On some homomorphsims

Find all homomorphisms $\varphi$ in the following cases:

1. $\varphi:\mathbb{Q} \rightarrow \mathbb{Q}$
2. $\varphi:\mathbb{Q}[\sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{3}]$
3. $\varphi:\mathbb{Z}[i]\rightarrow \mathbb{C}$

Solution 

 

1. Let us construct one. First of all we depict the unit of $\mathbb{Q}$ to the unit, that is $\varphi(1)=1$. It is also clear that $-1$ will be depicted to $-1$ also. Now suppose that $r \in \mathbb{Q}$. $r$ however can be written as $r=\frac{p}{q}, \;\; p \in \mathbb{Z}, \; q \in \mathbb{N}$. Let us see where $p$ is going to be depicted.
   
   $$\varphi(p)=\varphi\left ( \underbrace{1+1+\cdots_1}_{p \; \text{times}} \right )\overset{\text{homomorphism}}{=\! =\! =\! =\! =\! =\! =\! =\! =\!} \underbrace{\varphi(1)+\varphi(1)+\cdots+\varphi(1)}_{p \; \text{times}}= p \varphi(1)=p$$
   
   And let us see where we are depicting $\frac{1}{q}$. Well,
   
   $$1=\varphi(1)=\varphi\left ( q\cdot \frac{1}{q} \right )=\varphi\left ( q \right )\varphi\left ( \frac{1}{q} \right )=q \varphi\left ( \frac{1}{q} \right ) \Rightarrow \varphi \left ( \frac{1}{q} \right  )= \frac{1}{q}$$
   
   Thus all homomorphims $\varphi: \mathbb{Q} \rightarrow \mathbb{Q}$ are $\varphi(r)=r$. 


2. Before we deal with this case let us consider a very similar one, namely $\varphi:\mathbb{Q}[\sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{2}]$. It is known that
   
   $$\mathbb{Q}\left [ \sqrt{2} \right ]=\left \{ a+b \sqrt{2}, \; a , b \in \mathbb{Q} \right \}$$
   
   Again we are depicting the unit to the unit, that is $\varphi(1)=1$. Now , let us decide where $\sqrt{2}$ is going to be depicted. Well, let us make sure that $\varphi(2)=2$. That is immediate since $2=1+1=\varphi(1)+\varphi(1)=\varphi(1+1)=\varphi(2)$. Now, let's see where $\sqrt{2}$ is going to be depicted. We note that:
   
   $$2=\varphi(2)=\varphi\left ( \sqrt{2} \cdot \sqrt{2} \right )=\varphi^2 \left ( \sqrt{2} \right )\Rightarrow \varphi \left ( \sqrt{2} \right )= \pm \sqrt{2}$$
   
   Hence:
   
   $$2=\left ( a+b\sqrt{2} \right )^2 = a^2+2b^2 +2ab \sqrt{2}$$
   
   This in return means that $\left\{\begin{matrix}
   a^2+2b^2=2 & \\
   ab=0 &
   \end{matrix}\right.$. If $a=0$, then $b=\pm 1$ while if $b=0$ then $a=\pm \sqrt{2}$. In the first case we have two homomorphisms while in the second we have none.
   
   Hence all homomorphisms $\varphi: \mathbb{Q}[\sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{2}]$ are $\varphi \left ( r+s\sqrt{2} \right )=r \pm s \sqrt{2}$.
   
   __________________________________________
   Quite similarly for the homomorphisms $\varphi:\mathbb{Q}[\sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{3}]$ in the exact same way we deduce that there exists none. As strangly as it sounds these two sets have nothing in common, although someone would expect that somehow that $\mathbb{Q}[\sqrt{2}]$ could be embedded in $\mathbb{Q}[\sqrt{3}]$. We realize now, that numbers play a critical role in the existence of homomorphisms.


3. It is quite known that
   
   $$\mathbb{Z}[i]=\left \{ a+ib , \; a , b \in \mathbb{Z} \right \}$$
   
   As before we will decide where to depict $i$. But since $\varphi(1)=1$ and $\varphi(-1)=-1$ then:
   
   $$-1=\varphi(-1)=\varphi(i^2)= \varphi(i)\varphi(i)=\varphi^2 (i)$$
   
   Hence $\varphi(i)=\pm i$. Thus all homomorphisms are of the form $\varphi(a+ib)=a \pm bi $.