Answer to Question #104882 in Abstract Algebra for Sourav Mondal

Question #104882
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?
1
Expert's answer
2020-03-09T13:00:38-0400

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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