If a subalgebra ‎$‎K‎$‎ of a unital algebra ‎$‎A‎$‎ is a field and ‎$‎ K ‎\supseteq ‎F‎‎$‎ (i.e., ‎$‎K‎$‎ contains the unity of ‎$‎A‎$‎), then we call it a subfield of ‎$‎A‎$‎.‎

A subfield that is not properly contained in a larger subfield of ‎$‎A ‎$‎ is called a maximal subfield of ‎$‎A‎$‎.

Example : Each of ‎$‎\mathbb{C_{1}} = \mathbb{R} ‎\oplus‎ ‎‎\mathbb{R}i ‎‎$‎, $ \mathbb{C_{2}} =‎\mathbb{R} ‎\oplus‎ ‎‎\mathbb{R}j ‎‎$, $\mathbb{C_{3}}‎=\mathbb{R} ‎\oplus‎ ‎‎\mathbb{R} k ‎$ ‎is a maximal subfield of ‎$‎\mathbb{H}‎$‎, 2-dimensional and isomorphic to ‎$‎\mathbb{C}‎$‎. With some abuse of notation, we can therefore regard $\mathbb{H}‎$‎ a complex vector space (but not as a complex algebra).‎

Are there any other subfields for $\mathbb{H}$?

How do we get to these subfields $\mathbb{C_{1}}, \mathbb{C_{2}}, \mathbb{C_{3}}$?


Let's construct other subfields of $\mathbb H$ containing $\Bbb R$.

Fix any element $a \in \mathbb H \setminus \mathbb R$, and consider the ring homomorphism $f_a : \mathbb R[X] \longrightarrow \mathbb H$ defined as

$$f_a \left( p(X)\right) = p(a)$$

The image of this homomorphism is a commutative subring of $\mathbb H$, and it's denoted by $\mathbb R[a]$: it's the smallest subring of $\mathbb H$ containing $\mathbb R \cup \{ a \}$.

Clearly, $\mathbb R [a]$ is an integral domain, since it has no zero divisors. By the first isomorphism theorem $$\mathbb R [a] \cong \frac{\mathbb R [X]}{\ker f_a}$$ thus $P=\ker f_a$ is a prime ideal of $\mathbb R [X]$. You can also see that $P$ cannot be the zero ideal, otherwise $f_a$ would be injective, and this cannot hold since $\dim \mathbb R [X] = \infty$ while $\dim \mathbb H = 4$. Since every nonzero prime ideal of $\mathbb R [X]$ is maximal, we have that $\mathbb R [a]$ is a field.

Now, for $a=i$ you get $\mathbb R [i]=\mathbb C_1$, for $a=j$ you get $\mathbb R [j]=\mathbb C_2$, for $a=k$ you get $\mathbb R [k]=\mathbb C_3$.

If you take $a= i+j$ you get a new subfield: $\mathbb R [i+j]$ which is different from the previous three. This is isomorphic to $\mathbb C$ as well.

In general, all these subfields $\{ \mathbb R [a] : a \in \mathbb H \setminus \mathbb R\}$ are exactly all maximal subfields of $\mathbb H$ (containing $\Bbb R$), and they are all isomorphic to $\mathbb C$ (that's because they are proper extensions of $\mathbb R$).


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