What is the best way to understand that $D:=\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is a Dedekind domain?

I first noticed that $X^2+Y^2-1$ is irreducible in $\mathbb{Q}[X,Y]$ since it is $Y-1$ Eisenstein in $\mathbb{Q}[Y][X]$. It follows that $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is an integral domain. By Hilbert Basis theorem, $\mathbb{Q}[X,Y]$ is Noetherian, so this quotient is too.

I know that a Dedekind domain is precisely a Noetherian integral domain which is integrally closed in its fraction field, and has Krull dimension $1$. However, computing the fraction field and showing $D$ is integrally closed in it seems quite difficult, and showing the Krull dimension is 1 also seems difficult.

I'm aware of another result that a Noetherian integral domain is Dedekind domain if the localization at every prime is a discrete valuation ring. I think the prime ideals of $D$ are precisely the canonical images of the prime ideals containing $(X^2+Y^2-1)$ in $D$. But I'm stuck trying to get a general handle on $D_P$ and seeing it is a DVR. What is the best way to see this claim (preferably algebraically, not geometrically)? Thanks.

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Tiffany Hwang
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  • It's not difficult at all to show that the Krull dimension of this ring is one. – user26857 Apr 08 '14 at 10:54
  • @user26857 May I ask how? I guess I don't know the method. – Tiffany Hwang Apr 08 '14 at 14:51
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    $\dim\mathbb Q[X,Y]=2$ and therefore any prime ideal strictly containing $(X^2+Y^2-1)$ is necessarily maximal. – user26857 Apr 08 '14 at 21:04
  • If you want to prove that $D_P$ is a DVR, then you are already in the field of algebraic geometry since DVR = regular in dimension one and this is usually tested by using the Jacobian criterion. For a purely algebraic approach you should prove that $D$ is an integrally closed integral domain of dimension one. The argument for dimension is in my comment above and $D$ integrally closed it's proved now in [this topic](http://math.stackexchange.com/questions/745960/mathbbqx-y-x2y2-1-is-integrally-closed) via the isomorphism $\mathbb{Q}[X,Y]/(X^2+Y^2-1)\cong \mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]$. – user26857 Apr 09 '14 at 09:33

3 Answers3


The curve $f(x,y)=x^2+y^2-1=0$ is smooth by the jacobian criterion, just as in calculus: at each of its points one at least of the partial derivatives $\partial f/\partial x=2x$ or $\partial f/\partial y=2y$ is nonzero.
And a smooth affine curve has as coordinate ring a Dedekind ring. That's all (if you use geometry).

Georges Elencwajg
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  • Thanks Georges. I was hoping to see an algebraic proof, since I've studied algebra, but not geometry in any satisfactory sense, so I'd have a better chance of understanding an algebraic answer. I've no doubt your answer is elegant and correct, but "a smooth affine curve has as coordinate ring a Dedekind ring" makes little sense to me. – Tiffany Hwang Apr 07 '14 at 23:58
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    Dear Tiffany: first of all, thank you for your thoughtful comment. I more or less guessed what you explain in that comment. Yet I posted that geometric answer in the hope that it might at some point encourage you (or other users) to follow a course on introductory algebraic geometry. Once you know the basic jargon you will see that the arguments above are not so frightening as they appear! – Georges Elencwajg Apr 08 '14 at 00:11
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    In that case, do you have a reference for where it is proven that Dedekind rings arise as the coordinate rings of smooth affine curves? I've been googling around, but haven't found it. – Tiffany Hwang Apr 08 '14 at 00:19
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    Dear Tiffany, Chapter 2 of Morandi's [*An Invitation to Arithmetic Geometry*](http://books.google.fr/books/about/An_Invitation_to_Arithmetic_Geometry.html?id=c8vnGts4994C&redir_esc=y) seems to be ideally suited to your needs. The link I give will allow you to look at the beginning of that chapter. – Georges Elencwajg Apr 08 '14 at 00:46
  • Thanks, I'll look into this in the future. – Tiffany Hwang Apr 08 '14 at 00:51

This is another geometric proof, but it yields an algebraic proof at the end.

