Exercise 2.3.13 (Homomorphism preserves characteristic)

Question

This proof of Lemma 2.3.12 is quite abstract. Find
a more concrete proof, taking equation (2.2) as your definition of
characteristic. (You will still need the fact that $\phi$ is injective.)

Hassaan Naeem · Accepted Answer

\textbf{Solution:}
By (2.2) we have:

$$ char R = 
    \begin{cases} 
        \text{least}\ n>0: n*1_R = 0_R & ,\ \text{if such an n exists} \
        0 & ,\ \text{otherwise} \
    \end{cases}
$$
\
We know that $\phi(1_K) = 1_L$ and $\phi(0_K) = 0_L$, since $\phi$ is injective, then also
$\phi(n\cdot 1_K) = n \cdot 1_L \ \forall n \in \mathbb{N}$. We have two possible cases for the characteristic $c$ of $K$ ($char K$),
$c = 0$ or $c > 0$.\\
If $c=0$, then $\phi(0_K) = 0_L = 0$. Therefore $char L = c = char K$.\
If $c>0$, then $\phi(c \cdot 1_K) = c\cdot 1_L = 0$. Therefore $char L = c = char K$.

Richard Ganaye · Answer

\begin{proof} Let $\varphi : K 	o L$ a ring homomorphism between the fields $K,L$. We know that $\varphi$ is injective.

Write $n = \mathrm{char}\, K, m =\mathrm{char}\, L$.

Since $n\cdot 1_K = 0_K$,
$$0_L = \varphi(0_K) = \varphi(n\cdot 1_K) = n \varphi(1_K) = n 1_L.$$
$0_L = n 1_L$ implies $\mathrm{char}\, L = m \mid n$ (see p.28).

Moreover $\varphi(m1_K) = \varphi(0_K) = 0_L$. Since $\varphi$ is injective, $m1_K = 0_K$.
This shows that $n = \mathrm{char}\, K \mid m$.

Since $m \mid n$ and $n \mid m$, where $m \geq 0, n \geq 0$, we conclude $m = n$, that is $$\mathrm{char}\, K =\mathrm{char}\, L.$$
\end{proof}

Exercise 2.3.13 (Homomorphism preserves characteristic)

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Exercise 2.3.13 (Homomorphism preserves characteristic)

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