**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.8**

Describe $Z(\mathbb{H}), where \mathbb{H}$ denotes the Hamiltonian Quaternions. Prove that $\{a+bi \ |\ a,b \in \mathbb{R} \}$ is a subring of $\mathbb{H}$ which is a field but is not contained in $Z(\mathbb{H})$.

Solution: We claim that $Z(\mathbb{H}) = \{ x + 0i + 0j + 0k \ |\ x \in \mathbb{R} \}$. The ($\supseteq$) direction is clear. To see the ($\subseteq$) direction, let $\beta = x + yi + zj + wk \in Z(\mathbb{H})$.

Using $i\beta =\beta i$, we have $-zk+wj=zk-wj$. Hence $z=0$ and $w=0$. So $\beta=x+yi$. Using $j\beta=\beta j$, we have $yk=-yk$. Thus $y=0$. It follows from that $\beta \in \{ x + 0i + 0j + 0k \ |\ x \in \mathbb{R} \}$.

Now let $S = \{ a + bi \ |\ a,b \in \mathbb{R} \}$. Let $\alpha = a + bi$ and $\beta = c+di$. Since $\alpha – \beta = (a-c) + (b-d)i \in S$ and $0 + 0i \in S$, by the subgroup criterion $S \leq \mathbb{H}$. Since $$\alpha\beta = (ac-bd) + (ad + bc)i \in S,$$ $S$ is a subring of $\mathbb{H}$.

Now if $a + bi \neq 0 + 0i$, we see that $$(a+bi)(a-bi)/(a^2 + b^2) = 1;$$ thus $\alpha$ has an inverse in $S$, specifically $$\alpha^{-1} = (a-bi)/(a^2 + b^2).$$ Thus $S$ is a division ring. Moreover, $$\alpha \beta = (ac – bd) + (ad + bc)i = (ca – db) + (da + bc)i = \beta \alpha,$$ so that $S$ is a field.

Note, however, that $ij = k$ while $ji = -k$, so that $S \not\subseteq Z(\mathbb{H})$.