# Is there a non-commutative ring with a trivial automorphism group?

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.

• 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. Commented Apr 19, 2012 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. Commented Apr 20, 2012 at 0:04
• @plm: I do not see how Artin-Wedderburn rules out Artinian examples. Can you clarify? Commented Apr 23, 2012 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). Commented May 6, 2012 at 15:03
• @AsafKaragila Wow, thanks so much! Please award this bounty to George Lowther's answer.
– user23211
Commented May 6, 2012 at 15:36

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.

• 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. Commented Apr 26, 2012 at 22:12
• @GeorgeLowther Our minds run on similar tracks, then. I tried $yx=2xy$ right before this one. Commented Apr 26, 2012 at 22:20
• Thank you very much for the example!
– user23211
Commented Apr 27, 2012 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.

• 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
Commented Apr 26, 2012 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. Commented Apr 26, 2012 at 19:10
• @GeorgeLowther Please do post your example though. There can't be too many of them!
– user23211
Commented Apr 26, 2012 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
Commented Apr 27, 2012 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

• 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 $G$. Commented Apr 26, 2012 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? Commented Apr 27, 2012 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 Commented Apr 27, 2012 at 9:58
• Courtsey of ymar! Commented May 7, 2012 at 16:10
• @Asaf, ymar: Thanks! Commented May 7, 2012 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.

• 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
Commented Apr 20, 2012 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 $(ab-ba)z = 0$.

• 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! Commented Apr 27, 2012 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.

• 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. Commented Apr 20, 2012 at 1:00
• Nope...nevermind. Then $(1+x)(1+x+x^2)=1$. Commented Apr 20, 2012 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. Commented Apr 20, 2012 at 3:46