## Grobner Bases - Second question on D&F Proposition 24

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

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

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

(2) The ideal I has a Grobner Basis

The proof of Proposition 24 begins as follows:

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

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

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

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

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

... ... ...

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

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

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

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

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

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

Can someone please clarify this issue?

Peter