Understanding D&F Prop. 24: How Do We Know f in I => r in I?

[Math Amateur]

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 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 BasisThe 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

[This has also been posted on MHF]

 

[jvicens]

Dear Peter,

Thank you for your question regarding Proposition 24 in Dummit and Foote. I can understand why this step in the proof may be confusing. Let me try to clarify it for you.

First, let's recall the definition of an ideal. An ideal is a subset of a ring (in this case, the polynomial ring R) that is closed under addition and multiplication by elements of the ring. In other words, if f and g are in I, then f+g and fg must also be in I.

Now, let's look at the statement in the proof: "If f \in I, then r \in I." This is saying that if we can show that r is in I, then we have shown that the set {g_1, ..., g_m} generates the ideal I. Why is this the case?

Well, we know that f is in I. And we have assumed that g_1, ..., g_m generate I. This means that any element in I can be written as a linear combination of g_1, ..., g_m. So, if we can show that r is in I, then we can write it as a linear combination of g_1, ..., g_m. And since we already know that f can be written as a linear combination of g_1, ..., g_m, we can also write r as a linear combination of g_1, ..., g_m. Therefore, r is in I.

I hope this explanation helps to clarify the step in the proof that you were struggling with. If you have any further questions, please don't hesitate to ask.