A subring of a Noetherian ring need not be Noetherian-problem

• Dec 7th 2012, 01:22 AM
rayman
A subring of a Noetherian ring need not be Noetherian-problem
I have a problem with understanding the following solution to the problem:
A subring of a Noetherian ring is Noetherian? True or false?

Solution
let $A=k[X,Y]$ where k is a field, then A is Noetherian by Hilbert's basis theorem.
Now we want to construct a subring and then show that a sequence of ideals of this subring never becomes stationary thus the subring will not be Noetherian.

1)We start with letting $A_{1}=k+I$ where $I=(X)=AX$ then $A_{1}$ is a subring of $A=k[X,Y]$

as $\alpha,\alpha^{\prime}\in k$, $g,g^{\prime}\in I$ we have then that
$(\alpha+g)+(\alpha^{\prime}+g^{\prime})\in A_{1}$ and
$(\alpha+g)\cdot(\alpha^{\prime}+g^{\prime})\in A_{1}$

2) Important observation. If $f=\alpha+g\in k+I=A_{1}$ then $f(0,y)=\alpha=(0,0)$ as $g(0,y)=0$ and $y,y^{2},.....\not\in A_{1}$

3) Hence we may construct an infinite chain of ideals
$(X)\subset (X,XY)\subset (X,XY^2)\subset......$
this chain never stops $XY^{k+1}\not\in (X,XY^2,......XY^k)$
as
$XY^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(X,Y)XY^{ i}$ where all $f_{i}\in A_{1}$

$\Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,Y)Y^{i}$
$\Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,0)Y^{i}$ (contradition if they were contained).

Here are my questions and doubts....

Part 2) How do elements of $I=(X)=AX$ look like?
are they polynomials generated by element $X$ and then multiplied by any element from the ring? f ex let's take $p(X,Y)=3X-4XY\in k[X,Y]$ then $g=X(3X-4XY)=3X^2-4X^2Y\in I$ Right?

If we further assume that $k=\mathbb{R}$ we can take for ex $\alpha=5$ then if $f=\alpha+g=5+3X^2-4X^2Y$ then $f(0,y)=5=\alpha=f(0,0)$ and $g(0,y)=0$

but based on what do they conclude that $y,y^2,y^3,......\not\in A_{1}$??? is it because of the def of $A_{1}$ that it is a ring all polynomials that are all divisible by $X$??? and these ''alone'' terms with powers of y are not divisible by $X$

3) what does the last part of the solution say?

Could someone explain this solution to me step by step please?
Thanks
• Dec 7th 2012, 04:34 AM
Deveno
Re: A subring of a Noetherian ring need not be Noetherian-problem
your ideal of what AX looks like is essentially correct: every p(x,y) in AX factors as X(g(X,Y)) for some g(X,Y) in k[X,Y].

so all we do in A1 is add constant terms (from k) to these polynomials. it should be clear for f(X,Y) in A1, f(0,Y) is just an element of k (anything with a Y in it, also has an X with it, so setting X = 0 wipes these all out).

note that the polynomial Y in k[X,Y] does not have a constant term, so for it to be in A1, it would have to be in I = (X). but Y has no X terms, either (neither does any power of Y).

now let's look at why XY is not in (X): because if h(X,Y) is in (X), then h(X,Y) = Xf(X,Y) for some f(X,Y) in A1. but this means for h(X,Y) = XY, that f(X,Y) = Y (we can "cancel the X's" since k is a field, so k[X,Y] is an integral domain). and Y isn't in A1.

the same sort of reasoning shows XY2 isn't in (X,XY): we already know it's not in (X), and if it were in (XY), we could cancel a factor of XY to get Y in A1.

but maybe XY2 = f1(X,Y)X + f2(X,Y)XY for some f1,f2 in A1? we need to rule this out, too.

well, first we "cancel the X's" on both sides, and we get:

Y2 = f1(X,Y) + f2(X,Y)Y

now the left hand side doesn't have any X's at all that is if: h(X,Y) = Y2, then h(0,Y) = Y2 as well. so we evaluate both sides at (0,Y):

Y2 = f1(0,Y) + f2(0,Y)Y = a1 + a2Y (for some a1, a2 in k).

the LHS has degree 2, the RHS has degree 1.
• Dec 8th 2012, 06:06 AM
rayman
Re: A subring of a Noetherian ring need not be Noetherian-problem
thank you very much,that really helped me :)