Why is $k[x^2-1]\cap (x-1) = (x^2-1)$?

Question

let $k$ be a field, $k[x^2-1]\subset k[x]$, and $(x-1)$ the principal ideal in $k[x]$. Why does the equality in the title hold?

Answer 1

$k[x^2-1]$ means the polynomials in $x^2-1$, which are of the form $$ \sum_k a_k(x^2-1)^k.$$ $k[x^2-1]\cap(x-1)$ is the set of such polynomials that are divisible by $x-1$. Since $x^2-1=(x-1)(x+1)$, it is clear that any nonconstant polynomial in $x^2-1$ lies in $k[x^2-1]\cap (x-1)$. In particular, $$ k[x^2-1]\cap (x-1)=\{\sum_{k>0} a_k(x^2-1)^k\}=(x^2-1),$$ where I believe here $(x^2-1)$ is the ideal in $k[x^2-1]$ (rather than in $k[x]$). Indeed, in $k[x]$ $x^3-x=x(x^2-1)$, but $x^3-x\not\in k[x^2-1]$.