1. ## Grobner Bases

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

I have a problem understanding a step in the proof of Proposition 24, Page 322 of D&F

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 $r \in I$ ... ... etc etc

My question: How do we know $f \in I \Longrightarrow r \in I$?

Would appreciate some help

Peter

2. ## Re: Grobner Bases

Originally Posted by Bernhard
/snip/

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 $r \in I$ ... ... etc etc
My question: How do we know $f \in I \Longrightarrow r \in I$?
Since $g_i \in I$, using the definition of an ideal, we have $q_i g_i \in I$. Since an ideal is an additive group, and $f \in I$ we conclude $f - \sum q_i g_i \in I$.

3. ## Re: Grobner Bases

Thanks for the help

Peter