Let $I\subset K[x_1,\dots,x_n]$ be a monomial ideal, $t\ge 2$ an integer, and $\mathfrak p \in \operatorname{Ass}(R/I^t)$. Then one knows that $\mathfrak p=(I^t : c)$ for some monomial $c\in R$. Show that $c\in I^{t-1}$.
Since $\mathfrak p$ is necessarily a monomial prime ideal it is generated by a subset of the variables, and $x_ic \in I^t$ for a variable $x_i\in\mathfrak p$ then $c \in I^{t-1}$. I can't understand why.
Suppose $I$ is generated by the monomials $m_1,\dots,m_r$. Then $I^t$ is generated by all the products of $t$ (not necessarily distinct) monomials from the set $\{m_1,\dots,m_r\}$. The ideal $(I^t:c)$ is generated by the monomials $m_{i_1}\cdots m_{i_t}/\gcd(m_{i_1}\cdots m_{i_t},c)$ and since this ideal is also generated by a subset of variables, then there is an $i$ such that $$X_i=\frac{m_{i_1}\cdots m_{i_t}}{\gcd(m_{i_1}\cdots m_{i_t},c)}.$$ Now we get $\gcd(m_{i_1}\cdots m_{i_t},c)=\frac{m_{i_1}\cdots m_{i_t}}{X_i}$ $\Rightarrow$ $X_i\mid m_{i_1}\cdots m_{i_t}$ and $\frac{m_{i_1}\cdots m_{i_t}}{X_i}\mid c$ $\Rightarrow$ $c\in I^{t-1}.$
