Question

Let be the set of all polynomial functions from F to F.Show that T is a commutative ring with identity, with operations defined as in calculus: For each $\mathbf{r}\in \mathbf{F}$, $\left(\mathbf{f}\mathbf{+}\mathbf{g}\right)\left(\mathbf{r}\right)=\mathbf{f}\left(\mathbf{r}\right)+\mathbf{g}\left(\mathbf{r}\right)$and$\left(\mathbf{fg}\right)\left(\mathbf{r}\right)=\mathbf{f}\left(\mathbf{r}\right)\mathbf{g}\left(\mathbf{r}\right)$ . [Hint: To show that T is closed under addition and multiplication, use Exercise 23 to verify that $\mathbf{f}+\mathbf{g}$and $\mathbf{fg}$are the polynomial functions induced by the sum and product polynomials $\mathbf{f}\left(\mathbf{x}\right)+\mathbf{g}\left(\mathbf{x}\right)$and $\mathbf{f}\left(\mathbf{x}\right)\mathbf{g}\left(\mathbf{x}\right)$, respectively.]

Short answer

It is proved that T is a commutative ring with identity.

Step-by-step solution

Statement of Exercise 23

Exercise23states that $f\left(x\right),g\left(x\right),h\left(x\right)\in F\left[x\right]$and$r\in F$ .

1. If $f\left(x\right)=g\left(x\right)+h\left(x\right)$in$F\left[x\right]$, show that $f\left(r\right)=g\left(r\right)+h\left(r\right)$in$F$.
2. If $f\left(x\right)=g\left(x\right)h\left(x\right)$in $F\left[x\right]$, show that$f\left(r\right)=g\left(r\right)h\left(r\right)$ in F.

Show that is a commutative ring with identity, with operations defined in calculus

With each $f\in T$, there exists a polynomial $\overline{f}\left(x\right)\in F\left[x\right]$in which $f\left(r\right)=\overline{f}\left(r\right)$(it is noted that this might not be unique), and there is $\overline{f+g}\left(x\right)=\overline{f}\left(x\right)+\overline{g}\left(x\right)$, and $\overline{fg}\left(x\right)=\overline{f}\left(x\right)\overline{g}\left(x\right)$from Exercise 23. As a result, $T$would be closed under addition and multiplication.

As a result,$F\left[x\right]$ satisfies the commutative ring of identity axioms, therefore, doesT.

It concludes that$F\left[x\right]$ satisfies the commutative ring of identity axioms.

Hence, it is proved that T is a commutative ring with identity.