I don't understand the Quotient Space well I'm looking examples of problems a their solutions that involved Quotient Spaces and someone commented the following:
... Let be $V=\Bbb R[x]$ and if $U=\{\text{polynomials $p(x)$ such that $p(0)=0$}\}$, then $\dim(V/U)=1$...
Why the dimesion of $\dim(V/U)=1$?
Can someone explain, please.
Thanks
 A: To find the dimension of $\mathbb R[x]/U$, we must find a basis for it. This is actually pretty easy to do in this particular case just by starting with the standard basis for $\mathbb R[x]$:
$$
\{1, x, x^2, x^3, \ldots\}
$$
Of these basis vectors, nearly all of them are in $U$. For instance, $0^2 = 0$, and therefore $x^2 \in U$ by definition of $U$. So, other than $1$, all of these basis vectors become zero in the quotient $\mathbb R[x]/U$. As a consequence, $\overline 1$ (the image of $1$ in the quotient space) spans all of $\mathbb R[x]/U$. So $\dim(\mathbb R[x]/U)$ is at most $1$.
The possibilities, then, are that $\dim(\mathbb R[x]/U)$ is $0$ or $1$. It is $0$ precisely when $\mathbb R[x]/U = \{\overline 0\}$. This is equivalent to $\mathbb R[x] = U$. But of course this is not true! For instance, $1 \notin U$, so $\mathbb R[x] \neq U$. So we conclude that the dimension is exactly $1$.
A: The set of polynomials $p$ such that $p(0)=0$ are all of the polynomials whose constant term is $0$, i.e. $U=\{p: p(x)=a_nx^n+...+a_1x\}$. A general polynomial of the form $q(x)=b_mx^m+...b_1x+b_0$ gets mapped to $U+b_0$ where $b_0\in\mathbb R$.
Therefore, the set of cosets in the quotient space is isomorphic to $\mathbb R$ and the dimension of the quotient space is $1$.
A: The previous answers are correct. I will however propose another point of view. Consider the linear map:
$$
\mathrm{ev}_0:\left\{
\begin{matrix}
V & \longrightarrow & \mathbb R\\
p & \longmapsto & p(0).
\end{matrix}\right.
$$
Note that this linear map is surjective and you can identify its kernel to be exactly $U$ (it is actually the definition of $U$ to be the kernel of this linear map). This gives the isomorphism of linear spaces:
$$
V/U\simeq \mathbb R.
$$
Thus, of course $\mathrm{dim}(V/U)=1$.
