I am reading Dummit and Foote Section 9.6 Polynomials in Several Variables Over a Field and Grobner Bases.

I have a second problem (see previous post for first problem) understanding a step in the proof of Proposition 24, Page 322 of D&F

Proposition 24 reads as follows:

Proposition 24. Fix a monomial ordering on  R= F[x_1, ... , x_n] and let I be a non-zero ideal in R

(1) If  g_1, ... , g_m are any elements of I such that  LT(I) = (LT(g_1), ... ... LT(g_m) )

then  \{ g_1, ... , g_m \} is a Grobner Basis for I

(2) The ideal I has a Grobner Basis

The proof of Proposition 24 begins as follows:

Proof: Suppose  g_1, ... , g_m \in I with  LT(I) = (LT(g_1), ... ... LT(g_m) ) .

We need to see that  g_1, ... , g_m generate the ideal I.

If  f \in I use general polynomial division to write  f  = \sum q_i g_i + r where no non-zero term in the remainder r is divisible by any  LT(g_i)

Since  f \in I , also  r \in I , which means LT9r) is in LT(I).

But then LT(r) would be divisible by one of  LT(g_1), ... ... LT(g_m) , which is a contradiction unless r= 0 ... ... etc etc

... ... ...

My problem is with the last (bold) statement - as follows:

We have that  LT(I) = (LT(g_1), ... ... LT(g_m) )

Now since  r \in LT(I) , then we have that r is a finite sum of the form

 r = r_1 LT(g_1) + r_2 LT(g_2) + ... ... + r_m LT(g_m) ... ... ... (*)

where  LT(g_i) \in I and  r_i \in R

But surely (*) is NOT guaranteed to be divisible by  LT(g_i) for some i

Can someone please clarify this issue?