If $I\subseteq J$ are ideals in a polynomial ring over a field, how do I prove that $I = J$ if $\operatorname{in}_<(I)=\operatorname{in}_<(J)$?

Question

If $I\subseteq J$ are ideals in a polynomial ring of $n$ variables, how do I prove that $I = J$ if $\operatorname{in}_{\lt}(I)=\operatorname{in}_{\lt}(J)$, where $\lt$ is any monomial ordering?

Obviously it suffices to prove that $J \subseteq I$. I'm stuck with how to go forward once I pick an arbitrary element $f \in J$ and have that $\operatorname{in}_{\lt}(f) \in \operatorname{in}_{\lt} (J)=\operatorname{in}_{\lt}(I)$.

Answer 1

Proposition 2.2.6 of the book "Monomial Ideals" by "Herzog-Hibi"
If $I\neq J$ then $f\in J\setminus I$. Let $f'$ be the remainder of $f$ with respect to a Grobner basis of $I$. Then $f'\neq 0$, $f'\in J$ and $supp f' \nsubseteq in_<(I) $. So $ in_<f' \in in_<(J)-in_<(I)$

Answer 2

There is no need to use Gröbner bases, Dickson's lemma is enough.

Suppose there is $f\in J\setminus I$ and choose $f$ with $\operatorname{in}(f)$ minimal. Then $\operatorname{in}(f)\in\operatorname{in}(J)=\operatorname{in}(I)$, so there is $g\in I$ such that $\operatorname{in}(f)=\operatorname{in}(g)$. We have $\operatorname{in}(f-g)<\operatorname{in}(f)$, and $f-g\in J\setminus I$, a contradiction.