We've proven the following theorem in class:
Every monomial ideal has a presentation $$I = \bigcap_{i=1}^m Q_i,$$ where each $Q_i = (x_{i_1}^{a_1}, \dots , x_{i_k}^{a_k})$.
I've tried proving the result in question by observing that it suffices to show each $a_i=1.$ Then I'm thinking assuming to the contrary might be the right direction and so if there is some $t$ such that $a_t \ge 2$ and we might bring about a contradiction to the fact that $I$ is a squarefree monomial ideal. I'm stuck with how to do this though, and not even sure if my approach is the right one. Please help?
The idea is the following: if the generating set of $I$ contains a monomial which is not a power of a single variable, then write it as $m_1m_2$ with $\gcd(m_1,m_2)=1$, and notice that $I=(J,m_1)\cap (J,m_2)$, where $J$ is the monomial ideal having the same generators as $I$ excepting $m_1m_2$.