A question about initial ideals.

Question

Let $R = k[x_1, \dots, x_m]$ be a polynomial ring over a field $k$ and $I, J$ be ideals of $R$. Further assume that $J$ is generated by the polynomials $f_1, \dots, f_r$. Fix a monomial order $<$ on the monomials of $R$. Is it true that, with respect to the fixed order, $LT (I + J') \subset LT (I + J)$, where $J'$ is generated by the initial terms of $f_1, \dots, f_r$ and $LT (I)$ denote the usual initial ideal of $I$ with respect to $<$ for any ideal $I$?

Answer 1

No, this is not true, even in $k[x]$. Let $I = J = (1 +x)$. Then $J' = (x)$. We have $I + J' = k[x]$, and $I + J = (1+x)$.

The opposite inclusion also does not usually hold. Let $I = (x)$ and $J = (1+x)$. Then $I + J' = (x)$ and $I + J = k[x]$

Can you find an example where $LT(I +J')$ and $LT(I + J)$ are incomparable?