Showing that a subring $K$ of $\mathbb H$ contains a field which is isomorphic to $\mathbb C$ 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.
 A: Your starting point is wrong. What you know is that there exists a quaternion $a+bi+cj+dk$ such that at least one among $b,c,d$ is nonzero.
There is no reason why an elementary quaternion needs to be in $K$.
A simple example is $\mathbb{R}[q]$, where $q=(i+j+k)/\sqrt{3}$, which is actually a field isomorphic to $\mathbb{C}$ and does not contain any of $i,j,k$.

Let $u\in K$, $u\notin\mathbb{R}$. Then the quaternions $1,u,u^2,u^3,u^4$ are not linearly independent, because $\mathbb{H}$ has dimension four over $\mathbb{R}$. Therefore there exists a polynomial with real coefficients that vanishes at $u$. On the other hand, the polynomial can be factored into irreducible factors having degree one or two and, since the quaternions are a division algebra, one of the factors must vanish at $u$. Such a factor must have degree two, otherwise $u$ would be real.
Without loss of generality, the polynomial is monic. Thus there are $a,b\in\mathbb{R}$ such that $u^2+au+b=0$. We can now complete the square
$$
\Bigl(u-\frac{a}{2}\Bigr)^2+b-\frac{a^2}{4}=0
$$
Note that $b-a^2/4>0$, because $x^2+ax+b$ is by assumption an irreducible polynomial. Set $c=\sqrt{b-a^2/4}$ and $v=(u-a/2)/c$; it follows from the assumptions that $v\in K$. Then $c^2v^2+c^2=0$, hence $v^2=-1$.
Now show that $\mathbb{R}[v]$ is a field. Since it is algebraic over $\mathbb{R}$, it must be isomorphic to $\mathbb{C}$.
A: As is well-known, $\Bbb H$ is possessed of a basis consisting of
$1 \in \Bbb R \tag 1$
and $i$, $j$, $k$ such that
$ij = k, \; jk = i, \; ki = j, \tag 2$
$i^2 = j^2 = k^2 = -1; \tag 3$
of course, (2) and (3) together imply that $i$, $j$, $k$ anti-commute, viz:
$-j = i^2j = i(ij) = ik, \tag 4$
with similar arguments showing that
$ji = -k, \; kj = -i; \tag 5$
using (2)-(4) we compute $(ai + bj + ck)^2$, where $a, b, c \in \Bbb R$:
$(ai + bj + ck)^2 = (ai + bj + ck)(ai + bj + ck)$
$= a^2ii + b^2jj + c^2kk + abij + acik + abji + bcjk + acki + bckj$
$= -a^2 - b^2 - c^2 + ab(ij + hi) + ac(ik + ki) + bc(jk + kj)$
$= -(a^2 + b^2 + c^2) < 0, \tag 6$
provided at least one out of $a$, $b$, $c$ does not vanish.  This yields
$\left ( \dfrac{ai + bj + ck}{\sqrt{a^2 + b^2 + c^2}} \right )^2 = \dfrac{(ai + bj + ck)^2}{a^2 + b^2 + c^2} = -1. \tag 7$
Now if $K$ is a subring of $\Bbb H$ with
$\Bbb R \subsetneq K \subset \Bbb H, \tag 8$
then $K$ must contain an element $q \in\Bbb H$ of the form
$q = r + ai + bj + ck, \tag 9$
with
$r, a, b, c \in \Bbb R, \tag{10}$
and at least one of $a$, $b$, $c$ non-zero, a condition easily seen to be equivalent to
$a^2 + b^2 + c^2 > 0; \tag{11}$
since $K$ is a subring and (8) implies
$r \in K, \tag{12}$
(9) yields
$p = ai + bj + ck =  q - r  \in K, \tag{13}$
and from what we have seen above
$\left (\dfrac{p}{\sqrt{a^2 + b^2 + c^2}} \right )^2 = -1;  \tag{14}$
now in light of (8) and (10),
$\dfrac{1}{\sqrt{a^2 + b^2 + c^2}} \in K, \tag{15}$
and thus
$u = \dfrac{p}{\sqrt{a^2 + b^2 + c^2}} \in K \tag{16}$
with
$u^2 = -1, \tag{17}$
as shown above in (14); thus the field
$\Bbb R(u) \subset K, \tag{18}$
and using (17) it is easy to see that the elements of $\Bbb R(u)$ are all of the form $a + bu$, $a, b \in \Bbb R$, and thus the mapping
$\Bbb R(u) \ni a + bu \mapsto a + bi \in \Bbb C \tag{19}$
defines an isomorphism 'twixt $\Bbb R(u)$ and $\Bbb C$; we leave it to the sufficiently engaged reader to supply the simple details.
Nota Bene, Wednesday, 20 August 2020 11:24 PM PST:  We observe that the above demonstration indicates that there are many subalgebras of $\Bbb H$ containing $\Bbb R$ and isomorphic to $\Bbb C.$
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