Re: R[X,Y] polynomial ring
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 24 May 2005 16:21:11 +0000 (UTC)
In article <d6vh85$ppr$1@xxxxxxxxxxxxxxxxxx>,
Mike <Mike_jones@xxxxxxxxx> wrote:
>I want to show that R[x,y] is not a principal ideal domain,
>So i'm trying to construct an ideal which is not principal
Fair enough.
>i do not fully understand what my text book means by:
>
>R[X,Y]
>Let I=(x,y)={XP(X,Y) + YQ(X,Y)}
>={a_0 + a_1X _ a_2X^2+......... : a_0=0}
>( i dont understand why a_0=0?)
I'm not sure what their notation is, but:
The elements of R[X,Y] are polynomials in X and Y. They can be
expressed in many ways, for example, as:
a_0 + a_1X + a_2X^2 + ... + a_nX^n + b_1Y + b_2 XY + b_3 X^2Y + ....
etc.
The ideal in question, which we will prove is nonprincipal, is the
ideal I generated by X and Y. The elements of I are all elements of
the form X*P(X,Y) + Y*Q(X,Y), where P and Q are arbitrary elements of
R[X,Y]. Clearly, neither X*P(X,Y) nor Y*Q(X,Y) have constant terms, so
if you express an arbitrary element as
a_0 + a_1X + a_2X^2 + ...
then a_0, the constant term, must be equal to 0. And if the constant
term is 0, then we can take all factors that have an X in them, and
factor it out, and take all remaining factors (which must be monomials
in powers of Y) and factor out a Y, and thus express the polynomial as
an element of I.
So I is exactly the set of polynomial in X and Y which have constant
term equal to 0.
>then
>X + Y +X^3 + XY^14
"Then" what?
>If I=R[X,Y]s(x,y)
>X,Y elements of I
>s(x,y)|x and s(x,y)|y
>
>0 does not equal s(x,y)=a_0
>s(x,y) element of I Contradiction
>therefore Not PID
>
>If anyone can shed light on understanding this proof, or examples why R[X,Y]
>is not a PID it with be greatly appreciated
We want to show that I is not principal. That is, we want to show
that there does not exist a polynomial s(X,Y) in two variables such
that I = (s(X,Y)) = R[X,Y]s(X,Y) (that is, "all multiples of s(X,Y)").
(By the by, it is bad form to use lower case and upper case letters
for variables interchangeably).
We proceed by contradiction: assume such a polynomial s(X,Y)
exists. Then, since X and Y are clearly elements of I, both X and Y
must be multiples of s(X,Y); that is, s(X,Y) must divide X and must
divides Y. Since s(X,Y) must divide X, its degree is Y must be 0, and
its degree in X must be at most 1. Since s(X,Y) divides Y, by a
symmetric argument we know the degree in X is 0 and the degree in Y is
at most 1. Therefore, s(X,Y) must be a constant polynomiall,
s(X,Y)=a_0, for some a_0 in R.
However, we also know that the elements of I have zero constant
term. Since s(X,Y)=a_0 lies in I, it follows that its constant term,
namely a_0, must be equal to 0. That is, s(X,Y)=0. However, 0 does not
divide X and does not divide Y, leading to a contradiction (or,
alternatively, s(X,Y) cannot be the zero polynomial a priori, hence
a_0 is not zero, hence s(X,Y) is not in I, contradicting that it was a
->generator<- for I).
Thus, I cannot be principal, and so R[X,Y] cannot be a principal ideal
domain.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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