11

Is there a way to prove that any subring $R$ of the polynomial ring over a field $k$ such that $k\subset R$ is Noetherian without appealing to integral extensions, Eakin-Nagata, etc.?

The reason I ask is because I found this as an exercise 15.1.9 in Dummit and Foote, which well precedes any discussion of integral extensions.

It says to show that if $y\in R\setminus k$, then $k[x]$ is f.g. as a $k[y]$-module. Why is that? Since $k[y]$ is Noetherian, $k[x]$ is a Noetherian $k[y]$-module, but how does this show $R$ is Noetherian?

user26857
  • 53,190
Sanaa
  • 111

2 Answers2

8

Let $y \in k[x]$ be a non-constant polynomial.

Claim: $k[x]$ as a $k[y]$-module is generated by the polynomials $1,x,\ldots ,x^{d-1}$, where $d=\deg (y)$.

Proof: let $f\in k[x]$ be a polynomial of degree $n\geq d$. Let $e$ be the maximal integer such that $n-de\geq 0$. By the division-with-remainder algorithm one has $f=qy^e+r$, with $\deg (q)<d$ and $\deg (r)<de$ or $r=0$. Applying the same procedure to $r$ if necessary after finitely many steps leads to a linear combination $f=a_0+a_1x+a_2x^2+\cdots +a_{d-1}x^{d-1}$ with $a_j\in k[y]$.

If $y \in R$, then every ideal $I$ of $R$ is a submodule of the $k[y]$-module $k[x]$. If the latter is noetherian, then $I$ is noetherian too. In particular it is finitely generated as a $k[y]$- module and thus as an $R$-module.

darij grinberg
  • 19,173
Hagen Knaf
  • 9,387
  • 3
    Note: This claim is the polynomial analogue of the classical fact that if $b > 1$ is an integer, then any nonnegative integer can be written in the form $a_0 b^0 + a_1 b^1 + a_2 b^2 + \cdots + a_n b^n$ with $a_0, a_1, \ldots, a_n \in \left{0,1,\ldots,b-1\right}^{n+1}$. (Also known as base-$b$ representation.) – darij grinberg Oct 29 '19 at 18:27
1

An elementary messy argument:

Take an ideal I of R, take an element in I of minimum degree $p_I(x)$. Now in general it is not a generator (you can easily find examples with $K[x^2,x^5]$ and the ideal generated by $x^2,x^5$). But if $q(x)$ has minimal degree too, then there is $a \in K$ with $ap_I(x)-q(x)$ with lesser degree, so we have $ap_I(x)-q(x)=0$. So there is just a line of elements of minimal degree. Suppose now you enlarge I to I', putting inside some q(x). Suppose that $\deg q(x)>\deg p(x)$. Take in $I'-I$ a minimal in degree such q. Now you're allowed to take powers of $p$, so we have that $\deg p <\deg q <2\deg p$. So repeat this operation, taking in $I''$ (supposing always that p remains his lower in degree element), and taking q'' you get that $\deg p<\deg q''<2\deg p$ and moreover his remainder mod $\deg p$ is different from $\deg q$ (otherwise by linear combination as before you get a lesser term in degree). So if you iterate, after at most $\deg p$ steps you get an ideal $I^{n}$ with the term of lowest degree lesser then I. Iterating the reasoning you get to conclusion.

Hope it helps (and it works).

Community
  • 1
Donald
  • 224