The characteristics of a subfield of a field is same as that of the field. How to show that the characteristics of a subfield of a field is same as that of the field??
 A: Adding $1+1+\dots+1$ should work exactly the same way in the field that it does in the subfield.
A: Let $K$ be a field and $char(K)=p$. Let $\psi :F \longrightarrow K$ be a subfield of $K$. Suppose that $char(F)\neq char(K)$. As Reveillark pointed out, $char(F)$ could be less than or equal to $char(K)$. If $char(F)< char(K)$, call it $m$; then in $F$ you have $m1=0$; as the embedding $\psi$ is a ring homomorphism, then $\psi(m1)=m \psi(1)=m1=0$. But this contradict the minimality of $p$. 
A: The characteristic of a field $k$ is the positive generator of the kernel of the characteristic map $\chi_k:\mathbb{Z}\to k,\,1 \mapsto 1$. For $K|k$ you can factor  $\,\,\chi_K: \mathbb{Z} \overset{\chi_k}{\to} k \hookrightarrow K.$ Hence $\chi_k,\chi_K$ have the same kernel and therefore the same characteristic. 
A: If we suppose that $ char(F)=d $ and $char(K)=s $
Then   obviously $s<=d$
Now  if $ s<d $ then consider $ k= d-s$ ,a positive integer.
If we choose  an element  $b$ from the subfield $K$
Then it wil also belongs  to the field $F$
And $ k*b=0  $
Which imply that $ char( F)=k< d $
which is a contradiction  to our hypothesis that $ char(F)= d$
So we should have $s=d$
