I am looking for an answer to the following question.
If $k$ is a characteristic zero field and $I=(X^3+X^2,X^6)k[X]$, find a basis for the $k$-subspace $I$ of the vector space $k[X]$.
It is clear that since $k$ is a field $k[X]$ is a PID, and hence any ideal $I$ is principal, i.e. generated by a single polynomial. Now with a little bit of observation we can see that $$I=(X^3+X^2,X^6)k[X]=(X^2)k[X]$$ since $\gcd(X^3+X^2,X^6)=X^2$ in the polynomial ring $k[X]$. Now I want to find a $k$-basis for this principal ideal. But my question extends to a general question. In general, how can we figure out a basis for any given ideal $I$ of $k[X]$ generated by $p(X)$? What is the key thing that I am missing?
Like you said, $k[X]$ is a PID, hence for every ideal $I$ there exists some $p(X)\in k[X]$, with $\deg(p)=d$, such that $I=(p(X))k[X]$. Now $$\mathcal B=\{p(X),Xp(X), X^2p(X),\dots,X^np(X),\dots\},$$ give to you a basis for $I$ on the field $k$. It's obvious that they form a system of generators, i.e., $\mathrm{span}_k(\mathcal B)=I$, so let's verify that they are linearly independent. If there exists some $\lambda_0,\dots,\lambda_n\in k$ (assume that $\lambda_n\neq 0$) such that $$\lambda_0p(X)+\dots+ \lambda_nX^np(X)=0\implies \lambda_0p(X)+\dots+ \lambda_{n-1}X^{n-1}p(X)=-\lambda_{n}X^{n}p(X), $$ but now the degree of LHS is $\le n-1+d$ and the degree of RHS is equal to $n+d$, contradiction.
Edit: You can also prove directly the linear independence in this way $$ \lambda_0p(X)+\dots+ \lambda_nX^np(X)=0\iff p(x)(\lambda_0+\lambda_1X+\dots+\lambda_nX^n)=0,$$ but $k[X]$ is an integral domain hence $p(X)=0$ (in this case the problem is trivial) or $\lambda_0+\lambda_1X+\dots+\lambda_nX^n=0 \iff \lambda_i=0 \forall i=0,\dots,n.$