Consider a field $K$, a monomial $v_j \in K[x_1, \dots, x_n]$, and a monomial ideal $I=\langle u_1, u_2\dots, u_r \rangle$. I'm having trouble showing $I:\langle v_j \rangle \subseteq \langle \{ u_i / \gcd(u_i,v_j) \ : \ i=1,\dots, r \} \rangle$.
I fooled around with it myself for quite some time and wasn't really able to prove it completely (made some progress down multiple routes, but always seemed to eventually get stumped). So I succumbed to looking at what the book I'm using had done, but this is where my confusion begins. The book's proof starts with the following:
Let $u \in I : \langle v_j \rangle$. Then $uv_j \in I$, and hence there exists $i$ such that $u_i|uv_j$.
How do we know such an $i$ exists? Isn't $uv_j$ of the form $$uv_j = v_j \sum_{i=1}^r q_i u_i$$ where $q_i \in K[x_1, \dots, x_n]$?
It was probably intended that $u$ be a monomial. It suffices to consider monomials $u$ since the colon of two monomial ideals is a monomial ideal. Then $uv_j$ is a monomial, and you can use the fact that if a monomial $v$ belongs to the ideal $\langle u_1,\ldots,u_r\rangle$, then it is divisible by one of the $u_i$. If you don't know that fact, you should try to prove it, it is a fundamental property of monomial ideals.
As a starting point, write $v=\underline x^{\underline\alpha}$, and suppose $v=\sum q_iu_i$. Then in particular, the sum on the right hand side must have at least one nonzero $\underline x^{\underline\alpha}$ term. Where can it come from?