Why does $i$ have infinitely many conjugates in $\mathbb{H}$?

Question

Browsing this question: Why are the solutions of polynomial equations so unconstrained over the quaternions?, the pdf linked in the comments says that the infinitely many conjugates of $i$ in $\mathbb{H}$ are roots to $x^2+1$.

I get that they're roots, but how do we know that the conjugacy class of $i$ is in fact infinite?

Answer 1

This should be a comment but it came up too long.

I think there's a confusion here: the quaternion $\,i\,$ has one unique quaternionian conjugate in $\,\Bbb H\,$, as we can deduce from this definition

The sense in which the term "conjugate" seems to be used in the OP is that the minimal polynomial of $\,i\,$ over $\,\Bbb H\,$ has infinite other roots, all of which are "conjugate" to (i.e., roots of the same minimal polynomial of) $\,i\,$

So the question doesn't seem to be connected to conjugacy classes a la group theory but with roots of minimal polynomials.

Answer 2

$$(a+bi+cj+dk)^2+1=0$$ $$a^2+1 + b^2i^2+c^2j^2+d^2k^2+2abi+2acj+2adk+(bcij+cbji)+(cdjk+dckj)+(bdik+dbki)=0$$ The last three brackets are all $0$. The $i,j,k$ components must all be zero as well, which we can do by setting $a=0$. We're left with $$1-b^2-c^2-d^2=0$$ which can be satisfied for infinitely many $b,c,d$.

DonAntonio seems to have covered the issue about quaternion conjugates quite nicely, if that's more what you were asking.