$\textbf{Exercise 4.}$ Suppose $m$ is a positive integer. Is the set $\begin{equation*} \{0\} \cup \{p \in \mathcal{P}(\mathbb{F}) : \deg p = m\} \end{equation*}$ a subspace of $\mathcal{P}(\mathbb{F})$? $\textbf{Solution 4.}$ The set consisting of 0 and all polynomials with coefficients in $\mathbb{F}$ and of degree $m$ is not a subspace of $\mathcal{P}(\mathbb{F})$ because it is not closed under addition. Specifically, the sum of two polynomials of degree $m$ may be a polynomial of degree less than $m$. For example, suppose $m = 2$. Then $7 + 4z + 5z^2$ and $1 + 2z - 5z^2$ are both polynomials of degree 2 but their sum, which equals $8 + 6z$, is a polynomial of degree 1. ---  --- Commentary: $\textbf{Exercise 4.}$ Suppose $m$ is a positive integer. Is the set $\begin{equation*} \{0\} \cup \{p \in \mathcal{P}(\mathbb{F}) : \deg p = m\} \end{equation*}$ a subspace of $\mathcal{P}(\mathbb{F})$? $\textbf{Solution 4.}$ The set consisting of 0 and all polynomials with coefficients in $\mathbb{F}$ and of degree $m$ is not a subspace of $\mathcal{P}(\mathbb{F})$ because it is not closed under addition. Specifically, the sum of two polynomials of degree $m$ may be a polynomial of degree less than $m$. For example, suppose $m = 2$. Then $7 + 4z + 5z^2$ and $1 + 2z - 5z^2$ are both polynomials of degree 2 but their sum, which equals $8 + 6z$, is a polynomial of degree 1. $\textit{Commentary:}$ This exercise tests the understanding of the definition of a subspace. A subset of a vector space is a subspace if and only if it contains the zero vector and is closed under vector addition and scalar multiplication. Here, the given set contains 0 and is clearly closed under scalar multiplication (multiplying a polynomial of degree $m$ by a scalar gives another polynomial of degree $m$ or 0). However, it is not closed under addition, as the example in the solution demonstrates. This illustrates that a subset defined by a degree condition is not typically a subspace. $\textit{Example:}$ Let $\mathbb{F} = \mathbb{R}$ and $m = 1$. The set in question is $\begin{equation*} \{0\} \cup \{p \in \mathcal{P}(\mathbb{R}) : \deg p = 1\} = \{0\} \cup \{ax + b : a, b \in \mathbb{R}, a \neq 0\}. \end{equation*}$ This is not a subspace of $\mathcal{P}(\mathbb{R})$ because, for example, $(x + 1) + (x - 1) = 2x$ is not in the set (it's of degree 1 but it's not 0).