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

[Math Amateur]

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]

 

[Opalg]

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 $a-b\in L$.