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 $\mathbb Q$.

Question

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 $\mathbb Q$.

I'm wondering if you can help me with this one?

Answer 1

Any multiplicative homomorphism must send $1$ to an idempotent.

The only idempotents of $\mathbb{Q}$ are $0$ and $1$ (the roots of $x^2-x=x(x-1)$), so $\varphi(1)=0$ or $\varphi(1)=1$.

If $\varphi(1)=0$, then $\varphi(a) = \varphi(1a) = \varphi(1)\varphi(a)=0$, so $\varphi(x)=0$ for all $x$.

If $\varphi(1)=1$ and the map is additive, then prove inductively that $\varphi(n)=n$ for all positive integers, hence for all integers; deduce that $\varphi(q) = q$ for all $q$.

Answer 2

I understand that we're looking at rationals as a ring (as a group it is obviously false).

Pick an arbitrary homomorphism $\varphi:\mathbf Q\to \mathbf Q$. Put $e:=\varphi(1)$.

If $e=0$, then for any $x\in \mathbf Q$ we have $\varphi(x)=\varphi(1)\varphi(x)=0\cdot\varphi(x)=0$, so $\varphi$ is zero.

If $e\neq 0$, then $e\cdot e=\varphi(1)\varphi(1)=\varphi(1\cdot 1)=e$, so $e=1$ (since $e\neq 0$ and $0,1$ are the only solutions of the equation $x^2-x=0$). Furthermore, for any $\frac{p}{q}$ we have $1+\ldots+1=p\cdot 1= p=\frac{p}{q}+\ldots \frac{p}{q}=q\cdot \frac{p}{q}$, so $e+\ldots +e=p\cdot e= q\cdot \varphi(\frac{p}{q})$, so $\varphi(\frac{p}{q})=e\cdot \frac{p}{q}=\frac{p}{q}$ and we're done (for negative $p$ we can use $\frac{p}{q}=(-1)\cdot \frac{-p}{q}$ so $\varphi(\frac{p}{q})=-e\cdot\varphi(\frac{-p}{q})=\frac{p}{q}$).

Answer 3

If $f$ is a homomorphism of rings, we know the kernel of $f$ has to be an ideal of $\mathbb{Q}$, but as $\mathbb{Q}$ is a field, the only ideals of $\mathbb{Q}$ are $0$ and $\mathbb{Q}$ itself.

Answer 4

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 $f(n/m)=f(n)f(m)ˆ{-1}=n/m$.