now form the free product $G=G_1*G_2$ with canonical maps $\theta_i\colon G_i\to G$, which I will write multiplicatively. To be precise, $G$ is a (noncommutative) monoid generated by elements $\{\theta_i(x)\colon x\in G_i\}$ ($i=1,2$) subject to the relations that $\theta_i(0)=e$ is the identity and $\theta_i(x)\theta_i(y)=\theta_i(x+y)$. Every element $g\in G$ can be written uniquely as $$ g=\theta_{i_1}(x_1)\theta_{i_2}(x_2)\cdots\theta_{i_n}(x_n)\qquad\qquad{\rm(1)} $$ for $n\ge0$, $i_k\in\{1,2\}$, $x_k\in G_k\setminus\{0\}$ and $i_k\not=i_{k+1}$ (I'm taking the empty product to be the identity $e$, for the case $n=0$).

Construct the monoid ring $R=F_2[G]$, where $F_2$ is the field with two elements. Every element $a\in R$ can be written as $$ a = \sum_{i=1}^na_i g_i\qquad\qquad{\rm(2)} $$ for $n\ge0$, $a_i\in F_2$ and $g_i\in G$. Furthermore, this can be done so that $g_i$ are distinct and $a_i\not=0$ (equivalently, so that $n$ is minimal) in which case the representation is unique.

Then, $R$ is clearly noncommutative, as it contains the multiplicative and noncommutative monoid $G$. It also has no nontrivial automorphisms. I'll post the proof of this in a moment, but the idea is that any ring-automorphism of $R$ is given by a monoid-automorphism of $G$, and $G$ has no nontrivial automorphisms.

The proof that $R$ has no nontrivial automorphisms follows now.

First, I'll define a bit of notation denoting the 'degree' of elements of $G$ and $R$. For any $g\in G$, let $\vert g\vert$ denote the integer $n$ occuring in expansion (1). When we multiply two terms $g,h\in G$ then we just concatenate the expansions and, possibly combine the last term in the expansion for $g$ with the first term in the expansion for $h$. There can be no further cancellation, as $G_1,G_2$ have no nontrivial units. So, $$ \vert g\vert+\vert h\vert-1\le\vert gh\vert\le\vert g\vert+\vert h\vert. $$ Now, for any nonzero $a\in R$ let $\vert a \vert$ denote the maximum of $\vert g_i\vert$ as $g_i\in G$ runs through the terms in the minimal expansion (2) for $a$. If we have $a,b\in R$ then let $g\in G$ be a term in the expansion of $a$ maximizing $\vert g\vert$ and $h\in G$ be a term in the expansion of $b$ maximizing $\vert h\vert$. So, $\vert a\vert=\vert g\vert$ and $\vert b\vert=\vert h\vert$. Among the possible choices for $g$ choose one maximizing $x_n$ in expansion (1) and, among the possible choices for $h$, choose one maximizing $x_1$ in expansion (1). Then, expanding out $ab$, the term $gh$ occurs precisely once. So, $\vert ab\vert\ge\vert gh\vert$ and we get, $$ \vert a\vert+\vert b\vert-1\le\vert ab\vert\le \vert a\vert+\vert b\vert. $$ Now, we can prove the following.

If a nonzero element $a\in R$ has solutions to $b^n=a$ for infinitely many positive integers $n$, then $a=\theta_i(x)$ for some $i\in\{1,2\}$ and $x\in G_i$. Furthermore, in that case, the only solutions to $b^n=a$ are $b=\theta_i(x/n)$ and $x/n\in G_i$.

Proof: Assume $a$ is not the identity, for which the conclusion is immediate.

Suppose that $\vert b\vert\ge2$. Then, from the inequalities above, $\vert b^n\vert\ge n+1$. For large enough $n$, this will exceed $\vert a\vert$. So, we must have $\vert b\vert=1$ when $n$ is large. The only possibilities are $b\in\hat G_1\equiv{\rm Im}(\theta_1)$, $b\in \hat G_2\equiv{\rm Im}(\theta_2)$ and $b=g_1+g_2$ for $g_i\in G_i$. In the latter case, $g_1g_2g_1g_2\cdots$ occurs in the expansion for $b^n$, so $\vert b^n\vert=n$ which will exceed $\vert a\vert$ if $n$ is large. So, for large enough $n$, any solution to $b^n=a$ will be of the form $b=\theta_i(y)$ so $a=b^n=\theta_i(ny)$ as required.

So, we know that $a=\theta_i(x)$ for some nonzero $x\in G_i$ and, hence, $\vert a\vert=1$. If $b^n=a$ ($n > 1$) and $\vert b\vert\ge2$ then $\vert b^n\vert\ge n+1 > \vert a\vert$, giving a contradiction. As above, if $b$ is not of the form $\theta_j(y)$ for $y\in G_j$ then $\vert b^n\vert=n > \vert a\vert$. So, $b=\theta_j(y)$. As $\theta_j(ny)=b^n=a=\theta_i(x)$, we have $j=i$ and $x/n=y\in G_i$. QED.

Finally, for $i\in\{1,2\}$ and $x\in G_i$, the element $\theta_i(x)$ of $R$ is characterized purely by its algebraic properties, so must be fixed by every automorphism. This shows that the automorphism group of $R$ is trivial.

The element $a=\theta_i(x)$ of $R$ is uniquely determined by the following property: for postive integers $n$, $b^n=a$ has a solution in $R$ (for $b$) if and only if $x/n\in G_i$.

