I'm trying to prove that a local ring $(R,\mathfrak{m})$ contains a field if and only if $\mathrm{char}(R)$ and $\mathrm{char} (R/\mathfrak{m})$ are equal. To this we must relate the characteristic of $R$ to the characteristic of it's residue field $K=R/\mathfrak{m}$. If $\mathrm{char}(K)=0$, then, $\mathrm{char}(R)=0$. So $R$ contains a copy of $\mathbb{Z}$. Now, since $R/\mathfrak{m}$ has characteristic $0$, none of the images of non-zero integers in $R$ can be in $\mathfrak{m}$, else their images in $R/\mathfrak{m}$ will be zero, contradicting $\mathrm{char}(K)=0$. Then by the universal property of localization, $R$ contains $\mathbb{Q}$.
Now for the converse, I know that if $\mathrm{char}(K)=p$, then $\mathrm{char}(R)$ is $0$ a power of $p$, but I don't know how to completely prove this part. Why can't the characteristic be some non-prime integer? And furthermore, why can't $R$ contain a field in any case besides when $\mathrm{char}(R)= p$?