Linear & Abstract algebra

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Martin Sleziak

12:43 AM

in Mathematics, 1 hour ago, by MatheinBoulomenos

@MrCauchy it's automatically $\Bbb Q$-linear

@MrCauchy If I understand your question correctly, from additivity you get Q-linearity, so the answer is yes.

If you have f(x+y)=f(x)+f(y), you can easily get f(nx)=nf(x) for positive integers.

Using f(x)=0 you get also f(zx)=zf(x) for all integers.

And from this you can get f(rx)=rf(x) for any rational r.

This is explained, for example, on Wikipedia: Cauchy's functional equation.

in Mathematics, 1 hour ago, by MrCauchy

so i originally had two subfields with a field homomorphism, i have shown they are Q-subspaces. I want to show the the image of the homomorphism restricted to Q is exactly Q itself, i figured that if i could show it was a linear map then f(q) = qf(1) for any q in Q

I am not sure in which field you are working. (Two subfields of R? Two subfields of C?)

But perhaps this is related: Is an automorphism of the field of real numbers the identity map?

Q: Is an automorphism of the field of real numbers the identity map?

Is an automorphism of the field of real numbers $\mathbb{R}$ the identity map? If yes, how can we prove it? Remark An automorphism of $\mathbb{R}$ may not be continuous.