Question

If $\mathit{n}\mathbf{\ge }\mathbf{1}$is an integer, ${\mathit{R}}_{\mathbf{n}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$denote the set consisting of the constant polynomial 0 and all polynomial in ${\mathit{R}}_{\mathbf{n}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$of degree $\mathbf{\le }\mathit{n}$. Show that $\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$is a vector space over $\mathit{R}$.

Short answer

Answer:

$\mathit{R}\left[x\right]$is a vector space over $\mathit{R}$.

Step-by-step solution

Definition of space vector.

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

Showing that αf(x)=∑i=0nαaixi∈R[x].

Let $\mathit{R}\left[x\right]$be the ring of polynomial. Then the addition and the scaler multiplication be defined by $\mathit{f}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{,}\mathit{g}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{b}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{\in }\mathit{R}\left(x\right)$and $\mathit{\alpha }\mathbf{\in }\mathit{R}$.

$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{g}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\left(a_i+b_i\right){\mathit{x}}^{\mathbf{i}}$

And

$\mathit{\alpha }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\alpha }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{\in }\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$. Where $\mathit{m}\mathbf{=}\mathit{m}\mathit{a}\mathit{x}\left(n,p\right)$

Then apply the required condition to be a vector space over a field.

Step 3:Showing that α(fx+gx)=αf(x)+αg(x).

As $\mathit{R}\left[x\right]$is an abelian group with the respect to addition.

Let $\mathit{f}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{,}\mathit{g}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\rho }}{\mathit{b}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{\in }\mathit{R}\left[x\right]\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{\alpha }\mathbf{,}\mathit{\beta }\mathbf{\in }\mathit{R}$.

Then,

$$\begin{gathered} \mathit{\alpha }\left(f\left(x\right)+g\left(x\right)\right)\mathbf{=}\mathit{\alpha }\left(\sum _{i=0}^m\left(a_i+b_i\right)x^i\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{m}}\mathit{\alpha }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{+}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{m}}\mathit{\alpha }{\mathit{b}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{\alpha }\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{m}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{+}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{m}}{\mathit{b}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \end{gathered}$$

This gives,

$\mathit{\alpha }\mathbf{(}\mathbf{f}\left(x\right)\mathbf{+}\mathbf{g}\left(x\right)\mathbf{)}\mathbf{=}\mathit{\alpha }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{\alpha }\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Step 4:Showing that (α+β)f(x)=αf(x)+βf(x).

Third,

$$\begin{gathered} \left(\alpha +\beta \right)\mathit{f}\left(x\right)\mathbf{=}\left(\alpha +\beta \right)\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\left(\alpha +\beta \right){\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\alpha }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}}\mathbf{+}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\beta }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \end{gathered}$$

This gives,

$\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{\beta }\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Step 5:Showing that (αβ)f(x)=α[βfx].

$$\begin{gathered} \mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\left(\alpha \beta \right)\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\left(\alpha \beta \right){\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{\alpha }\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{\beta }{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \end{gathered}$$

Then gives

$\mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{\alpha }\mathbf{\left[}\mathbf{\beta f}\left(x\right)\mathbf{\right]}$

Step 6:Showing that R[x] is a space over R

Compatibility of scalar the multiplication with field multiplication.

That is $\left(\alpha \beta \right)\mathit{u}\mathbf{=}\mathit{\alpha }\left(\beta u\right)$here u are vectors and $\mathit{\alpha }\mathbf{,}\mathit{\beta }$is scalar.

$$\begin{gathered} \mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathit{u}\mathbf{=}\mathit{\alpha }\mathbf{\left(}\mathbf{\beta u}\mathbf{\right)}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}{\mathit{a}}_{\mathbf{i}}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{\left(}\mathbf{\alpha \beta }\mathbf{\right)}\mathit{a}\mathbf{,}{\mathit{x}}^{\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathit{\beta }\mathit{a}\mathbf{,}{\mathit{x}}^{\mathbf{i}} \end{gathered}$$

This gives

$\left(\alpha \beta \right)\mathit{f}\left(x\right)\mathbf{=}\mathit{\alpha }\left[\beta f\left(x\right)\right]$

All conditions for ${\mathit{R}}_{\mathbf{n}}\left[x\right]$to be a vector space are satisfied.

Therefore, polynomial ring role="math" localid="1656926814016" $\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ is a vector space over R.