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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.

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  • $\begingroup$ What's your preferred definition of the characteristic? $\endgroup$
    – Thorgott
    Sep 29, 2020 at 23:45
  • $\begingroup$ 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. $\endgroup$
    – integering
    Sep 29, 2020 at 23:49
  • $\begingroup$ 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. $\endgroup$
    – Darsen
    Oct 28, 2020 at 17:30

4 Answers 4

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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.

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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.

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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 $f$.

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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 $f(k1_A)=0_B=f(0_A)$.

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