Sean Eberhard

I am advertising a PhD position, with a start date of 1 October 2024, at Queen’s University Belfast. The ideal candidate is somebody with a strong undergraduate/Master’s-level background in algebra, combinatorics, or number theory, with a particular fascination for asymptotic group theory or additive combinatorics. Nominally the topic is “diameter bounds for irreducible linear groups generated by long root elements”, which refers to an important subquestion about diameter bounds for finite classical groups, but the ideal candidate would have a strong inclination to branch out beyond this specific goal into broader questions of asymptotic group theory. For the full official advertisement, see the link below!

https://www.qub.ac.uk/courses/postgraduate-research/phd-opportunities/diameter-bounds-for-irreducible-linear-groups-generated-by-long-root-elements.html

This semester I am lecturing the level-3 course Rings and Modules at QUB. While reviewing some material and preparing lecture notes I puzzled over something for a while before finally catching my mistake. How do you fare?

Question: Let $\mathbf{H}$ be the ring of quaternions — the four-dimensional real vector space with basis $1, i, j, k$ and the usual relations. Let $\mathbf{H}[X]$ be the ring of univariate polynomials with coefficients in $\mathbf{H}$ . What is $\mathbf{H}[X] / (X-i)$ ?

Take a moment to try to answer the question if you like before reading on.

The trap here is to think of quotienting by $(X-i)$ as meaning “substitute $X \mapsto i$ ”, which would suggest that the quotient ring is isomorphic to $\mathbf{H}$ . That’s the wrong answer. The problem is that the map defined by sending a polynomial $f(X) \in \mathbf{H}[X]$ to “ $f(i)$ ” is not a homomorphism! Indeed it is implicit in the definition of the ring of polynomials that $X$ commutes with the base ring, whereas the element $i$ does not.

In fact $\mathbf{H}[X] / (X-i)$ is the zero ring. To see this, let $I = (X-i)$ and observe that $I$ contains $-j(X - i) j = X + i$ and hence also $(X+i) - (X-i) = 2i$ , a unit, so $I = \mathbf{H}[X]$ .