Determining dimension and basis of a subspace of polynomial vector space. Then determining dimension and basis of subsequent quotient space. Denote by $X$ the vector space of all polynomials $p(x)$ of degree less than $n$. Denote by $Y$ the subset of $X$ containing polynomials that are zero at distinct $x_1, x_2, \dots , x_m \in K$ where $m<n$.
(i) Show that $Y$ is a subspace of $X$.
(ii) Determine dim $Y$ and find a basis of $Y$.
(iii) Determine dim$X/Y$ and find a basis of $X/Y$.
Proof.
$Y = \{q(x) \prod_{i=1}^m(x-x_i):q(x) \text{ is a polynomial of degree < n-m-1}\}$.
(i)  To show Y is closed under addition, consider $p_1,\, p_2\in Y$
\begin{align}
p_1(x)\in Y\Longrightarrow p_1(x) = q_1(x)\prod_{i=1}^m(x-x_i)\\
p_2(x)\in Y\Longrightarrow p_2(x) = q_2(x)\prod_{i=1}^m(x-x_i)\\
\end{align}
Then
\begin{align}
p_1(x)+p_2(x) &= q_1(x)\prod_{i=1}^m(x-x_i)+q_2(x)\prod_{i=1}^m(x-x_i)\\
&= (q_1(x) + q_2(x))\prod_{i=1}^m(x-x_i)\\
\end{align}
It follows that $p_1(x)+p_2(x) \in Y$.
For any $p \in Y$, $k\in K$ we have
\begin{align}
p(x) \in Y \Longrightarrow p(x) &= q(x)\prod_{i=1}^m(x-x_i)\\
kp(x)&= kq(x)\prod_{i=1}^m(x-x_i) \in Y
\end{align}
(ii) Guess: $\dim Y = n-m$. Possibly $\{1, x^{m+1},\dots,x^{n-1}\}$ could form a basis.
(iii) Guessing that $\dim X/Y = n - (n-m) = m$ from the theorem $\dim X/Y = \dim X - \dim Y$.
Here $X/Y$ represents the quotient space of $X$ mod $Y$. (Linear space of congruence classes). Looking for feedback on (i) and ideas for (ii) and (iii). Thanks for your time.
 A: I don't understand how you think you already showed closedness under addition. It should be
$$\begin{cases}p(x)\in Y\implies p(x)=p_1(x)\prod\limits_{i=1}^m(x-x_i)\\{}\\
g(x)\in Y\implies g(x)=g_1(x)\prod\limits_{i=1}^m(x-x_i)\end{cases}\;\;\implies p(x)+g(x)=(p_1(x)+g_1(x))\prod\limits_{i=1}^m(x-x_i)$$
and since $\;\deg(p+g)\le\max(\deg p,\,\deg g)\;$, we get that $\;p(x)+g(x)\in Y\;$ .
Multiplication by scvalar is trivial and you already made it...but still must show something similar as above, because not all depends on the degree but also it must vanish at the points $\;x_1,...,x_m\;$ .
Dimension: try to show the following set:
$$\;\left\{\,v_0=\prod\limits_{i=1}^m(x-x_i),\,v_1=x\prod\limits_{i=1}^m(x-x_i),\ldots,\,v_{n-m-2}=x^{n-m-1}\prod\limits_{i=1}^m(x-x_i)\,\right\}\;$$
is a basis of $\;Y\;$ .For example, if there are scalars $\;a_1,...,a_{n-m-2}\;$ s.t. that
$$\sum_{k=0}^{n-m-2}a_k \left(x^k\prod\limits_{i=1}^m(x-x_i)\right)=0\;,\;\;\text{then show that}\; a_0=a_1=\ldots=a_{n-m-2}=0$$
To show the above is a generator set of $\;Y\;$ , observe carefully what you wrote in your definition of $\;Y\;$, which is accurate...The rest is, I believe, easy for you.
