Question

Show that if :f Q->Q is a ring homomorphism, then f(x) = x ∀x ∈ Q. Would
this still be true if f were a group homomorphism? Why, or why not?

Answer 1

If $f\colon \mathbb Q\to\mathbb Q$ is a ring homomorphism, then $f(x)=f(x\cdot 1)=f(x)f(1)$ for any $x\in\mathbb Q$, so $f(1)=1$ or $f(1)=0$.

If $f(1)=0$, then $f(x)=f(1\cdot x)=f(1)f(x)=0$ for all $x\in\mathbb Q$.

So if $f$ is a nonzero homomorphism, then $f(1)=1$.

Prove by induction that $f(x)=x$ for any $x\in\mathbb N$

Indeed,

1)It is true for $n=1$: $f(1)=1$

2)Assume that $f(k)=k$ for some $k\ge 1$ and $k\in\mathbb N$

3)Then $f(k+1)=f(k)+f(1)$.

$f(k)=k$ by the inductive hypothesis, and $f(1)=1$, so $f(k+1)=k+1$

By the principle of mathematical induction we obtain that $f(x)=x$ for all $x\in\mathbb N$.

Moreover $f(x)=f(x+0)=f(x)+f(0)$ for all $x\in\mathbb Q$ , so $f(0)=0$ as additive identity. And $0=f(0)=f(x+(-x))=f(x)+f(-x)$, so $f(-x)=-f(x)$.

Then $f(-x)=-f(x)=-x$ for any $x\in\mathbb N$.

So we obtain that $f(x)=x$ for any $x\in\mathbb Z$.

Let $x\in\mathbb Q$, then $x=\frac{m}{n}$ for some $m,n\in\mathbb Z$. Then $m=f(m)=f(x\cdot n)=f(x)f(n)=f(x)n$, and we obtain that $f(x)=\frac{m}{n}=x$.

Therefore $f(x)=x$ for any $x\in\mathbb Q$.

If $f$ is a group homomorphism, then $f$ have not to be the identity homomorphism. For example, $f\colon\mathbb Q\to\mathbb Q$, were $f(x)=2x$, is a group homomorphism.