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.


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 xiyjzk are a basis for R.
We will call these the standard monomials.

For any f0 in R, write f=fij(z)xiyj. We will define the leading term of f to be the term fij(z)xiyj 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. .

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

Proof Every automorphism of k[z] is of the form zaz+b for ak. Since k has two elements, we must have σ:zz+b. Any automorphism of R descends to an automorphism of the abelianization, which is k[x,y,z]/(xy(z1)). Since z1 is a zero divisor in the abelianization, z+b1 must be a zero divisor as well, and this forces b to be zero. .

Lemma If f and g have leading terms fij(z)xiyj and gkl(z)xkyl, then the leading term of fg is zjkfij(z)gkl(z)xi+kyj+l.

Proof A computation.

Now, suppose that we have an automorphism xX, yY, zz of R. Let the leading terms of X and Y be f(z)xiyj and g(z)xkyl.

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

Proof We are supposed to have YX=zXY. Taking leading terms
So iljk=1 and det. \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.

Source : Link , Question Author : Community , Answer Author : David E Speyer

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