# Ring of Polynomial Forms over Field is Vector Space/Corollary

## Theorem

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

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

Let $S \subseteq F \sqbrk X$ denote the subset of $F \sqbrk X$ defined as:

- $S = \set {\mathbf x \in F \sqbrk X: \map \deg {\mathbf x} < d}$

for some $d \in \Z_{>0}$.

Then $S$ is an vector space over $F$.

## Proof

From Ring of Polynomial Forms over Field is Vector Space we note that $\struct {F, +, \times}$ is a vector space over $F$.

The remaining question is that $S$ remains closed under polynomial addition and scalar multiplication.

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

We have:

\(\ds \mathbf x + \mathbf y\) | \(=\) | \(\ds \sum_{j \mathop = 0}^m x_j X^j + \sum_{k \mathop = 0}^n y_k X^k\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {x_k + y_k} X^k\) | Definition of Polynomial Addition |

As both $m < d$ and $n < d$ by dint of them belonging to $S$, it follows that $\max \set {m, n} < d$.

Hence $\mathbf x + \mathbf y \in S$ and so $S$ is closed under polynomial addition.

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

Let $\lambda \in F$.

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

- $\mathbf x = \displaystyle \sum_{k \mathop = 0}^n x_k X^k$

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

Thus:

- $\lambda \times' \mathbf x := \displaystyle \lambda \times' \sum_{k \mathop = 0}^n x_k X^k = \displaystyle \sum_{k \mathop = 0}^n \paren {\lambda \times x_k} X^k$

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

By definition of polynomial multiplication:

\(\ds \map \deg {\lambda \times' \mathbf x}\) | \(=\) | \(\ds \map \deg \lambda + \map \deg {\mathbf x}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 0 + \map \deg {\mathbf x}\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds d\) | as $\mathbf x \in S$ |

The result follows.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $63$