This is an exercise in "Using Algebraic Geometry" by Cox, et al.
Let $V=V(f_1,\dots,f_s)$ be a variety containing the origin. Let $G=\{g_1,\dots,g_t\}$ be a standard basis for $I=\langle f_1,\dots,f_s\rangle k[x_1,\dots,x_n]_{\langle x_1,\dots,x_n\rangle}$ with respect to a degree-anticompatible order $>$. Show that $V(g_{1,min},\dots,g_{t,min})$ is the tangent cone of $V$ at the origin.
The tangent cone of $V$ at the origin is defined to be $$C(V)=V(f_{min}: f\in I(V)),$$ where $f_{min}$ is the homogeneous component of the lowest degree of $f$.
My attempt: It is easy to show the direction $\subseteq$. Let $u\in C(V)$. Then $f_{min}(u)=0$ for all $f\in I(V)$. Since $g_i\in I$, it is obvious that $g_i\in I(V)$. So $g_{i,min}(u)=0$. This proves the inclusion $\subseteq$. For the other direction, let $u\in V(g_{1,min},\dots,g_{t,min})$ so that $g_{i,min}(u)=0$ for all $i$. Let $f\in I(V)$ so that $f^m\in I$ for some $m$. We claim that if $g\in I$, then $g_{min}\in\langle g_{1,min},\dots,g_{t,min}\rangle$. If the claim is true, then $(f^m)_{min}(u)=0$. Then the proof is done.
My question:
The field $k$ has to be algebraically closed for the bold part to be true. I cannot see how to proceed if it is not.
To show the claim, I tried to use the following argument:
Let $S=\{\text{LT}(f_{min}): f_{min}\not\in \langle g_{1,min},\dots,g_{t,min}\rangle\}$ and choose the minimal element. If the minimal element exists, I can finish the proof. But for a degree-anticompatible order, I cannot see that the minimal element exists. The degree can be infinitely large so there is no lower bound. What should I do?
Thank you for any help!