2

Let $R = F[x,y]/(ax^2+by^2-1)$, where $F$ is a field with $\text{char}(F) \neq 2$ and $a, b \neq 0$. Then $R$ is Dedekind.

My attempt:

$R$ is Dedekind if and only if it is Noetherian, integrally closed, and if every prime ideal is maximal, so it suffices to show these conditions.

  1. Noetherian: I wasn't able to explicitly prove this, but I'm pretty sure that every polynomial in $F[x,y]$ of degree greater than 3 can be expressed as the sum of factors of $ax^2+by^2-1$ and polynomials of degree at most 2, so that $R$ consists of polynomials of degree at most 2. If this is true, then the possible non-units are the possible sums and products of $x, y, x^2, y^2$, and since these are the possible generators of nonzero proper ideals, every ideal is finitely generated, so $R$ is noetherian.

  2. Integrally closed: My idea was to show the stronger condition that $R$ is a UFD. Since $ax^2+by^2-1$ is irreducible over $F[x,y]$, $R$ is a domain. Since $ax^2 + by^2 = 1 \in R$, $f \in R^x$ if and only if $f \mid ax^2 + by^2$. But when I tried to use this to show that irreducible elements are prime, I got stuck, and I'm no longer sure $R$ is a UFD, but I can't think of a counterexample either.

  3. Every nonzero prime ideal is maximal: Let $P$ be a nonzero prime ideal of $R$. Then, there exists a nonzero ideal $P' \subset F[x,y]$ such that $P'/(ax^2+by^2-1) \cong P$, where this projection comes from the first isomorphism theorem and the canonical projection $\phi: P' \rightarrow P$, where $\text{ker}(\phi) = (ax^2+by^2-1)$. Since the preimage of a prime ideal is also a prime ideal, $P'$ is a prime ideal, and since $$R/P \cong \left(F[x,y]/(ax^2+by^2-1) \right) / \left(P' / (ax^2+by^2-1) \right) \cong F[x,y] / P'$$ by the third isomorphism theorem, it suffices to show that $P'$ is maximal. I feel like $P'$ should be maximal in $F[x,y]$ since $P'$ must contain polynomials of degree at most two, and since these polynomials are "small", the ideal containing them should be "large", but I'm stuck trying to turn this intuition into something formal.

user26857
  • 52,094
Sam Birns
  • 236

3 Answers3

1

To show that $R$ is a UFD is not a good idea. For instance, if $F=\mathbb R$, and $a=b=1$ this is not true; see here.

Instead, you can prove that $R$ is integrally closed. This follows from a general result which I proved here. See also the (solved) exercise 4.H from these notes.

For $\dim R=1$ you have to recall that $\dim F[X,Y]=2$.

Community
  • 1
user26857
  • 52,094
  • Thanks for the link. Am I on the right track in my attempt at showing $R$ is noetherian? – Sam Birns Jan 28 '16 at 08:45
  • 1
    @sam Since $F$ is a field $F[x,y]$ is noetherian by Hilbert's basis theorem. Then it's quotient by an ideal is noetherian too. – basket Jan 28 '16 at 09:02
1

This ring is clearly noetherian (as a quotient of a polynomial ring) and a $1$-dimensional domain, since we divide a principal prime ideal out of a two-dimensional domain.

For the last step, we can pass over to geometry:

For a one-dimensional noetherian domain, integrally closed is the same as regular, and we can test this with partial derivatives. Let $f=ax^2+by^2-1$, then $\partial_x f = 2ax, \partial_y f=2by$ and both vanish iff $x=y=0$ but the point $(0,0)$ is not contained on this ellipsoid, hence our ring is regular and thus integrally closed.

MooS
  • 31,390
0

Try to follow this hint...

$(i)$ $R$ is Noetherian since it is quotient of a Noetherian domain, i.e. the polynomial ring $F[x,y]$. Recall that this last fact follows from Hilbert's Basis Theorem.

$(ii)$ The field of fractions of $R$, which we denote by $k$, is given by $k=F(x)[y]/(ax^2+by^2-1)$, therefore you can check that $R$ is integrally closed since $F[x]$ it is.

$(iii)$ The canonical inclusion $F[x]\hookrightarrow R$ is integral, therefore $\dim R=\dim F[x]=1$, where $\dim$ is the Krull dimension.

Remark If $f\colon A\longrightarrow B$ is injective and integral then $\dim A=\dim B$, this follows by Going-Up Theorem.

user26857
  • 52,094
Vincenzo Zaccaro
  • 3,012
  • $(ii)$ : for convenience, omitting quotient notation, let $\sum f_n(x)y^n\in k$ be integral over $R$ where $f_n(x)\in F(x)$. To show it is in $R$, it suffices to show $f_n(x)\in R$ since integral closures are rings and $y\in R$. Let $f_n(x)^m+\sum g_i(x,y)f_n(x)^i=0$ where $g_i(x,y)\in F[x,y]$. By arranging terms with respect to $y$, we get $f_n(x)^m+\sum h_j(x)f_n(x)^j=0$ where $h_j(x)\in F[x]$.(in $F(x)[y]$, by having each coefficient of $y^r$ equal to $0$, the above polynoial is constant in $F(x)[y]$.) Since $F[x]$ is a UFD, it is normal so $f_n(x)\in F[x]\subset F[x,y]$. –  Aug 23 '19 at 22:00
  • How do you claim that $F[x] \to R$ is integral?. When $F=\mathbb{C}$, then $R \cong \mathbb{C}[x,\frac{1}{x}$ and it is easy to see that the element $\frac{1}{x}$ is not integral over $\mathbb{C}[x]$ – permutation_matrix May 28 '21 at 12:28