If $(T-\lambda)^n\in (q)$ and $(T-\lambda)^{n-1}\notin (q)$ then $(T-\lambda)^n=q$

Question

I look at the principle ideal domain $K[T]$ and I have an ideal $\mathcal{I} $ there exists exactly one $q\in K[T]$ such that the highest coefficient is $=1$ and $\mathcal{I}=(q)$. Now I know that $(T-\lambda)^n\in (q)$. I.e there exists a $r\in K[T]$ such that $rq=(T-\lambda=^n$ I also know that $(T-\lambda)^{n-1}\notin (q)$. Why can I say that $q=(T-\lambda)^n$. Now I understand that $(T-\lambda)^n$ is a polynomial of the degree $n$ where the highest coefficient is $=1$. And therefore if $q$ would be of degree $n$ then it would have to be equal to $(T-\lambda)^n$. I also know that $(T-\lambda)$ is a prime element in $K[T]$ and that every element in $K[T]$ is factorsiable. With that all in mind I was not able to deduce why the implication is true. Plese help me

Answer 1

Well, $(T-\lambda)^n\in (q)$ means that $q\mid(T-\lambda)^n$.
But we know that all divisors of $(T-\lambda)^n$ are $c(T-\lambda)^k$ where $c\in K$ (is a unit in $K[T]$) and $k\le n$, because of unique factorization.

Since $q\not\mid(T-\lambda)^{n-1}$, it follows that $q$ must be $(T-\lambda)^n$.