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 $\mathbb Q$.

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?

• Homomorphism of groups, rings, fields? – user29743 Jul 26 '12 at 15:02
• This might help you get started: Find every ring homomorphism between $\mathbb Z$ and $\mathbb Q$. (If the question is about ring homomorphisms.) – Martin Sleziak Jul 26 '12 at 15:09
• BTW, ring homomorphisms are often asked to be unital ($f(1) = 1$). If it is the convention you adopt, $0$ is not a morphism $\mathbb Q \to \mathbb Q$. And if it isn't, the property wou want to prove is false (see for example $x \mapsto 2x$.) – PseudoNeo Jul 26 '12 at 15:12
• @PseudoNeo: Wrong. $x\mapsto 2x$ does $1\cdot 1=1\mapsto 2\neq 2\cdot 2$. – tomasz Jul 26 '12 at 15:20
• @tomasz : You're right, I even read a question about non-unital morphisms $n\mathbb Z \to m\mathbb Z$ only few hours ago... – PseudoNeo Jul 26 '12 at 17:23

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 $q$.

• Thank you . It was of much help as well :) – Mirna Jul 27 '12 at 18:34

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 $\varphi(\frac{p}{q})=-e\cdot\varphi(\frac{-p}{q})=\frac{p}{q}$).

• As groups, it is also true. – user641 Jul 26 '12 at 17:07
• @tomasz I'm new at this "forum" and I do appreciate your answer, but I'm kind of confused, as I thought it would be easier to solve. Can you please tell if your answer belongs to the f(x)= 0 or f(x)=x? Thanks, – Mirna Jul 26 '12 at 17:36
• @Mirna: this deals with both cases at once. If $f(1) = 0$ then $f$ is identically 0 (sentence starting 'If $e = 0$, ...') - the rest shows that if $f(1) \neq 0$ then $f$ is the identity. – Kris Jul 26 '12 at 17:53
• @SteveD: How so? x→ax is a group homomorphism of Q into Q for any a∈Q. For a≠0 it is even an automorphism. Or, if you meant the multiplicative semigroup, then I believe the homomorphisms are induced by injections of the set of prime numbers within itself (as a set), and absolute value (so there's a lot of nontrivial ones around, many of them automorphisms, even many more than in the additive group case). – tomasz Jul 26 '12 at 23:29
• @tomasz: Yes, sorry, I tried to delete my comment after posting it, I simply meant that all (non-trivial) homomorphisms are isomorphisms. – user641 Jul 26 '12 at 23:33

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.

• It might be my ignorance in ring theory, but it seems to me you still need to show that there's no nontrivial embedding of $\mathbf Q$ into itself for this to be complete. For example, if instead of $\mathbf Q$ you used a field with nontrivial automorphisms, your argument would be clearly insufficient. – tomasz Jul 26 '12 at 15:24
• Noting that the prime field of $\mathbb{Q}$ is itself is sufficient, so if $\varphi$ is $1-1$ it is the $Id$ otherwise $\varphi\equiv 0$ because the kernel is an ideal. – Belgi Jul 26 '12 at 15:38

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 $f(n/m)=f(n)f(m)ˆ{-1}=n/m$.

• This is not an answer, but a comment. (But I think you can't comment yet, so I won't downvote it for now.) – tomasz Aug 7 '12 at 13:35