Showing that a subring $K$ of $\mathbb H$ contains a field which is isomorphic to $\mathbb C$

Question

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.

Answer 1

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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Answer 2

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 $i,j,k$.

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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 $\mathbb{C}$.