Let $R = \mathbf{Q}[x,y]/(x^2 + y^2 - 1)$.

Recall that any line through $(1,0)$ meets the circle in exactly one other point (the tangent line meets the circle twice, so the "other" point is also $(1,0)$). So we can identify points on the circle with slopes of lines through $(1,0)$.

Grinding through the algebra, we get the rational parametrization of the circle

$$t \mapsto \left( \frac{2t}{t^2+1}, \frac{t^2-1}{t^2+1} \right) $$

and correspondingly, we get an isomorphism of the fraction field of your ring with $\mathbf{Q}(t)$ given by

$$ (x,y) \mapsto \left( \frac{2t}{t^2+1}, \frac{t^2-1}{t^2+1} \right) \qquad \qquad t \mapsto \frac{y+1}{x} $$

Geometrically, your ring is missing its two points at infinity, with homogeneous coordinates $(1 : \pm i : 0)$. These correspond to the points $t = 1 \pm i$ on the projective line with parameter $t$.

We deal imaginary coordinates with the following artifice: $R[i]$ is an integral extension of $R$, so your ring is Dedekind iff $R[i]$ is.

Let $u = 1 / (t - 1 - i)$. Then $\mathbf{Q}(i, t) = \mathbf{Q}(i, u)$, but the ring $\mathbf{Q}[i, u]$ corresponds to the curve missing the point $t = 1+i$.

The other missing point is $t = 1-i$, or $u = i/2$. Therefore, I assert

$$ R[i] \cong \mathbf{Q}\left[i, u, \frac{1}{u - i/2} \right] $$

with the isomorphism given by

$$ (x,y) \mapsto \left( \frac{2u(1 + (1+i) u)}{(u+1)(1+(1+2i)u)} , \frac{(1+iu)(1+(2+i)u)}{(u+1)(1+(1+2i)u)} \right) $$ $$ u \mapsto \frac{x}{(-1-i) x+ y + 1} $$

and the latter ring is a localization of the Dedekind domain $\mathbf{Q}(i)[u]$.

  • Disclaimer: I've checked that the forward map kills $x^2 + y^2 - 1$, but I haven't verified that everything maps to things that are actually elements of the relevant rings and not the fraction fields; I'm lazy and I'm too used to letting a CAS do the grunt work for me, but I don't have handy and can't convince wolframalpha to do what I want. :( I would be unsurprised if I have an error in my derivation. –  Apr 08 '14 at 02:12

Here is a purely algebraic approach.

Think of the ring $\mathbf Q[x,y]/(x^2+y^2-1)$ as a quadratic extension of $\mathbf Q[x]$: since $y^2 = 1-x^2$ in this ring, we can regard the ring as $\mathbf Q[x][\sqrt{1-x^2}]$. This is analogous to $\mathbf Z[\sqrt{d}]$ as an extension of $\mathbf Z$, with $\mathbf Q[x]$ (a PID) playing the role of $\mathbf Z$ (a PID) and $1-x^2$ playing the role of $d$. When $d$ is squarefree, $\mathbf Z[\sqrt{d}]$ is integrally closed unless $d \equiv 1 \bmod 4$. Run through the analogue of that argument for $\mathbf Q[x][\sqrt{1-x^2}]$, noting $1-x^2 = (1+x)(1-x)$ is squarefree in $\mathbf Q[x]$, and in the polynomial setting the whole issue related to congruences mod $4$ doesn't occur since $2$ is a unit in $\mathbf Q[x]$ (unlike in $\mathbf Z$).

In fact, for all squarefree $f(x)$ in $\mathbf Q[x]$, the ring $\mathbf Q[x][\sqrt{f(x)}]$ is integrally closed.

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    Matsumura, Commutative Ring Theory, Example 4, page 65, shows that if $A$ is a UFD, 2 is invertible in $A$, and $f\in A$ is square-free, then $A[\sqrt f]$ is integrally closed. – user26857 May 02 '22 at 17:30