Proof: The previous lemma shows that if $a$ is of the required form then $b^n=a$ has a solution if and only if $x/n\in G_i$. Conversely, suppose that $a$ satisfies the required property. Then, there are infinitely many $n$ so that $x/n\in G_i$ and, by the previous lemma, $a=\theta_j(y)$ for some $j\in\{1,2\}$. By the previous lemma, a positive integer $n$ satisfies $y/n\in G_j$ if and only if $b^n=a$ which, by the hypothesis, is equivalent to $x/n\in G_i$. By the choice of $G_1,G_2$ this forces $i=j$ and $x=y$. QED

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Interestingly, although my example is rather more complicated than the one in David Speyer's answer, they are both of the form $F_2[G]$ for a noncommutative monoid $G$. – George Lowther Apr 26 '12 at 23:22
That's really elegant. Also, both proofs rely on a fairly similar notion of degrees and leading terms. Did you forget to write down the explanation for why $G$ has no nontrivial automorphisms? – David Speyer Apr 27 '12 at 5:00
Actually, I just realised that I missed out part of the proof of the first highlighted statement above. I'll come back and add this – George Lowther Apr 27 '12 at 9:58
Courtsey of ymar! – Asaf Karagila May 7 '12 at 16:10
@Asaf, ymar: Thanks! – George Lowther May 7 '12 at 19:08

Some thoughts. The basic observation is that if $r \in R$ is non-central and a unit, then $a \mapsto rar^{-1}$ is a nontrivial automorphism, so at a minimum the unit group of $R$ needs to be contained in its center. But it is not so easy to get rid of non-central units:

  • If $r$ is non-central and nilpotent, then $1 - r$ is non-central and a unit.
  • If $r$ is non-central and idempotent, then $1 - 2r$ is a unit (although it isn't necessarily non-central, e.g. if $2r = 0$).

I am in particular pessimistic about the possibility of finding a finite counterexample. If $r \in R$ is non-central, then by pigeonhole we have $r^n = r^m$ for some minimal $n \ge m$. If $r$ is a unit, we have failed; otherwise $r^m(r^{n-m} - 1) = 0$, and since $r^{n-m} - 1 \neq 0$ it follows that $r^m$ is a zero divisor (possibly zero). In this situation it is easy for $r$ to be nilpotent (guaranteed if $R$ is primary).

Even if $R$ isn't primary, it follows that there exists some $k$ (the smallest multiple of $n-m$ which is at least $n$) such that $r^k$ is idempotent. We have failed if this idempotent is $0$ or $1$, and probably we have failed if this idempotent is anything else as well (except in characteristic $2$).

Central idempotents are also bad! If $e$ is a central idempotent then $R$ breaks up into a direct product $Re \times R(1-e)$. Any automorphism of the subrings $Re$ and $R(1-e)$ extends to an automorphism of the entire ring, so at least in the finite case (also finite-dimensional over a field) we may assume that there are no nontrivial central idempotents.

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Thanks! It looks like even if there is no answer, I'm going to learn more from this question than from most other ones of mine. :) – user23211 Apr 20 '12 at 23:52

I thought of some properties such a ring should satisfy but don't know if they can be of any use to settle the question.

Let the Jacobson radical be $J(R)$ and the commutator ideal (the ideal generated by elements of the form $ab-ba$) be $C(R)$. First, $J(R)$ should be central because for $x \in J(R)$, $1-x$ is a unit and the units should be central, as mentioned before. From here, it follows that $C(R)J(R) = 0$. Indeed, given $a,b \in R$ and $z \in J(R)$ we have $$ (ab)z = a(bz) =(bz)a = b(za) = b(az) = (ba)z$$ so $(ab-ba)z = 0$.

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Thank you very much! – user23211 Apr 26 '12 at 17:41
Reading this suddenly made me realize that a local ring with trivial automorphism group would never be found: all units central+radical elements central = commutative ring! – rschwieb Apr 27 '12 at 18:46

Edit: This argument fails, but I think I'll leave it up in case the error I made is instructive.

Going off of Ted's example in the comments, I claim that $R = \mathbb{F}_2 \langle x, y \rangle / (xy)$ has no nontrivial automorphisms. Suppose otherwise and let $\phi : R \to R$ be such an automorphism. Then $\phi(x)$ must be a left zero divisor and $\phi(y)$ must be a right zero divisor, but a straightforward calculation shows that the only left zero divisors have the form $rx, r \in R$ and similarly the only right zero divisors have the form $ys, s \in R$.

Since $\phi$ is an automorphism, $r, s \neq 0$, and since $\phi$ is nontrivial, at least one of $r$ and $s$ cannot be equal to $1$. However, $R$ is graded, and if WLOG $r \neq 1$ then it has degree at least $1$, so the subring generated by $rx$ and $ys$ cannot contain $x$; thus $\phi$ cannot be surjective.

The above claim is false; $R$ is not graded because for example $(1 + yx)^2 = 1$. In fact $1 + yx$ is therefore invertible and non-central, so conjugation by it gives a nontrivial automorphism.

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Does it help to also mod out every word of length 3? That at least makes your example conjugation equal to the trivial automorphism. It would leave a finite ring of order $64$ if I've counted right. – alex.jordan Apr 20 '12 at 1:00
Nope...nevermind. Then $(1+x)(1+x+x^2)=1$. – alex.jordan Apr 20 '12 at 1:16
@alex: I am not optimistic about the possibility of finding a finite example. In the finite case left invertible implies right invertible, and if there are few invertible elements then there are many idempotents and that seems bad. – Qiaochu Yuan Apr 20 '12 at 3:46

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