Question

(a)Let$\mathit{k}\mathbf{\ge }\mathbf{1}$ be an integer. Show that the subset $\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{,}{\mathbf{x}}^{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{....}{\mathbf{x}}^{\mathbf{n}}\mathbf{\}}$ of $\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ is linearly independent over $\mathit{R}$.

(b)Show that $\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ is infinite dimensional over$\mathit{R}$ .

Short answer

(a)$\{1,x,x^2,\mathrm{.....}{x}^n\}$of$R\left[x\right]$ is linearly independent.

(b)$R\left[x\right]$is infinite dimension over$R$.

Step-by-step solution

definition of space. 

A space consisting of vectors, together with the associative and commutative operations of addition of vector and the associative operation of multiplication of vector by scalers.

Showing that {1,x,x2,.....xn}  of R[x]  is linearly independent.

(a)

Let $k\ge 1$ be an integer and ${\alpha }_0,{\alpha }_1\mathrm{.....}{\alpha }_k\in R$ are such that

${\alpha }_{0}1+{\alpha }_{1}x+\mathrm{........}+{\alpha }_k{x}^k=0$

By the definition of polynomial each coefficient must be zero. So,

${\alpha }_0={\alpha }_1=\mathrm{......}=0$

Thus ${\alpha }_i=0,1\le i\le k$

Therefore $\{1,x,x^2,\mathrm{.....}{x}^n\}$ of$R\left[x\right]$ is linearly independent over$R$.

Step-3: Showing that R[x]  is infinite dimensional over R . 

(b)

Let $\beta =\{1,x,x^2,\mathrm{.....}{x}^n\}$spans over$R\left[x\right]$.

So let $S$ be a finite subset of$\beta$ can span over$R\left[x\right]$ and $x^n$ be the highest power of $x$belonging to$S$ . Then

Consider the polynomial $x^{n+1}$

$x^{n+1}=\sum _{i=0}^n{a}_i{x}^i,a_0,a_1,\mathrm{.......}{a}_n\in R$

This implies $a_0+a_{1}x+\mathrm{........}+a_n{x}^n-x^{n+1}=0$. This is a zero polynomial so,

$a_0=a_1=\mathrm{.......}=-1$

This is a contradicitionas $0\ne -1$. Thus no finite subset of $\beta$ can be span$R\left[x\right]$.

Therefore$R\left[x\right]$ is infinite dimension over$R$.