Let $A$ be a commutative, Noetherian ring and let us define a monomial ordering, $\prec$ on $A[x_1, \ldots,x_n]$. My doubt is regarding the maximal chain of prime ideals in $A[x_1, \ldots,x_n]$. When we look at any one of the generators, $f$ of a prime ideal $P_i$ in $A[x_1, \ldots, x_n]$ such that $f\notin A$, is it enough to consider $f$ such that $\mathrm{lc}(f) = 1$? Will the coefficients increase the length of a chain of prime ideals?
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Why should a general prime ideal have one generator $f$? – Mohan Sep 03 '15 at 16:02

I meant if each of them is monic will we get a maximal chain. Can the proof go along the lines : if non monic generators give rise to a maximal chain this implies there exist two primes $P_i \subsetneq P_{i + 1}$ such that their leading monomial ideal is the same but not the leading coefficient ideal? This in turn means their contraction would be the same prime ideal in $A$ which is a contraction since that will be three prime ideals lying over the prime ideal in $A$. – Zoey Sep 03 '15 at 17:02

Monic makes sense only for polynomials in one variable. Are you suggesting that we treat one of the variable as special? – Mohan Sep 03 '15 at 17:04

Yes, based on some monomial ordering we order the monomials and therefore we have accordingly leading monomial, coefficient etc. – Zoey Sep 03 '15 at 17:15

My question is how to show that coefficients do not increase the length of the chain of prime ideals. – Zoey Sep 03 '15 at 17:16