Let $K$ be a subring of $\mathbb H$, the ring of the quaternions, with $\mathbb R \subseteq K$ and $\mathbb R \neq K$, there $\mathbb R$ is the ring of real numbers.
Show that there exists $x \in K$ such that $ x^2 = -1$. Use this fact to deduce that $K$ contains a field which is isomorphic to $\mathbb C$, the ring of complex numbers.
My reasonings:
Since $\mathbb R \subseteq K$ but $\mathbb R \neq K$, there should exists some $u \in \{i, j, k\}$, such that $u \in K$, where $i, j, k$ are the quaternion units and, in particular, satisfy
$i^2=j^2=k^2=-1$
This occured to me because, in order for $K$ to be different from $\mathbb R$, it has to contain at least one of these units. If $K$ actually contains $u$, then $u$ is a solution of
$x^2=-1$
At this point I showed, if everything is correct, that $K$ contains such $x$, but I don't know how to show the last part of the question.
I wondered that I could consider
$\mathbb R[u]=\{a+ub:a,b \in \mathbb R\}$
We have that $\mathbb R[u] \subseteq K$, since $\mathbb R \subseteq K$ and $u \in K$ and $K$ is a ring.
To show that $\mathbb R[u]$ is a field and that it is isomorphic to $\mathbb C$, it would be "easy" to use polynomials and quotients, in fact we have
$\mathbb R[u] \simeq \mathbb R[x]/(x^2+1)$
Where $\mathbb R[x]$ is the ring of polynomials over $\mathbb R$ and $(x^2+1)$ is the principal ideal generated by the polynomial $x^2+1$, which has no roots in $\mathbb R$, making it maximal. This isomorphism holds because $x^2+1$ is the minimal polyinomial of $u$ over $\mathbb R$.
But we also know that
$\mathbb C \simeq \mathbb R[x]/(x^2+1)$
Where we can actually see $\mathbb C$ as $\mathbb R[i]=\{a+ib:a,b \in \mathbb R\}$.
We conclude that
$\mathbb R[u] \simeq \mathbb C$
Now, this method might or might not be correct, but my real question is finding a way to do it without using quotients, maximal ideals and "advanced" properties of polynomials over a field, because this exercise is given, in my course, before all of them.