Let $R$ be an integrally closed local domain. Suppose there is a $y\in I^n$ such that $yI^n=I^{2n}$ for some $n$. I would like to prove that $I^n=(y)$.
Source: The above question comes from the proof of Lemma F on page 449 of this paper.
A proof is given, but i cannot understand it, I seem to be missing something very simple it seems.
I asked this question earlier and somebody downvoted, so i thought about it a little more, but still i feel i am missing something.
On
I believe their argument goes as follows:
Let $y \in I^n$ such that $yI^n = (I^n)^2$. We have a finite set of generators $x_1, \ldots, x_m$ for $I^n$.
Suppose we have some $t \in I^n/y$. We can then write $t x_i = \sum_{j=1}^m r_{ij} x_j$ for some $a_{ij} \in R$, since $I^{2n}/y = I^n$.
This implies $\sum_{j=1}^m (t \delta_{ij} - r_{ij}) x_j = 0$. If we multiply by the adjugate of the matrix $t \delta_{ij} - r_{ij}$ we get $\det(t \delta_{ij}-r_{ij}) x_j = 0$, and so $\det(t \delta_{ij}-r_{ij}) = 0$ (since the $x_i$ generate $I$ and so WLOG are non-zero, and $R$ is an integral domain). I believe this is what they are referring to when they say the "determinant trick".
In particular, this is a monic polynomial in $t$ with coefficients in $R$, so $t$ is integral over $R$ and (since $R$ is integrally closed) we have $t \in R$. Therefore $I^n/y \subseteq R$, so $I^n \subseteq (y)$. But since $y \in I^n$ we have $I^n = (y)$.
In such questions the Proposition 2.4 from Atiyah and Macdonald, Introduction to Commutative Algebra, is a kind of universal tool. (See also here.)
In the notation of the book, set $M=I^n/y$, $\mathfrak a= R$. Note that $M^2=M$ within the field of fractions of $R$. Now let $x\in M$ and define $\phi:M\to M$ by $\phi(a)=ax$. Then $\phi(M)\subseteq M$, and therefore there are $a_i\in R$ such that $$\phi^n+a_1\phi^{n-1}+\cdots+a_n=0.$$ Thus $a(x^n+a_1x^{n-1}+\cdots+a_n)=0$ for all $a\in M$. In particular, $x^n+a_1x^{n-1}+\cdots+a_n=0$, so $x$ is integral over $R$ hence $x\in R$, and we are done.