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

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?

• Yes, your proof is correct and well-written. – TPace Feb 25 '18 at 13:00

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 $\ker\chi_{K'}=\ker\chi_K$.