Quick Question concerning Field Homomorphism: If I have a Field Homomorphism $f$ between two Fields $(F, +, \cdot)$ and $(G, \oplus, \times)$. Then Addition and Multiplication must be preserved under $f$, i.e.
$\forall a,b \in F$
$f\left(a + b\right) = f(a) \oplus f(b)$
$f\left(a \cdot b\right) = f(a) \times f(b)$
My Question - In most cases the definition includes the axiom that the $f$ must map the multiplicative identity of F ($1_{F}$) to the multiplicative identity of G ($1_{G}$), i.e.
$f(1_{F}) = 1_{G}$
This condition does not seem necessary as if I've done things correctly it falls out from the preservation of Addition/Multiplication, i.e.
$f(1_{F}) = f(1_{F} \cdot 1_{F}) = f(1_{F}) \times f(1_{F})$
And so,
$f(1_{F})^{-1}\times f(1_{F}) = f(1_{F})^{-1}\times f(1_{F}) \times f(1_{F})$
$1_{G} = 1_{G} \times f(1_{F})$
Hence,
$f(1_{F}) = 1_{G}$
The exact same reasoning shows also that
$f(0_{F}) = 0_{G}$
So mapping between additive and multiplicative identities is not required within the definition as they are derived directly from them.
Is that correct? Have I missed something here?
In any field $E$ (integral domain, actually), there are only two elements $x \in E$ with $x^2=x$. They are clearly $x=0,1$ and these are the only ones by zero-divisor business.
In your case, you have $x^2=x$ in $G$ with $x=f(1_F)$, so $f(1_F)$ could be either $0$ or $1_G$. But, there is nothing preventing $f(1_F)=0$. In fact, if this is the case, then $f(x) =0$ identically as $f(x) = f(1_Fx)= f(1_F)f(x)=0f(x)=0$. And, this meets the more general definition of ring homomorphism. There is no problem with this.
On the other hand, if $f(1_F) \neq 0$ then $f(1_F) = 1_G$ by the result in my first paragraph. This explains Antoine Giard's first comment.