What is Homomorphic image of a field $F$?

Question

What is Homomorphic image of a field? How to define it? Can anyone please make me understand?

I was trying to prove the theorem $F$ can have only two homomorphic image. Then I got this doubt -- When ker $\phi $ is zero how the homomorphic image of $F$ is $F$?

Answer 1

Hint: what are the ideals of the field $F$? If $\phi: F \rightarrow R$ is a homomorphism and $I \subseteq R$ is an ideal, what can you say about $\phi^{-1}(I)$?

Answer 2

Well,

A homomorphic image of a field $F$ is $\phi(F)$, where

$\phi:F \to R \tag 1$

is a ring homomorphism 'twixt $F$ and some ring $R$; the definition really depends on the existence of such $R$, which is required for the existence of $\phi$. So to define $\phi(F)$, the existence of $R$ must be stipulated, as then the existence of such $\phi:F \to R$.

Since any such $\phi$ is a ring homomorphism, we have

$F/\ker \phi \simeq \phi(F), \tag 2$

so we inquire into the nature of $\ker \phi$; it is an ideal in $F$; but the only non-trivial ideal $I \subset F$ must satisfy

$1_F \subset I, \tag 3$

for if

$0 \ne s \in I, \tag 4$

then

$\exists s^{-1} \in F, \tag 5$

whence

$1_F = s^{-1}s \in I, \tag 6$

hence for every $f \in F$,

$f = f1_F \in I \Longrightarrow I = F; \tag 7$

since the only non-trivial ideal in $F$ is $F$ itself, if $\phi$ is non-trivial we must have

$\ker \phi =\{0\}; \tag 8$

otherwise

$\ker \phi = F \Longrightarrow \phi(f) = 0,\; \forall f \in F, \tag 9$

and $\phi$ is trivial. With (8), $\phi$ is injective, which in turn implies that

$\phi:F \simeq \phi(F) \subset R, \tag{10}$

and $R$ must contain a "copy" of $F$ as a subring.

Answer 3

Consider the homomorphism $\phi: F\rightarrow R$, where $F$ is a field and $R$ is a ring. The unit element $1_F$ of $F$ is mapped either to the zero element $0_R$ of $R$ or to a nonzero element $r\in R$. Note that $\ker\phi$ is an ideal of $F$. Note that $\phi(0_F)=0_R$ holds in each case.

In the first case, for any element $a\in F$, $\phi(a)=\phi(a\cdot 1_F) =\phi(a)\cdot\phi(1_F) = \phi(a)\cdot 0_R=0_R$, since the zero element of a ring is always absorbing; i.e., for each $r\in R$, $r\cdot 0_R = r\cdot (r+(-r)) = r^2-r^2=0_R$, similar $0_R\cdot r=0_R$. Thus $\ker\phi=F$ and hence the homomorphic image is $\{0_R\}$.

In the second case, let $\phi(1_F)=r\ne 0_R$. Then $\ker\phi$ is properly contained in $F$. But the only ideals of $F$ are the zero ideal $\{0_F\}$ and $F$ itself. Thus $\ker\phi = \{0_R\}$ and so the homomorphic image is $\phi(F)$ which is isomorphic to $F$.

Answer 4

The homomorphic image of $F$ under $\phi$ doesn't have to be $F$, even if the kernel of $\phi$ is trivial.

For example, let $I$ be the ideal in the integers generated by $5$, and let $J$ be the ideal in the Gaussian integers (that's ${\bf Z}[i]$) generated by $2+i$. Then $F_1={\bf Z}/I$ and $F_2={\bf Z}[i]/J$ are fields, and the map $\phi:F_1\to F_2$ induced by $\phi(1+I)=1+J$ is a surjective field homomorphism with trivial kernel, but $F_1\ne F_2$.

The two fields are isomorphic, though.