Question

Let $\mathit{F}$be a field and $\mathit{f}\mathbf{:}\mathit{F}\mathbf{\to }\mathit{R}$a homomorphism of rings.

(a) If there is a non-zero element $\mathit{c}$of$\mathit{F}$ such that $\mathit{f}\left(c\right)\mathbf{=}{\mathbf{0}}_{\mathbf{R}}$, prove that$\mathit{f}$ is the zero homomorphism (that is $\mathit{f}\left(x\right)\mathbf{=}{\mathbf{0}}_{\mathbf{R}}$, for every $\mathit{x}\mathbf{\in }\mathit{F}$).

(b) Prove that$\mathit{f}$ is either injective or zero homomorphism.

Short answer

It is proved that,

1. $F$is the zero homomorphism as $f\left(x\right)=0_R,\forall x\in F$.
2. $f$is either injective or zero homomorphism.

Step-by-step solution

a)Determine f is the zero homorphism

Consider that$x$ is an arbitrary element of field $F$, $f:F\to R$is a homorphism of rings and$c\in F$ so that$c\ne 0_F$ and $f\left(c\right)=0_R$.

As$F$ is a field and $c\ne 0_F$, and here$F$ has the identity elements$1_R$ and there existsrole="math" localid="1648206493778" $c^{-1}\left(c{c}^{-1}=c^{-1}c=1_R\right)$

Then,

$\begin{array}{rcl}f\left(x\right)& =& f\left(x{1}_R\right)=f\left(xc{c}^{-1}\right)=f\left(x\right)f\left(c\right)f\left(c^{-1}\right)=f\left(x\right)0_{R}f\left(c^{-1}\right)\\ & =& \left(f\left(x\right)0_R\right)f\left(c^{-1}\right)\stackrel{\mathrm{Theorem}3.5}{=}{0}_{R}f\left(c^{-1}\right)\stackrel{\mathrm{Theorem}3.5}{=}{0}_R\end{array}$

Therefore,$F$ is the zero homorphism as $f\left(x\right)=\begin{array}{rcl}& & 0\end{array}\begin{array}{rcl}& & \end{array}$_R,∀x∈F.

b)Determine f is either injective or the zero homomorphism

Let’s assume that $f:F\to R$, a homorphism of rings here$F$ is field and it not homorphism.

From part a), there is no non-zero element$c$ in$F$ so that,$f\left(c\right)=0_R,f\left(x\right)\ne 0_R,\forall x\in F$

so that $x\ne 0_F$.

Now, consider the$a$ and$b$ are the elements in$F$ so that $f\left(a\right)=f\left(b\right)$.

$f\left(a\right)=f\left(b\right)\Rightarrow f\left(a\right)-f\left(b\right)=0_R\Rightarrow f\left(a-b\right)=0_R\stackrel{\left(*\right)}{\Rightarrow }a-b=0_F\Rightarrow a=b$

Therefore,$f$ is an injection.

So,$f$ is either injective or zero homomorphism.