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If there is a unital ring homomorphism $f:K\to K'$ between two division rings such then prove that $\text{char}(K)=\text{char}(K').$ $\newcommand{\char}{\text{char}}$

My Attempt: Let $\char(K)=m$ and $\char(K')=n.$ First observe that $f(m\cdot 1_{K})=m\cdot 1_{K'}.$ Then $0=f(0)=m\cdot 1_{K'}$ and so $n\leq m.$ Now I claim that $f$ is injective. Let $x\in \ker(f)$ such that $x\not = 0.$ Then since $K$ is a division ring there exists $x'$ such that $xx'=x'x=1_K.$ Thus $f(xx')=f(x)f(x')=0f(x')=0$ implying that $f(1)=1_{K'}=0_{K'}.$ Assuming that $K'$ is not the zero ring we get a contradiction and so $f$ is injective. Thus $$f(n\cdot 1_{K})=n\cdot 1_{K'}$$ $$\implies f(n\cdot 1_{K})=0$$ $$\implies n\cdot 1_{K}=0$$ and thus $m\leq n.$ Thus we have that $m=n.$ Is this a valid proof?

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    $\begingroup$ Yes, your proof is correct and well-written. $\endgroup$ – TPace Feb 25 '18 at 13:00
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If you're not sure whether it is valid or not, where do you think some of your deduction could fail? Are you unsure, whether some steps are ok or are you just asking to make sure other people don't see a problem with it either?

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The proof is correct, but it can made simpler and more instructive.

If $R$ is a unital ring, there exists a unique ring homomorphism $\chi_R\colon\mathbb{Z}\to R$ (verify it). The kernel of $\chi_R$ has the form $n\mathbb{Z}$, for a unique $n\ge0$. The unique $n\ge0$ such that $\ker\chi_R=n\mathbb{Z}$ is the characteristic of $R$ (prove it).

If $f\colon K\to K'$ is a ring homomorphism of division rings, then $f$ is injective; since $\chi_{K'}=f\circ\chi_K$ (by uniqueness), we have $\ker\chi_{K'}=\ker\chi_K$.

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