A polynomial is a polynomial

## constituents

- a function \(p: \mathbb{F} \to \mathbb{F}\)
- coefficient \(a_0, \dots, a_{m} \in \mathbb{F}\)

## requirements

A polynomial is defined by:

\begin{equation} p(z)=a_0+a_1z+a_2z^{2}+\dots +a_{m}z^{m} \end{equation}

for all \(z \in \mathbb{F}\)

## additional information

### degree of a polynomial \(\deg p\)

A polynomial’s degree is the value of the highest non-zero exponent. That is, for a polynomial:

\begin{equation} p(z) = a_0+a_1z+\dots +a_{m}z^{m} \end{equation}

with \(a_{m} \neq 0\), the degree of it is \(m\). We write \(\deg p = m\).

A polynomial \(=0\) is defined to have degree \(-\infty\)

Of course, a polynomial with degree \(n\), times a polynomial of degree \(m\), has degree \(mn\). We see that:

\begin{equation} x^{n}x^{m} = x^{n+m} \end{equation}

### \(\mathcal{P}(\mathbb{F})\)

\(\mathcal{P}(\mathbb{F})\) is the set of all polynomials with coefficients in \(\mathbb{F}\).

### \(\mathcal{P}(\mathbb{F})\) is a vector space over \(\mathbb{F}\)

We first see that polynomials are functions from \(\mathbb{F}\to \mathbb{F}\). We have shown previously that F^s is a Vector Space Over F.

Therefore, we can first say that \(\mathcal{P}(\mathbb{F}) \subset \mathbb{F}^{\mathbb{F}}\).

Lastly, we simply have to show that \(\mathcal{P}(\mathbb{F})\) is a subspace.

- zero exists by taking all \(a_{m} = 0\)
- addition is closed by inheriting commutativity and distributivity in \(\mathbb{F}\)
- scalar multiplication is closed by distributivity

Having satisfied the conditions of subspace, \(\mathcal{P}(\mathbb{F})\) is a vector space. \(\blacksquare\)

### \(\mathcal{P}_{m}(\mathbb{F})\)

For \(m\geq 0\), \(\mathcal{P}_{m}(\mathbb{F})\) denotes the set of all polynomials with coefficients \(\mathbb{F}\) and degree at most \(m\).