Question

If $\mathrm{F}$ is an infinite field, prove that the polynomial ring$\mathbf{F}\left[x\right]$ is isomorphic to the ring$\mathrm{T}$ of all polynomial functions from F to F(Exercise 27). [Hint: Define a map $\phi :\mathbf{F}\left[\mathbf{x}\right]\to \mathbf{T}$by assigning to each polynomial $\mathbf{f}\left(\mathbf{x}\right)\in \mathbf{F}\left[\mathbf{x}\right]$its induced function in$\mathrm{T}$ ;$\phi$ is injective by Corollary 4.20.]

Short answer

It is proved that the polynomial ring$F\left[x\right]$ is isomorphic to the ring T of all polynomial functions from $F$to$F$ .

Step-by-step solution

Statement of Exercise 23

Exercise 23states that $f\left(x\right),g\left(x\right),h\left(x\right)\in F\left[x\right]$and $r\in F$.

1. When$f\left(x\right)=g\left(x\right)+h\left(x\right)$ in$F\left(x\right)$ , demonstrate that$f\left(r\right)=g\left(r\right)+h\left(r\right)$ in F.
2. When$f\left(x\right)=g\left(x\right)h\left(x\right)$ in $F\left(x\right)$, demonstrate that$f\left(r\right)=g\left(r\right)h\left(r\right)$ in F.

Show that the polynomial ring Fx is isomorphic to the ring T of all polynomial functions from F to F

Corollary 4.20states that consider $F$as an infinite field and$f\left(x\right),g\left(x\right)\in F\left[x\right]$. Then, $f\left(x\right)$and$g\left(x\right)$ provide the same function from $F$to$F$ such that if $f\left(x\right)=g\left(x\right)$in $F\left[x\right]$.

Consider that $\phi :F\left[x\right]\to T, f\left(x\right)\mapsto f,$with $f$as the corresponding polynomial function. According to exercise 23,$\phi$ is a homomorphism. The elements of $T$are described as the images of $\phi$, therefore, it is surjective. Then, according to corollary 4.20, $f\left(x\right)=g\left(x\right)$such that if $\phi \left(f\left(x\right)\right)=\phi \left(g\left(x\right)\right)$, as a result, $\phi$is injective. It concludes that $\phi$is an isomorphism.

Hence, it is proved that the polynomial ring$F\left[x\right]$ is isomorphic to the ring T of all polynomial functions from$F$ to $F$.