Proving injective ring homomorphism preserves characteristic

Let $$f: A \to B$$ be an injective ring (with multiplicative identity) homomorphism. I want to prove that R and S have the same characteristic. I know that $$f(1_A) = (1_B)$$, but I am unsure how to complete the proof.

• What's your preferred definition of the characteristic? Sep 29, 2020 at 23:45
• In a ring $R$, the smallest positive integer $n$ such that $1_R + ... + 1_R$ ($n$ times) = $0_R$. If no such $n$ exists, then characteristic is zero. Sep 29, 2020 at 23:49
• If one of the answers below answered your question, the way this site works works, you'd "accept" the answer, more here: What should I do when someone answers my question?. But only if your question really has been answered. If not, consider adding more details to the question. Oct 28, 2020 at 17:30

The characteristic is the natural number such that $$n\Bbb Z$$ is the kernel of the unique ring homomorphism from $$\Bbb Z$$ to $$R$$.

Let $$\varphi_A:\Bbb Z\to A$$ and $$\varphi_B \colon \Bbb Z \to B$$ be the unique homomorphisms. Then $$f\circ\varphi_A=\varphi_B$$, by uniqueness. Since $$f$$ is injective, the kernels are the same.

If $$k1_A=\underbrace{1_A+...+1_A}_{k\mathrm{ times}}=0$$ then $$k1_B=0$$ because $$f$$ is a homomorphism. Conversely if $$k1_B=0$$ but $$k1_A\ne 0$$ then the homomorphism $$f$$ is not injective since it maps $$k1_A$$ to $$0$$. So characteristics of $$A$$ and $$B$$ are the same.

For any ring $$R$$ there is a unique map $$i_R\colon \Bbb Z\to R$$ sending $$1$$ to $$1$$. The kernel of this map is an ideal in the PID $$\Bbb Z$$, and we define $$\operatorname{char}(R)$$ to be a non-negative generator of $$\ker i_R$$.

So if $$f\colon A\to B$$ takes $$1$$ to $$1$$, then the map $$\Bbb Z\xrightarrow{i_A} A\xrightarrow{f} B$$ also takes $$1$$ to $$1$$. Since this map is unique, we must have $$f\circ i_A=i_B$$. Now we are done if we can show $$\ker i_B=\ker i_A$$. For this, consider injectivity of $$f$$.

Assume your ring $$A$$ has charasteristic $$n>0$$. Then $$0_B=f(0_A)=f(n1_A)=f(\underbrace{1_A+\ldots+1_A}_\text{n times})=\underbrace{f(1_A)+\ldots+f(1_A)}_\text{n times}=\underbrace{1_B+\ldots+1_B}_\text{n times}=n1_B$$. If there was some $$m>0$$ smaller than $$n$$ such that $$m1_B=0_B$$, then we would have $$f(m1_A)=f(\underbrace{1_A+\ldots+1_A}_\text{m times})=\underbrace{f(1_A)+\ldots+f(1_A)}_\text{m times}=\underbrace{1_B+\ldots+1_B}_\text{m times}=m1_B=0_B$$, so, since $$f$$ is injective and $$f(0_A)=0_B$$, we conclude $$m1_A=0_A$$, which is a contradiction ($$n$$ was the smallest positive integer such that $$n1_A=0_A$$). Hence $$B$$ has charasteristic $$n$$ too.

Now suppose $$A$$ has charasteristic $$0$$ then $$B$$ must have it too, because otherwise there would be some $$k>0$$ such that $$k1_B=0_B$$, so $$f(k1_A)=f(\underbrace{1_A+\ldots+1_A}_\text{k times})=\underbrace{f(1_A)+\ldots+f(1_A)}_\text{k times}=\underbrace{1_B+\ldots+1_B}_\text{k times}=k1_B=0_B$$, but $$k1_A\neq 0_A$$ by hytothesis, so $$f$$ can't be injective since $$f(k1_A)=0_B=f(0_A)$$.