Question

Prove these parts of Theorem G. 1:

(a) addition in P is associative;

(b) both distributive laws hold in P;

(c) Pis commutative if R is.

Short answer

1. It is proved that$P$ is associative.
2. It is proved that both distributive laws hold in P.
3. It is proved that P is commutative if R is.

Step-by-step solution

Introduction of polynomial and theorem G.1

Consider the polynomial in $\mathit{x}$and $\mathbf{ℝ}$will be express as;

role="math" localid="1658911188375" $\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{a}}_{\mathbf{0}}{\mathbf{x}}^{\mathbf{0}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$

Polynomial is always finite but sequence can be finite and infinite.

According to the theorem G.1, let R is the ring with identity and P the set of polynomials with coefficients in R. then P is a ring with identity if R is commutative then P is also commutative.

Prove that addition in P is associative

(a)

Consider three infinite polynomial sequences $({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...}),({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})$and$({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})$ .

Now to prove addition in is associative, prove $\{({\mathrm{a}}_0,{\mathrm{a}}_1,\mathrm{...})\oplus ({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}=({\mathrm{a}}_0,{\mathrm{a}}_1,\mathrm{...})\oplus \{({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}$.

Now evaluate,$\{({\mathrm{a}}_0,{\mathrm{a}}_1,\mathrm{...})\oplus ({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}$ .

$\begin{array}{c}\{({\mathrm{a}}_0,{\mathrm{a}}_1,\mathrm{...})\oplus ({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\}\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})=\left\{({\mathrm{a}}_0+{\mathrm{b}}_0,{\mathrm{a}}_1+{\mathrm{b}}_1)\right\}\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\\ =({\mathrm{a}}_0+{\mathrm{b}}_0+{\mathrm{c}}_0,{\mathrm{a}}_1+{\mathrm{b}}_1+{\mathrm{c}}_1,\mathrm{...})\end{array}$

And,

$\begin{array}{c}({\mathrm{a}}_0,{\mathrm{a}}_1,\mathrm{...})\oplus \{({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}=({\mathrm{a}}_0,{\mathrm{a}}_1,\mathrm{...})\oplus \{{\mathrm{b}}_0+{\mathrm{c}}_0,{\mathrm{b}}_1+{\mathrm{c}}_1,\mathrm{...}\}\\ =({\mathrm{a}}_0+{\mathrm{b}}_0+{\mathrm{c}}_0,{\mathrm{a}}_1+{\mathrm{b}}_1+{\mathrm{c}}_1,\mathrm{...})\end{array}$

Therefore, role="math" localid="1658911388362" $\{(a_0,a_1,\mathrm{...})\oplus (b_0,b_1,\mathrm{...})\}\oplus (c_0,c_1,\mathrm{...})=(a_0,a_1,\mathrm{...})\oplus \{(b_0,b_1,\mathrm{...})\oplus (c_0,c_1,\mathrm{...})\}$.

Hence, it is proved that$P$ is associative.

Prove that both distributive laws hold in P

(b)

Consider the three infinite polynomial sequences $(a_0,\text{\hspace{0.17em}}{a}_1,\mathrm{...}),(b_0,b_1,\mathrm{...})$and $(c_0,c_1,\mathrm{...})$.

Now to prove distributive law show that

$({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes \{({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}=\{({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes ({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\}\oplus \{({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}$

Evaluate $({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes \{({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}$.

$\begin{array}{c}({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes \{({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\oplus ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}=({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes \left\{({\mathrm{b}}_0+{\mathrm{c}}_0,{\mathrm{b}}_1+{\mathrm{c}}_1,\mathrm{...})\right\}\\ =({\mathrm{a}}_0\cdot ({\mathrm{b}}_0+{\mathrm{c}}_0),{\mathrm{a}}_0\cdot ({\mathrm{b}}_1+{\mathrm{c}}_1),{\mathrm{a}}_1\cdot ({\mathrm{b}}_0+{\mathrm{c}}_0),{\mathrm{a}}_2\cdot ({\mathrm{b}}_1+{\mathrm{c}}_1),\mathrm{...})\\ =({\mathrm{a}}_0\cdot {\mathrm{b}}_0+{\mathrm{a}}_0\cdot {\mathrm{c}}_0,{\mathrm{a}}_0\cdot {\mathrm{b}}_1+{\mathrm{a}}_0\cdot {\mathrm{c}}_1,{\mathrm{a}}_1\cdot {\mathrm{b}}_0+{\mathrm{a}}_1\cdot {\mathrm{c}}_0,{\mathrm{a}}_1\cdot {\mathrm{b}}_1+{\mathrm{a}}_1\cdot {\mathrm{c}}_1,\mathrm{...})\end{array}$

Evaluate $\{({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes ({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})\}\oplus \{({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\otimes ({\mathrm{c}}_0,{\mathrm{c}}_1,\mathrm{...})\}$.

Therefore,

role="math" localid="1658911610678" $(a_0,\text{\hspace{0.17em}}{a}_1,\mathrm{...})\otimes \{(b_0,b_1,\mathrm{...})\oplus (c_0,c_1,\mathrm{...})\}=\{(a_0,\text{\hspace{0.17em}}{a}_1,\mathrm{...})\otimes (b_0,b_1,\mathrm{...})\}\oplus \{(a_0,\text{\hspace{0.17em}}{a}_1,\mathrm{...})\otimes (c_0,c_1,\mathrm{...})\}$

Hence, it is proved that both distributive laws hold in P.

Prove that P is commutative if R is

(c)

Consider two infinite polynomial sequences , $(a_0,\text{\hspace{0.17em}}{a}_1,\mathrm{...})$and $(b_0,b_1,\mathrm{...})$.

To prove that $P$is commutative if $R$is commutative then

$({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,\mathrm{...})\oplus ({\mathrm{b}}_0,{\mathrm{b}}_1,\mathrm{...})=({\mathrm{a}}_0+{\mathrm{b}}_0,{\mathrm{a}}_0+{\mathrm{b}}_1,{\mathrm{a}}_1+{\mathrm{b}}_0,{\mathrm{a}}_1+{\mathrm{b}}_1,\mathrm{...})$

And,

$({\mathrm{b}}_0,{\mathrm{b}}_1,...)\oplus ({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,...)=({\mathrm{b}}_0+{\mathrm{a}}_0,{\mathrm{b}}_0+{\mathrm{a}}_1,{\mathrm{b}}_1+{\mathrm{a}}_0,{\mathrm{b}}_1+{\mathrm{a}}_1,...)$

Therefore, $({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,...)\oplus ({\mathrm{b}}_0,{\mathrm{b}}_1,...)=({\mathrm{b}}_0,{\mathrm{b}}_1,...)\oplus ({\mathrm{a}}_0,\text{\hspace{0.17em}}{\mathrm{a}}_1,...)$if and only if$R$ is commutative.

Hence, it is proved that P is commutative if R is.