I need to show that If $L/K$ is a field extension then $\operatorname{char}(L) = \operatorname{char}(K)$. This is what I've tried:
Well, if $\operatorname{cha}(K) = p$, then $1_{K} + 1_{K} + \dots 1_{K} = 0$ (sum of $p$'s $1_{K}$). But $1_{K} = 1_{L}$ since $L/K$ is a field extension. So $\operatorname{char}(L)=p$.
Is this right? Thank you!
For every ring with identity $R$, there is a unique ring homomorphism $\chi_R\colon\mathbb{Z}\to R$, which sends $n$ to $n\mathbf{1}$ (the integral multiple of the identity; I'll use $\mathbf{1}$ to denote the identity in $R$, for better clarity).
There are two cases: either $\ker\chi_R=\{0\}$ or $\ker\chi_R=n\mathbb{Z}$, for some (unique) $n>0$. In the former case the characteristic of $R$ is $0$; in the latter case the characteristic is $n$, because $n\mathbf{1}=0$, but $m\mathbf{1}\ne0$ for every $m$ with $0<m<n$.
Thus if $R$ is a subring of $S$, the characteristic of $R$ and $S$ must be the same, because by uniqueness of $\chi_R$ we have that $\chi_R$ is the restriction to $R$ of $\chi_S$.