Prove a subring of $R=\mathbb{Q}[i]$ is equal to $R$ itself or $\mathbb{Q}$

Question

Consider the ring $R = \mathbb{Q}[i] = \{a + bi \mid a, b ∈ \mathbb{Q}\}$, the subring of $\mathbb{C}$ of all complex numbers with rational real and imaginary parts.

Let $T \subset R$ be a subring of R which contains $\mathbb{Q}$. Show that $T = \mathbb{Q}$ or $T = R$.

I am quite new to ring theory and not sure how to go about this. I have proved in an earlier question that R is a field, but I don't know if this helps. Can anyone provide me with a clue to point me in the right direction?

Answer 1

If

$T \ne \Bbb Q, \tag 1$

then for some $a, b \in \Bbb Q$, with $b \ne 0$,

$a + bi \in T; \tag 2$

but since

$\Bbb Q \subset T, \tag 3$

we have that

$i = b^{-1}((a + bi) - a) \in T, \tag 4$

and thus for all $c, d \in \Bbb Q$,

$c + di \in T \tag 5$

so $T = R$.

$OE\Delta$.