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Let $f : \mathbb{Q} → \mathbb{Q}$ be a ring homomorphism. Show that f is the identity.

Im having trouble with this problem. I started considering two elements in $\mathbb{Q}$ and the definition of ring homomorphism:

Let $x, y \in \mathbb{Q}$ such that $x = \frac{m}{n}, y=\frac{m'}{n'}$with $m,m',n,n'\in \mathbb{Z}$ then:

$f(\frac{m}{n}*\frac{m'}{n'}) = f(\frac{m}{n})*f(\frac{m'}{n'})$

and

$f(\frac{m}{n}+\frac{m'}{n'}) = f(\frac{mn'+m'n}{nn'})=f(\frac{m}{n})+f(\frac{m'}{n'})$

Clearly, my goal is to show that:

$f(\frac{m}{n}*\frac{m'}{n'}) =\frac{m}{n}*\frac{m'}{n'}$

or

$f(\frac{m}{n}+\frac{m'}{n'}) =\frac{m}{n}+\frac{m'}{n'}$

I've been thinking for like an hour and still achieve nothing, any ideas/hints?

Robert Lewis
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Juan Diego Chaves Sanguino
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3 Answers3

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Another way to see the answers above (which are all fine) is the following : Let $f$ be the ring morphism in question. For your question to be true, it has to be that you only allow $f(1)=1$ in your definition of ring morphism. Therefore that's what I will assume. So $f$ is also a field morphism.

Consider $R=\{x\in \Bbb{Q}\mid f(x) = x\}$. Then quite obviously, $R$ is not empty ($1,0\in R$) and it is closed under addition and multiplication and taking inverses for non-zero elements.

Therefore $R$ is a subfield of $\Bbb{Q}$.

But $\Bbb{Q}$ is a prime field and thus its only subfield is itself. Therefore $R=\Bbb{Q}$, and $f$ is the identity.

There are two advantages of this answer over the others : 1- it can be applied to any prime field, that is, the $\Bbb{F}_p$ for prime $p$, and $\Bbb{Q}$; 2- if you have already proved that $\Bbb{Q}$ is a prime field, then it can allow you not to redo the calculations that are done in other answers (but at some point you will have to do them)

Maxime Ramzi
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Consider $f(1)$. We have

$(f(1))^2 = f(1)f(1) = f(1 \cdot 1) = f(1), \tag{1}$

or

$f(1)(f(1) - 1) = 0, \tag{2}$

whence, if $f(1) \ne 0$, yields

$f(1) = 1. \tag{3}$

If $f(1) = 0, \tag{4}$

then for any $r \in \Bbb Q$,

$f(r) = f(r \cdot 1) = f(r)f(1) = f(r) \cdot 0 = 0, \tag{5}$

that is, $f:\Bbb Q \to \Bbb Q$ is the trivial homomorphism, which we set aside. Thus we look at the case (3). For any $n \in \Bbb Z_+ = \{z \in \Bbb Z: z > 0 \}$, we have

$f(n) = n, \tag{6}$

as may be seen by a simple induction: if

$f(k) = k, \tag{7}$

then

$f(k + 1) = f(k) + f(1) = f(k) + 1 = k + 1; \tag{8}$

thus $f$ fixes all of $\Bbb Z_+$. Also, for any $q \in \Bbb Q$,

$f(0) + f(q) = f(0 + q) = f(q), \tag{9}$

so

$f(0) = 0; \tag{10}$

if $m \in \Bbb Z_- = \{z \in \Bbb Z: z < 0 \}$, then

$f(m) + f(-m) = f(m + (-m)) = f(0) = 0, \tag{11}$

so

$f(m) = -f(-m) = -(-m) = m, \tag{12}$

where we have used the fact that $-m \in \Bbb Z_+$, hence is fixed by $f$. (12) shows that $f$ fixes $\Bbb Z_-$ as well; by (6), (10), and (12) we see that $f$ fixes all of $\Bbb Z$. Now if

$\dfrac{p}{q} \in \Bbb Q, \tag{13}$

where $p, q \in \Bbb Z$, $q \ne 0$, then

$q \cdot \dfrac{p}{q} = p, \tag{14}$

so

$q \cdot f(\dfrac{p}{q}) = f(q) \cdot f(\dfrac{p}{q}) = f(q \cdot \dfrac{p}{q}) = f(p) = p, \tag{14}$

whence

$f(\dfrac{p}{q}) = \dfrac{p}{q}; \tag{15}$

that is, $f$ fixes $\Bbb Q$, element-wise.

Robert Lewis
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A nonzero group homomorphism $f\colon\mathbb{Q}\to\mathbb{Q}$ is surjective.

Indeed, suppose $f(x)=a/b\ne0$. Then $f(bx)=a$, so consider $y=bx/a$; we have $$ af(bx/a)=f(abx/a)=f(bx)=a $$ and therefore $f(bx/a)=1$, because $\mathbb{Q}$ is torsionfree. Set $bx/a=y$ and choose $m/n\in\mathbb{Q}$ (so $n\ne0$). Then $$ nf(my/n)=f(my)=mf(y)=m=n\frac{m}{n} $$ Again this implies $f(my/n)=m/n$.

In your particular case, you can choose $y=1$. Thus $f(m/n)=m/n$.

egreg
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