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

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?

• Welcome to Math Stack Exchange. Did you mean R when you wrote R' ? – J. W. Tanner Feb 12 '19 at 3:17
• It's meant to be a quotation mark – user643891 Feb 12 '19 at 3:17
• Oh. T contains $\mathbb Q$, so either T = $\mathbb Q$ or T contains an element not in $\mathbb Q$, in which case you want to show T contains every element of R – J. W. Tanner Feb 12 '19 at 3:19

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 $$T = R$$.

$$OE\Delta$$.