MHB Proving L is an Ideal - Part of the Proof of Hilbert's Bass Theorem

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The discussion focuses on the proof of Hilbert's Basis Theorem as presented in Dummit and Foote's work, specifically regarding the ideal L formed by the leading coefficients of elements in an ideal I of R[x]. The proof establishes that if a and b are in L and r is in R, then the expression ra - b is also in L. A key question arises about how this implies that both a - b is in L and ra is in R. The clarification provided indicates that by setting b to 0, it follows that ra is in L, and by setting r to 1, it shows that a - b is in L. This reasoning confirms that L meets the criteria to be considered an ideal.
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I am reading Dummit ad Foote's proof of Hilbert's Basis Theorem (See attached for the theorem and proof)

In the proof I is an ideal in R[x] L is the set of all leading coefficients of elements of I

D&F then proceed to prove that L is an ideal of R

Basically they establish that if elements a and b belong to L and r belongs to L then ra - b belongs to L.

D&F claim that this shows that L is an ideal but for an ideal we need to show that for a, b \in L and r \in R we have:

a - b \in L and ra \in R

My question is how exactly does ra - b \in L \Longrightarrow a - b \in L and ra \in R??

Peter[This has also been posted on MHF]
 
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Peter said:
I am reading Dummit ad Foote's proof of Hilbert's Basis Theorem (See attached for the theorem and proof)

In the proof I is an ideal in R[x] L is the set of all leading coefficients of elements of I

D&F then proceed to prove that L is an ideal of R

Basically they establish that if elements a and b belong to L and r belongs to L then ra - b belongs to L.

D&F claim that this shows that L is an ideal but for an ideal we need to show that for a, b \in L and r \in R we have:

a - b \in L and ra \in R

My question is how exactly does ra - b \in L \Longrightarrow a - b \in L and ra \in R??
If you know that $ra-b\in L$ whenever $a,b\in L$ and $r\in R$ then in particular this will hold when $b=0$, so that $ra\in L$; and also when $r=1$ (the identity element of $R$) so that $a-b\in L$.
 
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