Question

Let a be a fixed element of F and define a map ${\mathit{\phi }}_{\mathbf{a}}\mathbf{:}\mathit{F}\left[x\right]\mathbf{\to }\mathit{F}$by ${\mathit{\phi }}_{\mathbf{a}}\left[f\left(x\right)\right]\mathbf{=}\mathit{f}\left(a\right)$. Prove that ${\mathit{\phi }}_{\mathbf{a}}$is a surjective homomorphism of rings. The map ${\mathit{\phi }}_{\mathbf{a}}$is called an evaluation homomorphism; there is one for each $\mathit{a}\mathbf{\in }\mathit{F}$.

Short answer

It is proved that${\phi }_a$ is surjective.

Step-by-step solution

Determine φa is a surjective

Let’s define, ${\phi }_a\left(f\left(x\right)\right):=f\left(a\right)$. Such that, ${\phi }_a\left(f\left(x\right)+g\left(x\right)\right)={\phi }_a\left(f\left(x\right)\right)+{\phi }_a\left(g\left(x\right)\right)$and ${\phi }_a\left(f\left(x\right)g\left(x\right)\right)={\phi }_a\left(f\left(x\right)\right){\phi }_a\left(g\left(x\right)\right)$.

Therefore, ${\phi }_a$is homomorphism.

Foe all $r\in F$contain polynomial $f\left(x\right)\in F\left[x\right]withf\left(x\right)=f_0=r$satisfy ${\phi }_a\left(f\left(x\right)\right)=r$Therefore,${\phi }_a$ is surjective.