Exercise Let $W=Span\{x^2-x^3,1+x^4\} \leqslant \mathbb{R}_4[x]$. Find a basis for $\mathbb{R}_4[x] / W$
So Let $W=Span\{x^2-x^3,1+x^4\} \leqslant \mathbb{R}_4[x]$.
Then $W = \{a_1(x^2-x^3) + a_2(1+x^4) \, \big| \, a_1,a_2 \in \mathbb{R} \}$
It follows that
$$V / W = \Big\{ v+W \Big | v \in V \Big\}$$
$$V / W = \Big\{ v + \big\{a_1(x^2-x^3)+a_2(1+x^4) \, \big| \, a_1,a_2 \in \mathbb{R}\big\} \, \Big| \, v \in \mathbb{R}_4[x] \Big\}$$
There are a lot of moving pieces here and I struggle to figure out how to find bases for quotient groups in general. Am I allowed to just take my basis to be $\{v+W \big| v \in \mathbb{R}_4[x]\}$?
How does one form a basis for the set of cosets?
Let $V={\mathbb R}_4[x]$. Intuitively, $W=Span\{x^2-x^3,1+x^4\}$ means that in $V/W$, we have $x^2-x^3=0$ (equivalently, $x^2=x^3$) and $1+x^4=0$ (equivalently, $x^4=-1$).
Formally, I will write $\bar{P}$ for the equivalent class of $P$ mod $W$ (so $\bar{P}=P+W$). We then have $\bar{x^3}=\bar{x^2}$ and $\bar{x^4}=\bar{-1}=-\bar{1}$. Now, every element of $V$ can be written as $P=a+bx+cx^2+dx^4$ so
$$ \bar{P}=a\bar{1}+b\bar{x}+c\bar{x^2}+d\bar{x^4}=(a-d)\bar{1}+(c+d)\bar{x^2} $$
so $U=\{\bar{1},\bar{x^2}\}$ spans $V/W$. Since $\dim(V/W)=\dim(V)-\dim(W)=4-2=2$, it is a basis.
Other "method" (more a rewriting of the previous one): Let $U$ be a subspace of $V$ such that $V=W\oplus U$. Then $V/W$ is isomorphic to $U$ via the morphism $\varphi$ induced by the inclusion of $U$ into $V$ (that is $\varphi(P)=\bar{P}$).
It is clear that $U=Span\{1,x^2\}$ satisfies $V=W\oplus U$ ($\{1,x^2,x^2-x^3,1+x^4\}$ is a basis of $V$). Therefore, $\{\varphi(1), \varphi(x^2) \}$ is a basis of $V/W$.