Is the set consisting of $0$ and all polynomials with coefficients in $\mathbb{F}$ and with degree $m$ a subspace of $P(\mathbb{F})$?
I want to say no, since it seems that it is not closed under addition.
Consider $p$,$q$ with $$\begin{align} p(z) = a_0+a_1z+ \cdots + a_mz^m,\\q(z) = b_0+b_1z+\cdots-a_mz^m,\\ \\ \text{where } a_m \neq 0. \end{align}$$
Then $p(z)+q(z) = (a_0+b_0)+(a_1+b_1)z+\cdots +(a_{m-1}+b_{m-1})z^{m-1}$, whose degree is less than or equal to $m-1$.
Best Answer
You are correct.
The space of all polynomials with coefficients in a field $\mathbb{F}$ is a vector space (better still, it is an algebra). For a subset of this vector space to be a subspace, it needs to contain the zero vector, be closed under multiplication by scalars (elements of $\mathbb{F}$), and be closed under addition. The proposed subset satisfies the first two conditions but not the third, as you have shown, so it is not a subspace. However, if we consider the space of all polynomials with coefficients in $\mathbb{F}$ and degree less than or equal to $m$, we do obtain a subspace - in general $\operatorname{deg}(f + g) \leq \max\{\operatorname{deg}(f), \operatorname{deg}(g)\}$.