$V$ is a vector space of polynomials of degree less than or equal to 50 and $W$ is the set of polynomials in $V$ which are divisible by $x^4$.
I have shown that $W$ is a subspace of $V$ and that the dimension of $V/W$ is four.
If $V$ was the space of all polynomials over $\mathbb{R}$, why would the dimension of $V/W$ remain the same?
Best Answer
Let $V$ be the space of polynomials of degree less than or equal to $N$, where $N \ge 3$, and, abusing notation slightly, we allow $N = \infty$ to mean 'all polynomials'.
Suppose $p \in V$, where $p(x) = \sum_{k=0}^n p_k x^k$. Then if $w(x) = \sum_{k=4}^n p_k x^k = x^4 \sum_{k=0}^n p_k x^{k-4}$, we see that $w \in W$. Hence $x \mapsto \sum_{k=0}^3 p_k x^k \in [p] = \{ p \} +W$. Since $p$ was arbitrary, it follows that $\dim V/W \le 4$.
Let $e_k(x) = x^k$ for $k=0,..,3$. Suppose $\sum_{k=0}^3 \alpha_k [e_k] = [0]$, or equivalently, $[\sum_{k=0}^3 \alpha_k e_k] = [0]$. This implies that $\sum_{k=0}^3 \alpha_k e_k = (x \mapsto \sum_{k=0}^3 \alpha_k x_k) \in W$. However, the only way $\sum_{k=0}^3 \alpha_k e_k$ can be divisible by $x \mapsto x^4$ is if $\alpha_k = 0$. Hence $\{[e_k] \}_{k=0}^3$ are linearly independent, and hence we have $\dim V/W = 4$.
If $N<3$, a slight modification of the above shows that $\dim V/W = \min(4,N+1)$.