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.
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