There is also a related result that can be useful (see Atiyah Macdonald): If $A$ is a field, then every homomorphism of $A$ into a nonzero ring  $B$ is injective. So, you can conclude that necessarily $f(1)=1$ (otherwise is the zero homomorphism), then $f(n)=n$, for $n$ integer, and the result follows, since $f(n/m)=f(n)f(m)ˆ{-1}=n/m$.

If $f: \mathbb Q\to \mathbb Q$ is a homomorphism, prove that $f(x)=0$ for all $x\in\mathbb Q$ or $f(x)=x$ for all $x$ in $\mathbb Q$.
Is it posiible to expose taxonomy term weight in node object ?

I need to sort nodes using EntityFieldQuery by term reference field, but weight is not exposed to node object so i can say for example 

-&gt;fieldOrderBy('field_custom_terms', 'weight', 'ASC');


and sort nodes by weight of terms.
