$\textbf{Exercise 5.}$ Is the set $\begin{equation*} \{0\} \cup \{p \in \mathcal{P}(\mathbb{F}) : \deg p \text{ is even}\} \end{equation*}$ a subspace of $\mathcal{P}(\mathbb{F})$? $\textbf{Solution 5.}$ The set consisting of 0 and all polynomials of even degree is not a subspace of $\mathcal{P}(\mathbb{F})$ because it is not closed under addition. Specifically, the sum of two polynomials of even degree may be a polynomial of odd degree. For example, $7 + 4z + 5z^2$ and $1 + 2z - 5z^2$ are both polynomials of degree 2 but their sum, which equals $8 + 6z$, is a polynomial of odd degree. ---  --- Commentary: Problem 5 (Chapter 4): $\textbf{Exercise 5.}$ Is the set $\begin{equation*} \{0\} \cup \{p \in \mathcal{P}(\mathbb{F}) : \deg p \text{ is even}\} \end{equation*}$ a subspace of $\mathcal{P}(\mathbb{F})$? $\textbf{Solution 5.}$ The set consisting of 0 and all polynomials of even degree is not a subspace of $\mathcal{P}(\mathbb{F})$ because it is not closed under addition. Specifically, the sum of two polynomials of even degree may be a polynomial of odd degree. For example, $7 + 4z + 5z^2$ and $1 + 2z - 5z^2$ are both polynomials of degree 2 but their sum, which equals $8 + 6z$, is a polynomial of odd degree. $\textit{Commentary:}$ This exercise is similar to the previous one and reinforces the understanding of subspaces. Again, the given set contains 0 and is closed under scalar multiplication (multiplying a polynomial of even degree by a scalar gives another polynomial of even degree or 0). However, it is not closed under addition, as the example in the solution shows. This illustrates that a subset defined by a parity condition on the degree is not typically a subspace. $\textit{Example:}$ Let $\mathbb{F} = \mathbb{R}$. The set in question is $\begin{equation*} \{0\} \cup \{p \in \mathcal{P}(\mathbb{R}) : \deg p \text{ is even}\} = \{0\} \cup \{a_0 + a_2x^2 + a_4x^4 + \cdots : a_i \in \mathbb{R}\}. \end{equation*}$ This is not a subspace of $\mathcal{P}(\mathbb{R})$ because, for example, $(1 + x^2) + (1 - x^2) = 2$ is not in the set (it's of degree 0, which is even, but it's not 0).