Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the three rings $\mathbb{C}[x,y] / \langle x^4 + xy -1\rangle$, $\mathbb{Z}[x,y] / \langle x^4 + xy -1\rangle$ and $\mathbb{F}_2[x,y] /\langle x^4- y^3 \rangle$. I am supposed to detect whether these are Dedekind domains or not. However I've no idea how to do this. I know that a Dedekind domain is normal, noetherian, and of dimension 1. So I can at least see that each of these is noetherian, because they are quotients of noetherian rings. But I have no idea how to verify the other things, or unverify them. Are there any standard methods or tricks to work this out? I would really appreciate any help on this, thank you.

share|cite|improve this question
Notice that for any $f \in \mathbb{Z}[x,y]$ the quotient $\mathbb{Z}[x,y]/(f)$ has no chance of being a Dedekind domain, since $\mathbb{Z}[x,y]$ is three-dimensional. – Justin Campbell May 21 '12 at 17:14
What are some examples of non-normal rings that you know? What about some higher dimensional rings? The point of this problem is that you are asked to recognize similarities with familiar counterexamples that you hopefully saw in class / whatever book you are reading. – Michael Joyce May 21 '12 at 17:16
@MichaelJoyce I see now that it is quite easy to find non-normal rings by constructing quotient rings with the right sort of relation for example $k[x,y] / \langle x^2 - y^3 \rangle$ where $\frac{x}{y}$ is integral but not in that ring! Similarly the final ring in my example is not normal. – Paul Slevin May 21 '12 at 22:31
up vote 3 down vote accepted

The trickiest part is probably proving that $f(x,y) = x^4 + xy - 1$ is irreducible over $\mathbb{C}$, and in particular over $\mathbb{Z}$. I'll leave this to you, but ask me if you have trouble.

Thus $A = \mathbb{C}[x,y]/(f)$ is a domain. You already saw that $A$ is noetherian. Now the prime ideal $(f)$ has height one (because it is principal) and $\mathbb{C}[x,y]$ is two-dimensional, so $A$ is one-dimensional. Geometrically, $A$ is the algebra of polynomial functions on the curve $X = \{ f = 0 \} \subset \mathbb{C}^2$. As for normality, take the partial derivatives $f_x(x,y) = 4x^3 + y$ and $f_y(x,y) = x$. The set $X \cap \{ f_x = f_y = 0 \}$ is empty, which means $X$ is nonsingular. This implies that $A$ is normal.

As for $B = \mathbb{Z}[x,y]/(f)$, again $(f)$ has height one, but $\mathbb{Z}[x,y]$ is three-dimensional, so $B$ is two-dimensional and in particular cannot be a Dedekind domain.

Finally, $C = \mathbb{F}_2[x,y]/(x^4 - y^3)$ is not normal because $y/x$, which lies in the fraction field of $C$ but not $C$ itself, satisfies the monic polynomial $t^3 - x$.

share|cite|improve this answer
Since $\mathbb{C}$ is characteristic zero and we're working in dimension 2, I think it is enough to show that $X$ is smooth – then it is automatically irreducible. (Bézout's theorem.) – Zhen Lin May 21 '12 at 18:06
@Zhen Lin: I was thinking along the same lines and about to add something like that to my answer. But this is only true in the projective plane: $f(x) = x(x-1)$ is a counterexample in $\mathbb{A}^2$. – Pete L. Clark May 21 '12 at 18:18
However, if you mean that it's enough to check smoothness of the homogenized polynomial in $\mathbb{P}^2$, then yes: that's a nice approach to the problem, albeit rather openly algebraic-geometric rather than commutative algebraic. – Pete L. Clark May 21 '12 at 18:20
Indeed. Ever since I learned of this from Qiaochu I haven't been too fussed about checking irreducibility properly for plane curves... (also, the lecturer for this particular commutative algebra course is an algebraic geometer through and through!) – Zhen Lin May 21 '12 at 18:22
To someone who knows more about this than I: does any subtlety arise here because $C$ has characteristic $2$? – Justin Campbell May 21 '12 at 22:05

As said in the comments, $\mathbb{Z}[x,y]/(f)$ cannot have dimension $1$ so cannot be a Dedekind domain.

For any field $k$ and any nonzero polynomial $f(x,y) \in k[x,y]$, the quotient $k[x,y]/(f)$ is the coordinate ring of the plane algebraic curve $f(x,y) = 0$. This ring is a domain iff $f$ is irreducible. It is always be Noetherian and one-dimensional. It is a Dedekind domain iff it is nonsingular (nonsingular $\implies$ normal $\implies$ nonsingular in codimension one).

Assuming that the ground field $k$ is perfect, nonsingular is equivalent to smooth, and the latter is much easier to detect. The way to do it goes back essentially to multivariable calculus: the curve defined by $f(x,y) = 0$ is smooth at a point $P = (x,y)$ (such that $f(P) = 0$!) iff at least one of $\partial f / \partial x, \partial f / \partial y$ is nonzero at $P$.

Thus for instance in your third example both partial derivatives vanish at $P = (0,0)$. What about the first example?

share|cite|improve this answer
I understand that $\mathbb{Z}[x,y]$ has dimension 3, but why does that imply that the quotient $\mathbb{Z}[x,y] / f$ cannot have dimension 1? – Paul Slevin May 22 '12 at 2:26
Rather than coming at this from the standpoint of general dimension theory of commutative rings (which, as you seem to be discovering, is somewhat tricky and nonelementary), why not try to exhibit a prime ideal of height $2$? – Pete L. Clark May 22 '12 at 2:31
I'm really having no luck on this I am afraid. – Paul Slevin May 22 '12 at 10:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.