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Let $K$ be a (possibly non-commutative) ring extension of $\mathbb{R}$ and let $i \in K$ be a root of $x^2 + 1$.

$i$ commutes with the rational numbers. Does $i$ need to commute with the irrational numbers as well? If not, what is an example of a ring $K$ where that happens?

$K$ being a ring extension of $\mathbb{R}$ means that $\mathbb{R}$ is a subring of $K$

Hussein Aiman
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    It depends on what you mean by "ring extension of $\mathbb{R}$." If you mean an $\mathbb{R}$-algebra, the elements of $\mathbb{R}$ in $K$ commute with every element of $K$ by definition. If you mean something else you'll have to clarify. – Qiaochu Yuan Jun 07 '24 at 07:05
  • @QiaochuYuan I meant that $\mathbb{R}$ is a subring of $K$. Edited the question to add that – Hussein Aiman Jun 07 '24 at 08:02
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    Just take the free product $K=\mathbb R*_{\mathbb Q} \mathbb Q(i)$, which is the coproduct of $\mathbb R$ and $\mathbb Q(i)$ in the category of associative $\mathbb Q$-algebras. – Just a user Jun 07 '24 at 13:43
  • @Justauser How do you prove that the irrational numbers don't commute with $i$ in that ring? This seems equally hard as the original question. – Adayah Jun 08 '24 at 09:23

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Counterexample

Let $K = \operatorname{End}(\mathbb{R}, +)$ be the endomorphism ring of the additive group of the reals. The ring $(\mathbb{R}, +, \cdot)$ embeds in $K$ via $r \mapsto \varphi_r$, $\varphi_r(x) = r \cdot x$. It is easy to prove that if $f \in K$ commutes with (the embedded copy of) $\mathbb{R}$, then $f \in \mathbb{R}$. Thus it suffices to construct an endomorphism $i \in K$ satisfying $i^2 = -1$.

Fix a basis $B$ of $\mathbb{R}$ as a $\mathbb{Q}$-vector space . Let $B = \{ b_j : j \in J \} \cup \{ c_j : j \in J \}$, where all $b_j$ and $c_k$ are distinct. Then there exists a unique $\mathbb{Q}$-linear map $i \in K$ such that $i(b_j) = c_j$ and $i(c_j) = -b_j$. The identity $i(i(x)) = -x$ holds on $B$, so it holds on $\mathbb{R}$. Thus $i^2 = -1$.

Adayah
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    An equivalent way to construct $i \in K$ is by writing $\mathbb{R} \cong \bigoplus_{j \in J} \mathbb{Q}^2$ (as $\mathbb{Q}$-vector spaces) and defining $i$ on each $\mathbb{Q}^2$ as a rotation by $\frac{\pi}{2}$. – Adayah Jun 07 '24 at 13:06