Let $R=K[x_1,...,x_n]$ and $I$ be an ideal of $R$, $K$ being a field
Given $h\in I$, $g\in \sqrt{I}$ and $f\in\sqrt{I}$
Where $in_<(f)=in_<(h)$ and $g=f-h$. So $in_<(g) < in_<(f)=in_<(h)$ w.r.t the term order $<$
How can I show that $g\in I$
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Are you sure what you want to show is true? Take $K[x,y]$ and $<$ to be lex order with $ x>y$. Now let $I=\langle x^2,y^2\rangle$. Then $\sqrt{I}=\langle x,y\rangle$. Now take $h=x^2+y^2\in I$ and $f=x^2+y\in\sqrt{I}$. Then $g=f-h\in\sqrt{I}$ but $g\not\in I$. – Casteels Mar 03 '15 at 12:41
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what if I add the constraint that $in_<(I)$ is radical. Is it possible for $g\in I$ – Andrew Brick Mar 03 '15 at 13:19
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If $in_<(I)$ is radical, then $I$ is radical, so yes. – Casteels Mar 03 '15 at 13:21
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As written, perhaps the OP meant that $h\in\sqrt{I}$ and $f\in\sqrt{I}$ and let $g=f-h$. In that case, the problem would be to show that $g\in\sqrt{I}$.
Since $f\in\sqrt{I}$, there is some $m$ such that $f^m\in I$ and since $h\in\sqrt{I}$, there is some $n$ such that $h^n\in I$. Then, consider $g^{n+m-1}$. In this case, you have that $$g^{n+m-1}=\sum_{i=0}^{n+m-1}\begin{pmatrix}n+m-1\\i\end{pmatrix}(-1)^if^{n+m+1-i}h^i.$$
Notice that for all choices of $i$, either $i\geq n$ or $n+m-i\geq m$. In these cases, the RHS is either a multiple of $h^n\in I$ or $f^m\in I$. Therefore, $g^{n+m+1}$ is a sum of elements of $I$ and is in $I$. Hence $g\in\sqrt{I}$.
Michael Burr
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