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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?

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    $\begingroup$ Don't forget about the zero morphism. It's actually the only case where $1_F$ isn't map to $1_G$. $\endgroup$ – Antoine Giard May 14 '18 at 0:41
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    $\begingroup$ What @AntoineGiard said. All you've shown is that $f(1_F)$ is an idempotent. This doesn't mean that it's the identity. Luckily, in a field there are only two idempotents, so.... $\endgroup$ – Randall May 14 '18 at 0:42
  • $\begingroup$ @Randall Does the identity come from the $f(1_{F})^{-1} \times f(1_{F})$ ? $\endgroup$ – user150203 May 14 '18 at 0:44
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    $\begingroup$ But now you've assumed that $f(1_F)$ has an inverse (isn't zero). $\endgroup$ – Randall May 14 '18 at 0:46
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    $\begingroup$ Roughly, yes, because cancellation is ALWAYS valid under addition. Note that, in a field, cancellation across multiplication is always valid EXCEPT for $0$. This explains your issue. The "standard" proof for $f(0)=0$ is that $f(0)= f(0+0)=f(0)+f(0)$ and now we just cancel an $f(0)$ from each side (additively!). $\endgroup$ – Randall May 14 '18 at 1:04
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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.

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  • $\begingroup$ Thank you! Yes I now understand - had forgotten the zero element not having a multiplicative inverse! $\endgroup$ – user150203 May 14 '18 at 1:04

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