I have to prove the following statement: Let $I$ be an ideal and < be any monomial order. If the initial monomial ideal $in_<(I)$ is square free, then $I$ is a radical ideal.
If I can prove that $in_<(I)$ is radical then the proof would be done because of this: Showing that if the initial ideal of I is radical, then I is radical.
I am a bit stuck on proving that though, any help or hints would be greatly appreciated. Thank you!
As you correctly pointed out, this is a consequence of two results. First, monomial ideals are radical iff they are squarefree (see this question). As hinted by @Mariano Suárez-Álvarez, it helps to know that the radical of a monomial ideal is also a monomial ideal (see this question) to prove the implication "squarefree $\implies$ radical". Second, if the initial ideal $\text{in}_<(I)$ of an ideal $I$ is radical, then $I$ is also radical (see this question). Therefore, \begin{equation*} \text{in}_<(I) \text{ squarefree (and monomial)} \implies \text{in}_<(I) \text{ radical} \implies I \text{ radical}. \end{equation*}
