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.

**Answer**

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 f≠0 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 z↦az+b for a∈k∗. Since k has two elements, we must have σ:z↦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. ◻.

**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 x↦X, y↦Y, z↦z 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

zilf(z)g(z)xi+kyj+l=zjk+1f(z)g(z)xi+kyj+l.

So il−jk=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.

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