# Nullstellensatz for quaternionic plane curves?

By a quaternionic plane curve I mean the zero locus of a noncommutative polynomial in two variables, $x$ and $y$ say, over ${\Bbb H}$, Hamilton's quaternions. It is evidently well-known that, after identifying ${\Bbb H}$ with ${\Bbb R^4}$, one can capture with a single quaternionic polynomial any locus that occurs as the set of solutions of four real polynomials. (One needs a trick: if $Q=a+bi+cj+dk$, $(-Q+iQi+jQj+kQk)/2=a$; and similarly for $b$,$c$ and $d$).

Is there a criterion in terms of ${\Bbb H}<x,y>$ for two polynomials to cut out the same zero locus? Of course in principle one could unwind everything and then appeal to some appropriate realnullstellensatz, but I'm interested in a solution that is idiomatically quaternionic.

And a softer question. Capturing a real 4-fold in ${\Bbb R}^4$ as a specific quaternionic plane curve singles out a way to produce the 4-fold as the intersection of 4-hypersurfaces. That imposes a great deal of extra structure on the 4-fold. Indeed presumably too much, seeing as certain natural quaternionic manipulations (multiplication by non-zero constants on either side) change the intersecting polynomials but not the locus. What is the right level of extra structure in order to move to the quaternionic picture?

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This is quite a neat trick! Could you refer to a publication that would elaborate on non-commutative quaternionic polynomials? Also there seems to be a typo. I think the formulae you refer to are these: a = (Q -iQi -jQj -kQk)/4 b = (Q -iQi +jQj +kQk)/4 c = (Q +iQi -jQj +kQk)/4 d = (Q +iQi +jQj -kQk)/4 This begs for a generalization to algebras with sufficiently many inner automorphisms. – Michael Sep 5 '13 at 17:26