Do subrings of $\mathbb{H}$ containing 1 contain the conjugates of its elements?

Question

Let $x\in\mathbb{H}$ which is the ring of real quaternions such that $$x=a+bi+cj+dk$$ where $a,b,c,d\in\mathbb{R}$. The quaternion conjugate of $x$ is $$x^* = a-bi-cj-dk.$$

I want to show (or find a counterexample) that if $R$ is a subring of $\mathbb{H}$ such that $1\in R$, then for any $x\in R$, $x^*\in R$.

I have tried some rings (finitely generated) but they either satisfy the property above or I can't show either. For example, I can't show whether $\pi -i$ is in the subring generated by $\{1,\pi+i\}$. But the property is easy to show when I included $j$ and $k$, that is, the property above is true for the subring generated by $\{1,\pi+i,j,k\}$.

Is property always satisfied? Any hint will be appreciated.

Answer 1

Hint: If $\pi-i$ is in the subring generated by $\pi+i$, then so is $2i$.

Stronger hint:

Now get a contradiction using the fact that $\pi+i$ is transcendental.

More details:

Since $\pi+i$ is transcendental, the ring it generates is isomorphic to a polynomial ring $\mathbb{Z}[x]$. This ring contains no square root of $-4$, so it cannot contain $2i$.