Question

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

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.