Prove that the unity element in a subfield of a field must be the unity of the whole field The solution I was given says: 
Let $F$ be a field and suppose $u^2=u$ for some nonzero $u$ in $F$.  By multiplying each side by $u^{-1}$ it is clear that $0$ and $1$ are the only solutions of $x^2=x$ in a field.  Now let $K$ be a subfield of $F$.  The unity of $K$ satisfies the equation $x^2=x$ in $K$ and hence also in $F$ and thus must be the unity $1$ of $F$.  
I am really confused why they used $u^2=u$ and what does this have to do with $x$ in a field?  And from there how can you just say that the unity of $K$ satisfies the $x^2=x$ relationship?  
An explanation here would be really helpful
 A: If $u$ is any element of $F$ satisfying $u^2=u$, then either $u=0$ or $u=1$.
If $K$ is a subring of $F$ having unity $e$, possibly different from $1\in F$, then $e^2=ee=e$ by definition of unity. So either $e=0$ or $e=1$, because $e\in F$. The case $e=0$ is disallowed if $K$ is a subfield, because in a field it is required that the unity is nonzero.
The authors of that solutions use a confusing mix of letters; what they want to say is that the equation $x^2=x$ in $F$ has only the solutions $0$ and $1$. A unity in $K$ must be a solution of that equation.
A: I would say things slightly differently. Given the equation $x^2=x$ to
be solved in a field, it is clear that the additive identity 0 is a
solution. Then, if you consider that $x\neq 0$, then for any solution
$u$ in the field (other than $0$), you have $u^2=u$. But since $u\neq
0$, $u$ has an inverse $u^{-1}$. Multiplying both sides by $u^{-1}$,
you get $u^2\times u^{-1}=u\times u^{-1}$, thus $u\times 1=1$, and
finally $u=1$, since $u\times u^{-1}=1$ for any $u\neq 0$. Thus, this equation in a field has exactly two solutions: $0$ and $1$.
All the solutions in the subfield must be solutions in the field, and in this case they both give exactly two solutions, which are the identities of the two operations. So these must be the same. Now in both the field and the subfield, the sum of the two solutions is the mutiplication identity. So the multiplication identity is the same for both.
Note: I would not normally write the first paragraph in such a lengthy
and detailed way, with both $x$ and $u$. But I tried to stay very
close to your question and mimic as much as possible its presentation.
As I remember, such a detailed presentation is used only in
introductory courses on equations resolution. It is
used in formal theorem provers, and known as skolemization: the $x$ in the equation is originally a existentially quantified variable, which is replaced by a constant $u$.
