$\mathbb{Q}[i]$ is the smallest subfield of $\mathbb{C}$ that contains the ring $\mathbb Z[i]$ We have that $\mathbb{Z}[i]=\{a+bi,a,b\in \mathbb{Z}\}\subseteq \mathbb{C}$ and $\mathbb{Q}[i]=\{a+bi,a,b\in \mathbb{Q}\}\subseteq \mathbb{C}$, where $\mathbb{Z}$ is a subring of $\mathbb{C}$ and $\mathbb{Q}$ is a subfield of $\mathbb{C}$. I am asked to show that $\mathbb{Q}[i]$ is the smallest subfield of $\mathbb{C}$ that contains the ring $\mathbb Z[i]$.
How could I do that? Do I have to suppose that there is an other field that is smaller than $\mathbb{Q}$?? But how could I get a contradiction??
 A: Note that any subfield $\Bbb F \subset \Bbb C$ with $Z[i] \subset \Bbb F$ contains $\Bbb Z$, hence contains $\Bbb Q$.  $\Bbb F$ also contains $i$, so it in fact contains every $a + bi \in \Bbb Q[i]$.  This shows that $Q[i] \subset \Bbb F$ for any such  field $\Bbb F$, including $Q[i]$ itself.  Thus $Q[i]$ is the smallest such subfield of $\Bbb C$ with respect the the partial ordering "$\subset$".
Addendum in response to comment of Mary Star, Wednesday 8 October 2014 6:39 PM PST:  We have $\Bbb Z \subset \Bbb F$; but since $\Bbb F \subset \Bbb C$ is a field,
it must also contain $z^{-1}$ for every $0 \ne z \in \Bbb Z$.  Then for every $y \in \Bbb Z$, we have $yz^{-1} \in \Bbb F$ as well.  But the set of all complex numbers of the form $yz^{-1}$, $0 \ne z \in \Bbb Z$, $y \in \Bbb Z$ is clearly the rationals $\Bbb Q$.  Thus $\Bbb Q \subset \Bbb F$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: The fraction field of an integral domain is the smallest field in which it can embedded. Now the fraction field of $\mathbb{Z}[i]$ clearly is $\mathbb{Q}[i]$, see the above link. Hence $\mathbb{Q}[i]$ is the smallest field containing $\mathbb{Z}[i]$.
A: The standard way to show "$X$ is the smallest $Y$ which has property $Z$" is to show that


*

*$X$ is actually a $Y$ which has property $Z$; and

*any $Y$ which has property $Z$ must contain $X$.


Here, you already have the first part: $\mathbf{Q}[i]$ is a field which contains $\mathbf{Z}[i]$. You now need to show the second part: any field which contains $\mathbf{Z}[i]$ must contain $\mathbf{Q}[i]$.
A: Hint: a field contains $\mathbf{Z}[i]$ if and only if it contains $\mathbf{Z}$ and also contains $i$.
Work one piece at a time rather than trying to use everything at once.
