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

as $\displaystyle \alpha,\alpha^{\prime}\in k$, $\displaystyle g,g^{\prime}\in I$ we have then that

$\displaystyle (\alpha+g)+(\alpha^{\prime}+g^{\prime})\in A_{1}$ and

$\displaystyle (\alpha+g)\cdot(\alpha^{\prime}+g^{\prime})\in A_{1}$

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

3) Hence we may construct an infinite chain of ideals

$\displaystyle (X)\subset (X,XY)\subset (X,XY^2)\subset......$

this chain never stops $\displaystyle XY^{k+1}\not\in (X,XY^2,......XY^k)$

as

$\displaystyle XY^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(X,Y)XY^{ i}$ where all $\displaystyle f_{i}\in A_{1}$

$\displaystyle \Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,Y)Y^{i}$

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

Here are my questions and doubts....

Part 2) How do elements of $\displaystyle I=(X)=AX$ look like?

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

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

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

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

Could someone explain this solution to me step by step please?

Thanks

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 A_{1} is add constant terms (from k) to these polynomials. it should be clear for f(X,Y) in A_{1}, 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 A_{1}, 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 A_{1}. 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 A_{1}.

the same sort of reasoning shows XY^{2} 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 A_{1}.

but maybe XY^{2} = f_{1}(X,Y)X + f_{2}(X,Y)XY for some f_{1},f_{2} in A_{1}? we need to rule this out, too.

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

Y^{2} = f_{1}(X,Y) + f_{2}(X,Y)Y

now the left hand side doesn't have any X's at all that is if: h(X,Y) = Y^{2}, then h(0,Y) = Y^{2} as well. so we evaluate both sides at (0,Y):

Y^{2} = f_{1}(0,Y) + f_{2}(0,Y)Y = a_{1} + a_{2}Y (for some a_{1}, a_{2} in k).

the LHS has degree 2, the RHS has degree 1.

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

thank you very much,that really helped me :)