$\bullet~$ Consider the vector space $K_{n}[x]$ over $K, n \geq 2$. Let $W_{1}$ and $W_{2}$ be subspaces of $K_{n}[x]$ defined by:
\begin{array}{l} W_{1}=\left\{p(x) \in K_{n}[x] \mid p(0)=0\right\}\\ \quad \quad \quad \quad \quad~\text{ and } \\ \left.W_{2}=\left\{p(x) \in K_{n} [ x\right] \mid p(1)=0\right\} \end{array}
Find subsets $S_{1}$ and $S_{2}$ such that. $W_{1} \cap W_{2}=\left\langle S_{1}\right\rangle,$ and $W_{1}+W_{2}=\left\langle S_{2}\right\rangle .$ Verify that the set $W_{1} \cup W_{2}$ is not a subspace.
$\bullet~$My attempt:
$\circ~$For $\quad p(x) \in W_{1} \cap W_{2}$,
$p(x)$ Should satisfy $\quad p(x)=0, p(1)=0$
Let $~ p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{2} x^{2}+a_{1} x+a_{0}$
\begin{array}{l} p(0)=0 \Rightarrow a_{0}=0 \\ p(1)=0 \Rightarrow a_{n}+a_{n-1}+\cdots+a_{2}+a_{1}=0 \end{array}
$\Rightarrow \quad p(x)=a_{n}\left(x^{n}-x\right)+a_{n-1}\left(x^{n-1}-x\right)......,+a_{2}\left(x^{2}-x\right)$ $\Rightarrow \quad W_{1} \cap W_{2}=\left\langle x^{n}-x_{1}, \ldots, x^{2}-x\right\rangle$ $\Rightarrow S_{1}=\left\{x^{n}-x, x^{n-1}-x, \dots, x^{2}-x\right\}$
We know $ W_1\cup W_2 $ is subspace of $V$ iff $W_1\subset W_2$ or $W_2\subset W_1$
But neither $W_1\subset W_2$ nor $W_2\subset W_1$. Thus $W_1\cup W_2$ is not a subspace of $K_n[x]$
For $S_2$:
$W_1+W_2=\langle W_1\cap W_2 \rangle $
For this first I find generating sets of $W_1$ and $W_2$ it comes out to be $W_1=\langle x,x^2,....x^n \rangle$
$W_2=\langle x-1,x^2-1,x^3-1,.....x^n-1\rangle$
So , $W_1\cup W_2 = \langle x-1,x^2,x^3..... , x^n\rangle$
$\because$ every element of the generating set of $W_2$ can be split as
$x^k-1=(x-1)(x^{k-1}+x^{k-2}+\cdots+x+1)$
Except $(x-1)$ other terms are consumed by the generator of $W_{1}$
I want you to verify my solution if it has any error Any other approach to find $S_2$ will be appreciated, thankyou