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I am trying to answer this question:

Let $k$ be a field and $k[x,y] \cong k^{[2]}.$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.

And I got the following hint:

Hint: Consider the ideal $I = (x, xy, xy^2, xy^3, ...).$ Assume $xy^{n+1}= f_{0}x + f_1xy + \dots + f_n xy^n$ for $f_i \in A.$ Divide by $x$ and evaluate at $x=0.$}

But I have the following questions:

1- What are the elements of this ideal $I = (x, xy, xy^2, xy^3, ...)$ look like in general? what confuses me while trying to write an element of this ideal, is that this ideal is generated by the following infinite set $x, xy, xy^2, xy^3, ...$, could anyone help me in figuring this out please?

  • I think that elements of that ideal are all polynomials with 0 coefficients along y^1,y^2,y^3,.... – robin3210 Dec 06 '20 at 18:49
  • @MaciejFicek why do you think that? and could you please express your idea mathematically? –  Dec 06 '20 at 18:55
  • every element of that ideal is of the form a_1el_1 +a_2el_2 ... +a_k*el_k ewhere el_1, ..., el_k are from generators of the ideal and a_1,...,a_k are from ring. All generators of ideal have x, so it is impossible to obtain a summand with only power of y. It is easy to see that all other polynomials can be obtained. – robin3210 Dec 06 '20 at 19:04
  • @MaciejFicek so $a_1, ... , a_k$ are polynomials with coefficients in $k$ and independent variables $x, xy, xy^2, ... $? if so can you express them mathematically to me please? –  Dec 06 '20 at 19:42
  • Which ring are $a_1, a_2, ....,a_k$ from? do you mean they are elements from the field $k$?@MaciejFicek –  Dec 07 '20 at 10:22

1 Answers1

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Let $K[x,y] = R$ for convenience. So you know that any element of $I$ is a linear combination of any of the generators, which as you rightly point out there are infinitely many of them. Nonetheless, they are very regular, so we can get some idea of what elements of this ideal look like. For example, it is clear that any element of $I$ must have degree in $x$ of at least $1$, since $x$ divides every generator. We can also see that there are no elements which have pure powers of $y$, since no such elements appear among the generators. These two facts tell you that the elements of $I$ are precisely polynomials with zero constant term and no terms composed just of powers of $y$ (including of course $0$, which is in every ideal).

To make this a bit more formal, we can write any element in $I$ as an $R$ linear combination of the generators, i.e. $$\forall f(x,y) \in I, f(x,y) = f_1(x,y)x + f_2(x,y)xy+f_3(x,y)xy^2+...+f_n(x,y)xy^n = x(f_1(x,y) + f_2(x,y)y+f_3(x,y)y^2+...+f_n(x,y)y^n)$$ where $f_1(x,y),...,f_n(x,y)$ are polynomials from the original ring $R$. In this form, one can see that there can be no non-zero constant terms and no pure powers of $y$, since each term in the sum is divisible by $x$

Noah Solomon
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  • why zero is the constant term in this type of ideals? –  Dec 06 '20 at 19:34
  • can you express the elements of this ideal mathematically then? –  Dec 06 '20 at 19:37
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    @Confusion I edited my answer to try and provide some more clarity, does this clear up your question? – Noah Solomon Dec 06 '20 at 22:56
  • what is the ring $R$? –  Dec 07 '20 at 00:32
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    @Confusion I just mean the original polynomial ring, $K[x,y]$ – Noah Solomon Dec 07 '20 at 00:42
  • So that brings the question I asked here https://math.stackexchange.com/questions/3937679/are-the-elements-of-this-ideal-i-x-xy-xy2-xy3-polynomials-with-p to my mind. Do you have any answer to it? –  Dec 07 '20 at 00:44
  • Also, I want to clarify this in my mind , is not $I$ subring of $A$ which is subring of $k[x,y]$? –  Dec 07 '20 at 00:46
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    Yes, $I$ is a subring of $A$ which is a subring of $K[x,y]$ – Noah Solomon Dec 07 '20 at 00:48
  • Ok. so could you please take a look at my question in the link? –  Dec 07 '20 at 00:50
  • @Confusion I think the other user gave a sufficient answer to that question in the other page. Also it is not good practice to post the same question multiple times, especially on the same day. – Noah Solomon Dec 07 '20 at 01:17
  • I am not asking about the same thing in the other question. Also, the answer given there does not actually answers my question even though it clarified many things in my mind. –  Dec 07 '20 at 01:22
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    If $f_i(x)$ are polynomials from the original ring .... I think they should be functions of $y$ as well i.e. $f_i(x,y)$ .... do you agree? –  Dec 07 '20 at 10:29
  • @Math yeah you are totally right, I edited my answer to reflect that – Noah Solomon Dec 07 '20 at 16:24