# 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?

-

13
Homomorphism of groups, rings, fields? – countinghaus Jul 26 '12 at 15:02
– 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

2
@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

## 4 Answers

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. – Steve D 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

1
@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 *semi*group, 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. – Steve D 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.

-

3
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

2
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
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 79, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9343393445014954, "perplexity_flag": "head"}