Ring of Polynomial Forms over Field is Vector Space

Theorem
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.

Let $F \sqbrk X$ be the ring of polynomials over $F$.

Then $F \sqbrk X$ is an vector space over $F$.

Proof
Let the operation $\times': F \to F \sqbrk X$ be defined as follows.

Let $x \in F$.

Let $\mathbf y \in F \sqbrk X$ be defined as:


 * $\mathbf y = \displaystyle \sum_{k \mathop = 0}^n y_k X^k$

where $n = \map \deg {\mathbf y}$ denotes the degree of $\mathbf y$

Thus:
 * $x \times' \mathbf y := \displaystyle x \sum_{k \mathop = 0}^n y_k X^k = \displaystyle \sum_{k \mathop = 0}^n \paren {x \times y_k} X^k$

We have that $\times': F \to F \sqbrk X$ is an instance of polynomial multiplication where the multiplier $x$ is a polynomial of degree $0$.

Hence, let the supposed vector space over $F$ in question be denoted in full as:
 * $\mathbf V = \struct {F \sqbrk X, +', \times'}_F$

where:
 * $+': F \sqbrk X \to F \sqbrk X$ denotes polynomial addition
 * $\times': F \to F \sqbrk X$ denotes the operation as defined above.

We already have that $F \sqbrk X$ is an integral domain.

Thus vector space axioms $\text V 0$ to $\text V 4$ are fulfilled.

By definition of $\times'$, it is seen that the remaining vector space axioms are fulfilled as follows:

Let $\lambda, \mu \in F$.

Let $\mathbf x, \mathbf y \in F \sqbrk X$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.

All vector space axioms are hence seen to be fulfilled.

Hence the result.